Class 7 math ex-5.2 solution | ৭ম শ্রেণি গণিত অনু ৫.২ সমাধান
সূত্রের সাহায্যে গুণফল নির্ণয় করঃ
১. (4x+3)(4x-3)
সমাধানঃ
(4x+3)(4x-3)
=(4x)2-(3)2
=16x2-9
২. (13-12p), (13+12p)
সমাধানঃ
(13-12p)(13+12p)
=(13)2-(12p)2
=169-144p2
৩. (ab+3), (ab-3)
সমাধানঃ
(ab+3)(ab-3)
=(ab)2-32
=a2b2-9
৪. (10-xy),(10+xy)
সমাধানঃ
(10-xy)(10+xy)
=(10)2-(xy)2
=100-x2y2
৫. (4x2+3y2), (4x2-3y2)
সমাধানঃ
(4x2+3y2)(4x2-3y2)
=(4x2)2-(3y2)2
=16x4-9y4
৬. (a-b-c), (a+b+c)
সমাধানঃ
(a-b-c)(a+b+c)
={(a-(b+c)}{a+(b+c))
=(a2-(b+c)2
=a2-(b2+2bc+c2)
=a2-b2-c2-2bc
৭. (x2-x+1), (x2+x+1)
সমাধানঃ
(x2-x+1)(x2+x+1)
={(x2+1)-x}{(x2+1)+x}
=(x2+1)2-x2
=(x2)2+2✕x2✕1+12-x2
=x4+2x2+1-x2
=x4+x2+1
৮. (x-\[\frac{a}{2}\]), (x-\[\frac{5a}{2}\])
সমাধানঃ
(x-\[\frac{a}{2}\])(x-\[\frac{5a}{2}\])
=(x)2+(-\[\frac{a}{2}\]– \[\frac{5a}{2}\])x+(-\[\frac{a}{2}\])✕( \[\frac{5a}{2}\])
=x2+(-3a)x+\[\frac{5a^2}{4}\]
=x2-3ax+\[\frac{5a^2}{4}\]
৯. (\[\frac{x}{4} – \frac{y}{3}\]), (\[\frac{x}{4} + \frac{y}{3}\])
সমাধানঃ
(\[\frac{x}{4} – \frac{y}{3}\]), (\[\frac{x}{4} + \frac{y}{3}\])
= \[(\frac{x}{4})^2 – (\frac{y}{3})^2\]
= \[\frac{x^2}{16} – \frac{y^2}{9}\]
১০. (a4+3a2x2+9x4), (9x4-3a2x2+a4)
সমাধানঃ
(a4+3a2x2+9x4)(9x4-3a2x2+a4)
={(a4+9x4)+3a2x2}{(a4+9x4)-3a2x2}
=(a4+9x4)2-(3a2x2)2
=(a4)2+2✕a4✕9x4+(9x4)2-9a4x4
=a8+18a4x4+81x8-9a4x4
= a8+9a4x4+81x8
১১. (x+1), (x-1), (x2+1)
সমাধানঃ
(x+1)(x-1)(x2+1)
={(x)2-(1)2}(x2+1)
=(x2-1)(x2+1)
=(x2)2-12
=x4-1
১২. (9a2+b2), (3a+b)(3a-b)
সমাধানঃ
(9a2+b2)(3a+b)(3a-b)
=(9a2+b2){(3a)2-(b)2}
=(9a2+b2) (9a2+b2)
=(9a2)2+(b2)2
=81a4-b4
