JSC/class 8 Math Exercise 5.1 Solution Part 2

JSC/class 8 math exercise 5.1 solution

(ঘ) $\frac{x + y}{(x - y)^2}, \frac{x - y}{x^3 + y^3}, \frac{y - z}{x^2 - y^2}$

সমাধান : প্রদত্ত ভগ্নাংশ গুলো $\frac{x + y}{(x - y)^2}, \frac{x - y}{x^3 + y^3}, \frac{y - z}{x^2 - y^2}$

এখানে, ১ম ভগ্নাংশের হর = (x – y)²

২য় ভগ্নাংশের হর = x³ + y³ = (x + y) (x² – xy + y²)

৩য় ভগ্নাংশের হর = x² – y² = (x + y)(x – y)

 হরগুলোর ল.সা.গু. = (x – y)² (x + y)(x² – xy + y²)

অতএব, $\frac{x + y}{(x - y)^2} = \frac{(x + y) (x - y)2 (x + y)(x2 - xy + y2)×}{(x - y)^2× (x - y)2 (x + y)(x2 - xy + y2)} = \frac{(x + y)(x^3 + y^3)}{(x - y)^2 × (x^3 + y^3)}$

$\frac{x + y}{(x - y)^2} = \frac{(x + y) × (x - y)2 (x + y)(x2 - xy + y2)}{(x - y)^2× (x - y)2 (x + y)(x2 - xy + y2)} = \frac{(x + y)(x^3 + y^3)}{(x - y)^2 × (x^3 + y^3)}$ ;

$\frac{x - y}{x^3 + y^3} = \frac{(x - y) × (x - y)2 (x + y)(x2 - xy + y2)}{ (x^3 + y^3) × (x - y)2 (x + y)(x2 - xy + y2)} = \frac{(x - y)^3}{(x - y)^2 × (x^3 + y^3)}$

এবং $\frac{y - z}{x^2 - y^2} = \frac{(y - z) × (x - y)2 (x + y)(x2 - xy + y2)}{ (x^2 - y^2) × (x - y)2 (x + y)(x2 - xy + y2)} = \frac{(y - z)(x - y)(x^2 - xy + y^2)}{(x - y)^2 × (x^3 + y^3)}$

অতএব, সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো :

$\frac{(x + y)(x^3 + y^3)}{(x - y)^2 × (x^3 + y^3)}, \frac{(x - y)^3}{(x - y)^2 × (x^3 + y^3)}, \frac{(y - z)(x - y)(x^2 - xy + y^2)}{(x - y)^2 × (x^3 + y^3)}$

(ঙ) $\frac{a}{a^3 + b^3}, \frac{b}{a^2 + ab + b^2}, \frac{c}{a^3 - b^3}$

সমাধান : প্রদত্ত ভগ্নাংশগুলো $\frac{a}{a^3 + b^3}, \frac{b}{a^2 + ab + b^2}, \frac{c}{a^3 - b^3}$

এখানে, ১ম ভগ্নাংশের হর = = a3 + b3

= (a + b)(a² – ab + b²)

২য় ভগ্নাংশের হর = a² + ab + b²

৩য় ভগ্নাংশের হর = a³ – b³

= (a – b)(a² + ab + b²)

∴ হরগুলোর ল.সা.গু = (a + b)(a – b)(a² – ab + b²)(a² + ab + b²)

= (a3 + b3)(a3 – b3)

অতএব, $\frac{a}{a^3 + b^3} = \frac{a × (a³ + b³)(a³ - b³)}{(a^3 + b^3) × (a³ + b³)(a³ - b³)} = \frac{a(a^3 - b^3)}{(a³ + b³)(a³ - b³)}$

$\frac{b}{ a^2 + ab + b^2} = \frac{b × (a³ + b³)(a³ - b³)}{( a^2 + ab + b^2) × (a³ + b³)(a³- b³)} = \frac{b(a - b)(a^3 + b^3)}{(a3 + b3)(a3 - b3)}$

এবং $\frac{c}{a³ - b³} = \frac{c × (a³ + b³)(a³ - b³)}{( a^2 + ab + b^2) × (a³ + b³)(a³ - b³)} = \frac{c(a^3 + b^3)}{(a³ + b³)(a³ - b³)}$

অতএব, সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো :

$\frac{a(a^3 - b^3)}{(a³+ b³)(a³ - b³)}, \frac{b(a - b)(a^3 + b^3)}{(a³ + b³)(a³ - b³)}, \frac{c(a^3 + b^3)}{(a³ + b³)(a³ - b³)}$

(চ) $\frac{1}{x^2 - 5x + 6}, \frac{1}{x^2 - 7x + 12}, \frac{1}{x^2 - 9x + 20}$

সমাধান : প্রদত্ত ভগ্নাংশগুলো

$\frac{1}{x^2 - 5x + 6}, \frac{1}{x^2 - 7x + 12}, \frac{1}{x^2 - 9x + 20}$

এখানে, ১ম ভগ্নাংশের হর = x² – 5x + 6

= x² – 3x – 2x + 6

= x(x – 3) – 2(x – 3)

= (x – 3)(x – 2)

২য় ভগ্নাংশের হর = x² – 7x + 12

= x² – 3x – 4x + 12

= x(x – 3) – 4(x – 3)

= (x – 3)(x – 4)

৩য় ভগ্নাংশের হর = x² – 9x + 20

= x² – 4x – 5x + 20

= x(x – 4) – 5(x – 4) = (x – 4)(x – 5)

 হরগুলোর ল.সা.গু. = (x – 2)(x – 3)(x – 4)(x – 5)

অতএব, $\frac{1}{x^2 - 5x + 6} = \frac{1}{(x - 2)(x - 3)} = \frac{1 × (x - 2)(x - 3)(x - 4)(x - 5)}{(x - 2)(x - 3) × (x - 2)(x - 3)(x - 4)(x - 5)} = \frac{(x - 4)(x - 5)}{(x - 2)(x - 3)(x - 4)(x - 5)}$

এবং $\frac{1}{ x² - 7x + 12} = \frac{1}{(x - 3)(x - 4)} = \frac{1 × (x - 2)(x - 3)(x - 4)(x - 5)}{ (x - 3)(x - 4)× (x - 2)(x - 3)(x - 4)(x - 5)} = \frac{(x - 2)(x - 5)}{(x - 2)(x - 3)(x - 4)(x - 5)}$

এবং $\frac{1}{ x² - 9x + 20} = \frac{1}{(x - 4)(x - 5)} = \frac{1 × (x - 2)(x - 3)(x - 4)(x - 5)}{ (x - 3)(x - 4)× (x - 2)(x - 3)(x - 4)(x - 5)} = \frac{(x - 2)(x - 3)}{(x - 2)(x - 3)(x - 4)(x - 5)}$

অতএব, সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো :

$\frac{(x - 4)(x - 5)}{(x - 2)(x - 3)(x - 4)(x - 5)}, \frac{(x - 2)(x - 5)}{(x - 2)(x - 3)(x - 4)(x - 5)}, \frac{(x - 2)(x - 3)}{(x - 2)(x - 3)(x - 4)(x - 5)}$

(ছ) $\frac{a - b}{a^2b^2}, \frac{b - c}{b^2c^2}, \frac{c - a}{c^2a^2}$

সমাধান : প্রদত্ত ভগ্নাংশগুলো $\frac{a - b}{a^2b^2}, \frac{b - c}{b^2c^2}, \frac{c - a}{c^2a^2}$

এখানে, ১ম ভগ্নাংশের হর = a²b²

২য় ভগ্নাংশের হর = b²c²

৩য় ভগ্নাংশের হর = c²a²

 হরগুলোর ল.সা.গু. = a²b²c²

অতএব, $\frac{a - b}{a^2b^2} = \frac{(a - b) × a²b²c²}{ a²b²× a²b² c²} = \frac{c^2(a - b)}{ a²b²c²}$

$\frac{b - c}{b^2c^2} = \frac{(b - c) × a²b²c²}{ b²c²× a²b²c²} = \frac{a^2(b - c)}{a²b²c²}$

এবং $\frac{c - a}{c^2a^2} = \frac{(c - a) × a²b²c²}{ a²c²× a²b²c²} = \frac{b^2(c - a)}{a²b²c²}$

অতএব, সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো :

$\frac{c^2(a - b)}{ a²b²c²}, \frac{a^2(b - c)}{a²b²c²} ,\frac{b^2(c - a)}{a²b²c²}$

(জ) $\frac{x - y}{x + y}, \frac{y - z}{y + z} ,\frac{z - x}{z + x}$ সমাধান : প্রদত্ত ভগ্নাংশগুলো $\frac{x - y}{x + y}, \frac{y - z}{y + z} ,\frac{z - x}{z + x}$

এখানে, ১ম ভগ্নাংশের হর = x + y

২য় ভগ্নাংশের হর = y + z

৩য় ভগ্নাংশের হর = z + x

 হরগুলোর ল.সা.গু. = (x + y)(y + z)(z + x)

অতএব, $\frac{x - y}{x + y}, \frac{( x - y) × (x + y)(y + z)(z + x)}{(x + y) × (x + y)(y + z)(z + x)} = \frac{( x - y) (y + z)(z + x)}{(x + y)(y + z)(z + x)}$

$\frac{y - z}{y + z}, \frac{( y - z) × (x + y)(y + z)(z + x)}{(x + y) × (x + y)(y + z)(z + x)} = \frac{( y - z) (x + y)(z + x)}{(x + y)(y + z)(z + x)}$

এবং $\frac{z - x}{z + x}, \frac{( z - x) × (x + y)(y + z)(z + x)}{(x + y) × (x + y)(y + z)(z + x)} = \frac{( z - x) (x + y)}{(x + y)(y + z)(z + x)}$

অতএব, সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো :

$\frac{( x - y) (y + z)(z + x)}{(x + y)(y + z)(z + x)}, \frac{( y - z) (x + y)(z + x)}{(x + y)(y + z)(z + x)}, \frac{( z - x) (x + y)}{(x + y)(y + z)(z + x)}$

প্রশ্ন \ ৩ \ যোগফল নির্ণয় কর :

(ক) $\frac{a - b}{a} + \frac{a + b}{b}$

সমাধান : $\frac{a - b}{a} + \frac{a + b}{b}$

= $\frac{b(a - b) + a(a + b)}{ab}$

= $\frac{ab - b^2 + a^2 + ab}{ab}$

= $\frac{ a^2 + 2ab - b^2 }{ab}$

(খ) $\frac{a}{bc},\frac{b}{ca}, \frac{c}{ab}$

সমাধান : $\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab}$

= $\frac{a . b + b .b + c. c}{abc}$

= $\frac{a^2 + b^2 + c^2}{abc}$

(গ) $\frac{x - y}{x} + \frac{y - z}{y} + \frac{z - x}{z}$

সমাধান : $\frac{x - y}{x} + \frac{y - z}{y} + \frac{z - x}{z}$

= $\frac{yz(x - y) + xz(y - z) + xy(z - x)}{xyz}$

= $\frac{xyz - y^2z + xyz - xz^2 + xyz - x^2y}{xyz}$

= $\frac{3xyz - y^2z - xz^2 - x^2y}{xyz}$

= $\frac{3xyz - x^2y - y^2z - xz^2 }{xyz}$

(ঘ) $\frac{x+y}{x-y}+\frac{x-y}{x+y}$

সমাধান : $\frac{x+y}{x-y}+\frac{x-y}{x+y}\\=\frac{(x\;+\;y)(x+y)+(x-y)(x-y)}{(x-y)(x+y)}\\=\frac{{(x+y)}^2+{(x-y)}^2}{(x-y)(x+y)}\\=\frac{x^2+2xy+y^2+x^2-2xy+y^2}{(x-y)(x+y)}\\=\frac{2x^2+2y^2}{x^2-y^2}\\=\frac{2(x^2+y^2)}{x^2-y^2}$

(ঙ) $\frac1{x^2-3x+2}+\frac1{x^2-4x+3}+\frac1{x^2-5x+4}$

সমাধান : $\frac1{x^2-3x+2}+\frac1{x^2-4x+3}+\frac1{x^2-5x+4}\\=\frac1{x^2-2x-x+2}+\frac1{x^2-3x-x+3}+\frac1{x^2-4x-x+2}\\=\frac1{x(x-2)-1(x-2)}+\frac1{x(x-3)-1(x-3)}+\frac1{x(x-4)-1(x-4)}\\=\frac1{(x-2)(x-1)}+\frac1{(x-3)(x-1)}+\frac1{(x-4)(x-1)}\\=\frac{(x-3)(x-4)+(x-2)(x-4)+(x-3)(x-2)}{(x-1)(x-2)(x-3)(x-4)}\\=\frac{x^2-7x+12+x^2-6x+8+x^2-5x+6}{(x-1)(x-2)(x-3)(x-4)}\\=\frac{3x^2-18x+26}{(x-1)(x-2)(x-3)(x-4)}\\$

(চ) $\frac1{a^2-b^2}+\frac1{a^2+ab+b^2}+\frac1{a^2-ab+b^2}$

সমাধান : $\frac1{a^2-b^2}+\frac1{a^2+ab+b^2}+\frac1{a^2-ab+b^2}\\=\frac1{(a+b)(a-b)}+\frac1{a^2+ab+b^2}+\frac1{a^2-ab+b^2}\\=\frac{(a^2+ab+b^2)(a^2-ab+b^2)+(a+b)(a-b)(a^2-ab+b^2)+(a+b)(a-b)(a^2+ab+b^2)}{(a+b)(a-b)(a^2+ab+b^2)(a^2-ab+b^2)}\\=\frac{{(a^2+b^2)}^2-{(ab)}^2+(a-b)(a^3+b^3)+(a+b)(a^3-b^3)}{(a+b)(a-b)(a^2+ab+b^2)(a^2-ab+b^2)}\\=\frac{a^4+b^4+2a^2b^2-a^2b^2+a^4+ab^3-a^3b-b^4+a^4-ab^3+a^3b-b^4}{(a^3-b^3)(a^3+b^3)}\\=\frac{3a^4+a^2b^2-b^4}{(a^3-b^3)(a^3+b^3)}$

(ছ) $\frac1{x-2}+\frac1{x+2}+\frac4{x^2-4}$

সমাধান : $\frac1{x-2}+\frac1{x+2}+\frac4{x^2-4}\\\\=\;\frac1{x-2}+\frac1{x+2}+\frac4{x^2-2^2}\\\\=\frac1{x-2}+\frac1{x+2}+\frac4{(x-2)(x+2)}\\\\=\frac{x+2+x-2+4}{(x-2)(x+2)}\\\\=\frac{2x+4}{(x-2)(x+2)}\\\\=\frac{2(x+2)}{(x-2)(x+2)}\\\\=\frac2{(x-2)}$

এখানে, ১ম ভগ্নাংশের হর = x + y

২য় ভগ্নাংশের হর = y + z

৩য় ভগ্নাংশের হর = z + x

 হরগুলোর ল.সা.গু. = (x + y)(y + z)(z + x)

অতএব, $\frac{x - y}{x + y}, \frac{( x - y) × (x + y)(y + z)(z + x)}{(x + y) × (x + y)(y + z)(z + x)} = \frac{( x - y) (y + z)(z + x)}{(x + y)(y + z)(z + x)}$

$\frac{y - z}{y + z}, \frac{( y - z) × (x + y)(y + z)(z + x)}{(x + y) × (x + y)(y + z)(z + x)} = \frac{( y - z) (x + y)(z + x)}{(x + y)(y + z)(z + x)}$

এবং $\frac{z - x}{z + x}, \frac{( z - x) × (x + y)(y + z)(z + x)}{(x + y) × (x + y)(y + z)(z + x)} = \frac{( z - x) (x + y)}{(x + y)(y + z)(z + x)}$

অতএব, সাধারণ হরবিশিষ্ট ভগ্নাংশগুলো হলো :

$\frac{( x - y) (y + z)(z + x)}{(x + y)(y + z)(z + x)}, \frac{( y - z) (x + y)(z + x)}{(x + y)(y + z)(z + x)}, \frac{( z - x) (x + y)}{(x + y)(y + z)(z + x)}$

প্রশ্ন \ ৩ \ যোগফল নির্ণয় কর :

(ক) $\frac{a - b}{a} + \frac{a + b}{b}$

সমাধান : $\frac{a - b}{a} + \frac{a + b}{b}$

= $\frac{b(a - b) + a(a + b)}{ab}$

= $\frac{ab - b^2 + a^2 + ab}{ab}$

= $\frac{ a^2 + 2ab - b^2 }{ab}$

(খ) $\frac{a}{bc},\frac{b}{ca}, \frac{c}{ab}$

সমাধান : $\frac{a}{bc} + \frac{b}{ca} + \frac{c}{ab}$

= $\frac{a . b + b .b + c. c}{abc}$

= $\frac{a^2 + b^2 + c^2}{abc}$

(গ) $\frac{x - y}{x} + \frac{y - z}{y} + \frac{z - x}{z}$

সমাধান : $\frac{x - y}{x} + \frac{y - z}{y} + \frac{z - x}{z}$

= $\frac{yz(x - y) + xz(y - z) + xy(z - x)}{xyz}$

= $\frac{xyz - y^2z + xyz - xz^2 + xyz - x^2y}{xyz}$

= $\frac{3xyz - y^2z - xz^2 - x^2y}{xyz}$

= $\frac{3xyz - x^2y - y^2z - xz^2 }{xyz}$

(ঘ) $\frac{x+y}{x-y}+\frac{x-y}{x+y}$

(ঙ) $\frac1{x^2-3x+2}+\frac1{x^2-4x+3}+\frac1{x^2-5x+4}$

(চ) $\frac1{a^2-b^2}+\frac1{a^2+ab+b^2}+\frac1{a^2-ab+b^2}$

(ছ) $\frac1{x-2}+\frac1{x+2}+\frac4{x^2-4}$

(জ) $\frac1{x^2-1}+\frac1{x^4-1}+\frac1{x^8-1}$

সমাধান : $\frac1{x^2-1}+\frac1{x^4-1}+\frac1{x^8-1}\\\\=\frac1{x^2-1}+\frac1{{(x^2)}^2-1}+\frac1{x^8-1}\\\\=\frac1{x^2-1}+\frac1{(x^2+1)(x^2-1)}+\frac1{x^8-1}\\\\=\frac{(x^2+1)+1}{(x^2+1)(x^2-1)}+\frac1{{(x^4)}^2-1}\\\\=\frac{x^2+2}{x^4-1}+\frac1{(x^4+1)(x^4-1)}\\\\=\frac{(x^2+2)(x^4+1)+1}{(x^4+1)(x^4-1)}\\\\=\frac{x^6+x^2+2x^4+2+4}{(x^4+1)(x^4-1)}\\\\=\frac{x^6+2x^4+x^2+6}{x^8-1}\\\\$

প্রশ্ন \ ৪ \ বিয়োগফল নির্ণয় কর :

(ক) $\frac a{x-3}-\frac{a^2}{x^2-9}$

সমাধান : $\frac a{x-3}-\frac{a^2}{x^2-9}\\\\=\frac a{x-3}-\frac{a^2}{x^2-3^2}\\\\=\frac a{x-3}-\frac{a^2}{(x-3)(x+3)}\\\\=\frac{a(x+3)-a^2}{(x-3)(x+3)}\\\\=\frac{ax+3a-a^2}{x^2-3^2}\\\\=\frac{ax+3a-a^2}{x^2-9}$

(খ) $\frac{x+1}{y(x-y)}-\frac1{x\left(x+y\right)}$

সমাধান : $\frac1{y(x-y)}-\frac1{x\left(x+y\right)}\\\\=\frac{x(x+y)-y(x-y)}{xy(x-y)(x+y)}\\\\=\frac{x^2+xy-xy+y^2}{xy(x^2-y^2)}\\\\=\frac{x^2+y^2}{xy(x^2-y^2)}$

(গ) $\frac{x+1}{1+x+x^2}-\frac{x-1}{1-x+x^2}$

সমাধান : $\frac{x+1}{1+x+x^2}-\frac{x-1}{1-x+x^2}\\\\=\frac{(x+1)(1-x+x^2)-(x-1)(1+x+x^2)}{(1+x+x^2)(1-x+x^2)}\\\\=\frac{x^3+1-(x^3-1)}{{(1+x^2)}^2-x^2}\\\\=\frac{x^3+1-x^3+1}{1+2x^2+x^4-x^2}\\\\=\frac2{1+x^2+x^4}\\\\=\frac2{x^4+x^2+1}\\\\$

(ঘ) $\frac{a^2+16b^2}{a^2-16b^2}-\frac{a-4b}{a+4b}$

সমাধান : $\frac{a^2+16b^2}{a^2-16b^2}-\frac{a-4b}{a+4b}\\\\=\frac{a^2+16b^2}{a^2-{(4b)}^2}-\frac{a-4b}{a+4b}\\\\=\frac{a^2+16b^2}{(a+4b)(a-4b)}-\frac{a-4b}{a+4b}\\\\=\frac{a^2+16b^2-(a-4b)(a-4b)}{(a+4b)(a-4b)}\\\\=\frac{a^2+16b^2-{(a-4b)}^2}{(a+4b)(a-4b)}\\\\=\frac{a^2+16b^2-(a^2-8ab+16b^2)}{a^2-{(4b)}^2}\\\\=\frac{a^2+16b^2-a^2+8ab-16b^2}{a^2-16b^2}\\\\=\frac{8ab}{a^2-16b^2}$

(ঙ) $\frac1{x-y}-\frac{x^2-xy+y^2}{x^3+y^3}$

সমাধান : $\frac1{x-y}-\frac{x^2-xy+y^2}{x^3+y^3}\\\\=\frac1{x-y}-\frac{x^2-xy+y^2}{(x+y)(x^2-xy+y^2)}\\\\=\frac1{x-y}-\frac1{x+y}\\\\=\frac{x+y-(x-y)}{(x+y)(x-y)}\\\\=\frac{x+y-x+y}{x^2-y^2}\\\\=\frac{2y}{x^2-y^2}$

প্রশ্ন \ ৫ \ সরল কর :

(ক) $\frac{x-y}{xy}+\frac{y-z}{yz}+\frac{z-x}{zx}$

সমাধান : $\frac{x-y}{xy}+\frac{y-z}{yz}+\frac{z-x}{zx}\\\\=\frac{z(x-y)+x(y-z)+y(z-x)}{xyz}\\\\=\frac{zx-yz+xy-yz+yz-xy}{xyz}\\\\=\frac0{xyz}\\\\=0$

(খ) $\frac{x-y}{(x+y)(y+z)}+\frac{y-z}{(y+z)(z+x)}+\frac{z-x}{(z+x)(x+y)}$

সমাধান : $\frac{x-y}{(x+y)(y+z)}+\frac{y-z}{(y+z)(z+x)}+\frac{z-x}{(z+x)(x+y)}\\\\=\frac{(x-y)(z+x)+(y-z)(x+y)+(z-x)(y+z)}{(x+y)(y+z)(z+x)}\\\\=\frac{zx+x^2-yz-xy+xy+y^2-zx-yz+yz+z^2-xy-zx}{(x+y)(y+z)(z+x)}\\\\=\frac{x^2+y^2+z^2-xy-yz-zx}{(x+y)(y+z)(z+x)}$

(গ) $\frac y{(x-y)(y-z)}+\frac x{(z-x)(x-y)}+\frac z{(y-z)(z-x)}$

সমাধান : $\frac y{(x-y)(y-z)}+\frac x{(z-x)(x-y)}+\frac z{(y-z)(z-x)}\\\\=\frac{y(z-x)+x(y-z)+z(x-y)}{(x-y)(y-z)(z-x)}\\\\=\frac{yz-xy+xy-zx+zx-yz}{(x-y)(y-z)(z-x)}\\\\=\frac0{(x-y)(y-z)(z-x)}\\\\=0$

(ঘ) $\frac1{x+3y}+\frac1{x-3y}+\frac{2x}{x^2-9y^2}$

সমাধান : $\frac1{x+3y}+\frac1{x-3y}-\frac{2x}{x^2-9y^2}\\\\=\frac{x-3y+x+3y}{(x+3y)(x-3y)}-\frac{2x}{x^2-9y^2}\\\\=\frac{2x}{x^2-9y^2}-\frac{2x}{x^2-9y^2}\\\\=\frac{2x-2x}{x^2-9y^2}\\\\=\frac0{x^2-9y^2}\\\\=0\\\\$

(ঙ) $\frac1{x-y}-\frac2{2x+y}+\frac1{x+y}-\frac2{2x-y}$

সমাধান : $\frac1{x-y}-\frac2{2x+y}+\frac1{x+y}-\frac2{2x-y}\\\\=\frac1{x-y}+\frac1{x+y}-\frac2{2x+y}-\frac2{2x-y}\\\\=\frac{x+y+x-y}{(x-y)(x+y)}-\left(\frac{2(2x-y)+2(2x+2)}{(2x+y)(2x-y)}\right)\\\\=\frac{2x}{x^2-y^2}-\frac{4x-2y+4x+2y}{4x^2-y^2}\\\\=\frac{2x}{x^2-y^2}-\frac{8x}{4x^2-y^2}\\\\=\frac{2x(4x^2-y^2)-8x(x^2-y^2)}{(x^2-y^2)(4x^2-y^2)}\\\\=\frac{8x^3-2xy^2-8x^3+8xy^2}{(x^2-y^2)(4x^2-y^2)}\\\\=\frac{6xy^2}{(x^2-y^2)(4x^2-y^2)}$

(চ) $\frac1{x-2}-\frac{x-2}{x^2+2x+4}+\frac{6x}{x^3+8}$

সমাধান : $\frac1{x-2}-\frac{x-2}{x^2+2x+4}+\frac{6x}{x^3+8}\\\\=\frac{1(x^2+2x+4)-(x-2)(x-2)}{(x-2)(x^2+2x+4)}+\frac{6x}{x^3+8}\\\\=\frac{x^2+2x+4-(x^2-2x-2x+4)}{(x-2)(x^2+2x+4)}+\frac{6x}{x^3+8}\\\\=\frac{x^2+2x+4-x^2+4x-4}{(x-2)(x^2+2x+4)}+\frac{6x}{x^3+8}\\\\=\frac{6x}{x^3-8}+\frac{6x}{x^3+8}\\\\=\frac{6x(x^3+8)+6x(x^3-8)}{(x^3-8)(x^3+8)}\\\\=\frac{6x^4+48x+6x^4-48x}{{(x^3)}^2-8^2}\\\\=\frac{12x^4}{x^6-64}$

(ছ) $\frac1{x-1}-\frac1{x+1}-\frac2{x^2+1}+\frac4{x^4+1}$

সমাধান : $\frac1{x-1}-\frac1{x+1}-\frac2{x^2+1}+\frac4{x^4+1}\\\\=\frac{x+1-x+1}{(x-1)(x+1)}-\frac2{x^2+1}+\frac4{x^4+1}\\\\=\frac2{x^2-1}-\frac2{x^2+1}+\frac4{x^4+1}\\\\=\frac{2(x^2+1)-2(x^2-1)}{(x^2-1)(x^2+1)}+\frac4{x^4+1}\\\\=\frac{2x^2+2-2x^2+2}{x^4-1}+\frac4{x^4+1}\\\\=\frac4{x^4-1}+\frac4{x^4+1}\\\\=\frac{4(x^4+1)+4(x^4-1)}{(x^4-1)(x^4+1)}\\\\=\frac{4x^4+4+4x^4-4}{{(x^4)}^2-1}\\\\=\frac{8x^4}{x^8-1}$

(জ) $\frac{x-y}{(y-z)(z-x)}+\frac{y-z}{(z-x)(x-y)}+\frac{z-x}{(x-y)(y-z)}$

সমাধান : $\frac{x-y}{(y-z)(z-x)}+\frac{y-z}{(z-x)(x-y)}+\frac{z-x}{(x-y)(y-z)}\\\\=\frac{(x-y)(x-y)+(y-z)(y-z)+(z-x)(z-x)}{(x-y)(y-z)(z-x)}\\\\=\frac{{(x-y)}^2+{(y-z)}^2+{(z-x)}^2}{(x-y)(y-z)(z-x)}\\\\=\frac{x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2}{(x-y)(y-z)(z-x)}\\\\=\frac{2x^2+2y^2+2z^2-2xy-2yz-2zx}{(x-y)(y-z)(z-x)}\\\\=\frac{2(x^2+y^2+z^2-xy-yz-zx)}{(x-y)(y-z)(z-x)}$

(ঝ) $\frac1{a-b-c}+\frac1{a-b+c}+\frac1{a^2+b^2-c^2-2ab}$

সমাধান : $\frac1{a-b-c}+\frac1{a-b+c}+\frac a{a^2+b^2-c^2-2ab}\\\\=\frac{a-b+c+a-b-c}{(a-b-c)(a-b+c)}+\frac a{a^2+b^2-c^2-2ab}\\\\=\frac{2a-2b}{(a-b-c)(a-b+c)}+\frac a{a^2+b^2-c^2-2ab}\\\\=\frac{2a-2b}{{(a-b)}^2-c^2}+\frac a{a^2+b^2-c^2-2ab}\\\\=\frac{2a-2b}{a^2-2ab+b^2-c^2}+\frac a{a^2-2ab+b^2-c^2}\\\\\\=\frac{2a-2b+a}{a^2-2ab+b^2-c^2}\\\\=\frac{3a-2b}{a^2-2ab+b^2-c^2}$

(ঞ) $\frac1{a^2+b^2-c^2+2ab}+\frac1{b^2+c^2-a^2+2bc}+\frac1{c^2+a^2-b^2+2ca}$

সমাধান : $\frac1{a^2+b^2-c^2+2ab}+\frac1{b^2+c^2-a^2+2bc}+\frac1{c^2+a^2-b^2+2ca}\\\\=\frac1{a^2+b^2+2ab-c^2}+\frac1{b^2+c^2+2bc-a^2}+\frac1{c^2+a^2+2ca-b^2}\\\\=\frac1{{(a+b)}^2-c^2}+\frac1{{(b+c)}^2-a^2}+\frac1{{(c+a)}^2-b^2}\\\\=\frac1{(a+b+c)(a+b-c)}+\frac1{(a+b+c)(b+c-a)}+\frac1{(a+b+c)(c+a-b)}\\\\=\frac{(b+c-a)(c+a-b)+(a+b-c)(c+a-b)+(a+b-c)(b+c-a)}{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\\\\=\frac{(c+a-b)(b+c-a+a+b-c)+ab+ca-a^2+b^2+bc-ab-bc-c^2+ca}{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\\\\=\frac{2b(c+a-b)+2ca-a^2+b^2-c^2}{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\\\\=\frac{2bc+2ab-2b^2+2ca-a^2+b^2-c^2}{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}\\\\=\frac{2bc+2ab+2ca-a^2-b^2-c^2}{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$

লক্ষ কর : বীজগণিতীয় ভগ্নাংশর যোগ/বিয়োগ করার সময় প্রয়োজনে প্রদত্ত ভগ্নাংশগুলোকে লঘিষ্ঠ আকারে প্রকাশ করে নিতে হয়।

JSC/class 8 Math Exercise 5.1 Solution Part 1

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JSC/class 8 math exercise 5.1 solution part 2