Question

Cho biểu thức :

A = \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\left(\dfrac{1-x}{\sqrt{x}}\right)^2\)

với x > O , x#1.

a) Rút gọn.

b) Tìm giá trị lớn nhất của A

Answer 1

a)\(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\dfrac{1-x}{\sqrt{x}}\right)^2\)

=\(\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)^2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(x-1\right)\left(\sqrt{x}+2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}\right]\left(\dfrac{1-x}{\sqrt{x}}\right)^2\)

=\(\left[\dfrac{x\sqrt{x}-2x+2x-4\sqrt{x}+\sqrt{x}-2-x\sqrt{x}+\sqrt{x}-2x+2}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}\right]\)

=\(\left[\dfrac{-2\sqrt{x}-2x}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}\right]\left(\dfrac{1-x}{\sqrt{x}}\right)^2\)

=\(\left[\dfrac{-2\sqrt{x}\left(1+\sqrt{x}\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)^2}\right]\left(\dfrac{1-x}{\sqrt{x}}\right)^2\)

=\(\dfrac{-2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}\left(\dfrac{1-x}{\sqrt{x}}\right)^2\)

=\(\dfrac{-2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)=\(\dfrac{-2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{-2\sqrt{x}-2}{\sqrt{x}}\)

ý b) bn tự lm nha