class-eight_math_img

class eight বীজগনিত (অনুশীলনী-৪.১ )(৫ নং)

Class-eight_math_img

\displaystyle x-\frac{1}{x} =3\ \হলে x\displaystyle ^{2}+\displaystyle \frac{1}{x^{2}} এর মান নির্নয় কর ।
দেওয়া আছে,,
\displaystyle x-\frac{1}{x} =3
এখন,
\displaystyle \ x^{2} +\frac{1}{x^{2}}
=\displaystyle ( x)^{2} +\left(\frac{1}{x}\right)^{2}
=\displaystyle \left( x-\frac{1}{x}\right)^{2} +2.x.\frac{1}{x} [ আমরা জানি,, a\displaystyle ^{2}+b\displaystyle ^{2}=\displaystyle ( a-b)^{2} +2ab ]
=(3)\displaystyle ^{2}+2
=9+2
=11

class-eight_math_img

\displaystyle x-\frac{1}{x} =3\ \হলে x\displaystyle ^{2}+\displaystyle \frac{1}{x^{2}} এর মান নির্নয় কর ।
দেওয়া আছে,,
\displaystyle x-\frac{1}{x} =3
\displaystyle \left( x-\frac{1}{x}\right)^{2} =( 3)^{2}
\displaystyle ( x)^{2} -2.x.\frac{1}{x} +\left(\frac{1}{x}\right)^{2} =9
\displaystyle x^{2} -2+\frac{1}{x^{2}} =9
\displaystyle x^{2} +\frac{1}{x^{2}} =9+2
\displaystyle x^{2} +\frac{1}{x^{2}}=11 (ans):

Post Author: showrob

1 thought on “class eight বীজগনিত (অনুশীলনী-৪.১ )(৫ নং)

    soikot

    (March 28, 2020 - 3:47 pm)

    nc math

Leave a Reply

Your email address will not be published. Required fields are marked *

80 ÷ = 16