Câu hỏi của Kiều Vũ Minh Đức

Question

Chứng minh:

$\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1$    với a,b>0; a$\ne$b

Phạm Lan Hương · Accepted Answer

Ta có $\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1$

$\Leftrightarrow\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{ab}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}+1=0$

$\Leftrightarrow\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}+1=0$

$\Leftrightarrow\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}+1=0$

$\Leftrightarrow-1+1=0\Leftrightarrow0=0$(luôn đúng với mọi a ;b >0 : a$\ne b$)

đpcmCâu trả lời: Ta có $\frac{2}{\sqrt{ab}}:\left(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}\right)^2-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}=-1$

$\Leftrightarrow\frac{2}{\sqrt{ab}}:\frac{\left(\sqrt{b}-\sqrt{a}\right)^2}{ab}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}+1=0$

$\Leftrightarrow\frac{2\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)^2}-\frac{a+b}{\left(\sqrt{a}-\sqrt{b}\right)^2}+1=0$

$\Leftrightarrow\frac{-\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}+1=0$

$\Leftrightarrow-1+1=0\Leftrightarrow0=0$(luôn đúng với mọi a ;b >0 : a$\ne b$)

đpcm

Violympic toán 9