Đáp án:
Giải thích các bước giải:
$m=\frac{\frac{1-{{x}^{3}}}{1-x}-x}{\frac{1-{{x}^{2}}}{1-x-{{x}^{2}}+{{x}^{3}}}}$
Đặt:
$A=\frac{1-{{x}^{3}}}{1-x}-x$
$B=\frac{1-{{x}^{2}}}{1-x-{{x}^{2}}+{{x}^{3}}}$
$\to m=\frac{A}{B}$
$A=\frac{1-{{x}^{3}}}{1-x}-x$
$A=\frac{\left( 1-x \right)\left( 1+x+{{x}^{2}} \right)}{1-x}-x$
$A=1+x+{{x}^{2}}-x$
$A=1+{{x}^{2}}$
$B=\frac{1-{{x}^{2}}}{1-x-{{x}^{2}}+{{x}^{3}}}$
$B=\frac{1-{{x}^{2}}}{\left( 1-{{x}^{2}} \right)-\left( x-{{x}^{3}} \right)}$
$B=\frac{1-{{x}^{2}}}{\left( 1-{{x}^{2}} \right)-x\left( 1-{{x}^{2}} \right)}$
$B=\frac{1-{{x}^{2}}}{\left( 1-{{x}^{2}} \right)\left( 1-x \right)}$
$B=\frac{1}{1-x}$
$m=\frac{A}{B}=\frac{1+{{x}^{2}}}{\frac{1}{1-x}}=\left( 1+{{x}^{2}} \right)\left( 1-x \right)$
