Câu hỏi của Love Mon

Question

$\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2=1\left(a\ge0,a\ne1\right)$

Chứng minh đẳng thức trên

KAl(SO4)2·12H2O · Accepted Answer

$\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2$

$=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}.\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}$

$=\left(1-a\sqrt{a}+\sqrt{a}-a\right)\frac{1-\sqrt{a}}{\left(1-a\right)^2}$

$=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a.\left(\sqrt{a}\right)^2-\left(\sqrt{a}\right)^2+a\sqrt{a}}{\left(1-a\right)^2}$

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

$=\left(\frac{a-1}{1-a}\right)^2=\left(-1\right)^2=1=VP\left(ĐPCM\right)$

Câu trả lời:

$\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2$

$=\frac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}.\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}$

$=\left(1-a\sqrt{a}+\sqrt{a}-a\right)\frac{1-\sqrt{a}}{\left(1-a\right)^2}$

$=\frac{1-a\sqrt{a}+\sqrt{a}-a-\sqrt{a}+a.\left(\sqrt{a}\right)^2-\left(\sqrt{a}\right)^2+a\sqrt{a}}{\left(1-a\right)^2}$

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

$=\left(\frac{a-1}{1-a}\right)^2=\left(-1\right)^2=1=VP\left(ĐPCM\right)$