Question

সমাধান সেট নির্ণয় কর: $\frac{\left(x+1\right)^3-\left(x-1\right)^3}{\left(x+1\right)^2-\left(x-1\right)^2}=2$

Answer 1

$\frac{\left(x+1\right)^3-\left(x-1\right)^3}{\left(x+1\right)^2-\left(x-1\right)^2}=2$

বা, $\frac{\left(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3\right)-\left(x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\right)}{\left(x^2+2\cdot x\cdot1+1^2\right)-\left(x^2-2\cdot x\cdot1+1^2\right)}=2$

বা, $\frac{\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)}{\left(x^2+2x+1\right)-\left(x^2-2x+1\right)}=2$

বা, $\frac{x^3+3x^2+3x+1-x^3+3x^2-3x+1}{x^2+2x+1-x^2+2x-1}=2$

বা, $\frac{6x^2+2}{4x}=2$

বা, $\frac{2\left(3x^2+1\right)}{4x}=2$

বা, $\frac{3x^2+1}{2x}=2$

বা, $3x^2+1=4x$

বা, $3x^2-4x+1=0$

বা, $3x^2-3x-x+1=0$

বা, $3x\left(x-1\right)-1\left(x-1\right)=0$

বা, $\left(x-1\right)\left(3x-1\right)=0$

হয়,
$\left(x-1\right)=0$

$\therefore x=1$

অথবা,
$\left(3x-1\right)=0$

বা, $3x=1$

$\therefore x=\frac13$

সুতরাং নির্ণেয় সমাধান সেট: $\left\lbrace1,\frac13\right\rbrace$ [Answer]