Question

Cho biểu thức 

A= \(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)

a, Rút gọn A

b, Tìm x để A = 2

Answer 1

A=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{2+5\sqrt{x}}{x-4}\)

A=\(\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{x-4}\)

A=\(\frac{3x-6\sqrt{x}}{x-4}=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

b)

để A=2 thì 

\(\frac{3\sqrt{x}}{\sqrt{x}+2}=2\) => 3\(\sqrt{x}\)=2\(\sqrt{x}+4\)

=>\(\sqrt{x}\)=4

=> x = 16