Class 9-10 math ex 11.1 solution: Part 1

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(i) দেওয়া আছে, a : b = b : c,

বা, \frac{a}{b} = \frac{b}{c}

\therefore b^2 = ac

বামপক্ষ  = \frac{a}{c}

ডানপক্ষ = \frac{a^2 + b^2}{b^2 + c^2}

          = \frac{a^2 + ac}{ac + c^2}

          = \frac{a (a + c)}{c (a + c)}

          = \frac{a}{c}

অর্থাৎ, \frac{a}{c} = \frac{a^2 + b^2}{b^2 + c^2}

Class 9-10 math ex 11.1 solution

(ii) দেওয়া আছে, a : b = b : c,

বা, \frac{a}{b} = \frac{b}{c}

\therefore b^2 = ac

বামপক্ষ = a^2b^2c^2 (\frac1{a^3} + \frac1{b^3} + \frac1{c^3})

= \frac{a^2b^2c^2}{a^3} + \frac{a^2b^2c^2}{b^3} + \frac{a^2b^2c^2}{c^3}

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

= \frac{ac.c^2}{a} + \frac{(b^2)^2}{b} + \frac{a^2.ac}{c}             [\Since b^2 = ac]

= \frac{a.c^3}{a} + \frac{b^4}{b} + \frac{a^3c}{c}

= c^3 + b^3 + a^3

= a^3 + b^3 + c^3

= ডানপক্ষ

 

অর্থাৎ, a^2b^2c^2 (\frac1{a^3} + \frac1{b^3} + \frac1{c^3}) = a^3 + b^3 + c^3

 

(iii) দেওয়া আছে, a : b = b : c,

বা, \frac{a}{b} = \frac{b}{c}

\therefore b^2 = ac

 

বামপক্ষ = \frac{abc (a + b + c)^3} {(ab + bc + ca)^3}

= \frac{b. b^2 (a + b + c)^3}{(ab + bc + b^2)^3}   [\because b^2 = ac]

= \frac{b^3 (a + b + c)^3}{{b (a + c + b)}^3}

= \frac{b^3 (a + b + c)^3}{b^3 (a + b + c)^3}

= 1 = ডানপক্ষ

(iv) দেওয়া আছে, a : b = b : c,

          বা, \frac{a}{b} = \frac{b}{c}

          \therefore b^2 = ac

১ম পক্ষ = a – 2b + c

২য় পক্ষ = \frac{(a - b)^2}a

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

                   = \frac{a^2 - 2ab + ac}{a} [\since b^2 = ac]

                   = \frac{a(a - 2b + c)}{a}

                   = a – 2b + c

          ৩য় পক্ষ = \frac{(b - c)^2}{c}

                   = \frac{b^2 - 2bc + c^2}{c}

                   = \frac{ac - 2bc + c^2}{c} [\since b^2 = ac]

                   = \frac{c(a - 2b + c)}{c}

                   = a – 2b + c

অর্থাৎ, a – 2b + c = \frac{(a - b)^2}a =  \frac{(b - c)^2}{c}

       সমাধান : (i) \frac{1 - \sqrt{1 - x}}{1 + \sqrt{1 - x}} = \frac13
বা,  \frac{1-\sqrt{1-x}+1+\sqrt{1-x}}{1-\sqrt{1-x}-1-\sqrt{1-x}} = \frac{1 + 3}{1 - 3} [যোজন-বিয়োজন করে]
বা, \frac{2}{ - 2(\sqrt{1 - x}} = \frac{4} {- 2}
বা, \frac{1}{\sqrt{1 - x}} = 2               [ – 1 দ্বারা উভয়পক্ষকে গুণ করে]
বা, 2 \sqrt{1 - x} = 1           [আড়গুণন করে]
বা,  4 (1– x) = 1           [উভয়পক্ষকে বর্গ করে]
বা,  4 – 4x = 1
বা,  – 4x = 1 – 4
বা,  – 4x = – 3
∴ x = \frac{- 3}{- 4} = \frac34
নির্ণেয় সমাধান, x = \frac34
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