Question

Cho biểu thức \(A=\left(1+\frac{1}{\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x}-x}\right)+\frac{5}{\sqrt{x}}\) với \(x>0,x\ne1\)

a) Rút gọn A

b) Tìm x để A = 5

c) Tìm x để A > 4

Answer 1

a, \(A=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{1-\sqrt{x}-1}{\sqrt{x}\left(1-\sqrt{x}\right)}+\frac{5}{\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}}.\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}+\frac{5}{\sqrt{x}}=\frac{x-1}{\sqrt{x}}+\frac{5}{\sqrt{x}}=\frac{x+4}{\sqrt{x}}\)

b, \(A=5\Leftrightarrow\frac{x+4}{\sqrt{x}}=5\Leftrightarrow x-5\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\left(l\right)\end{matrix}\right.\Rightarrow x=16\)

c, \(A>4\Leftrightarrow\frac{x+4}{\sqrt{x}}>4\Leftrightarrow x-4\sqrt{x}+4=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)

\(\Leftrightarrow x=4\)