Bài 4:
a: ĐKXĐ: x>=0; x<>1
\(P=\left(\frac{x+2\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\frac{x+2\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{x-1}\)
\(=\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)}\cdot\frac{1}{x-1}=\frac{1}{x-1}\)
b: \(x=4+2\sqrt3\)
=>\(P=\frac{1}{4+2\sqrt3-1}=\frac{1}{2\sqrt3+3}=\frac{2\sqrt3-3}{12-9}=\frac{2\sqrt3-3}{3}=\frac{6\sqrt3-9}{9}\)
=>\(\sqrt{P}=\sqrt{\frac{6\sqrt3-9}{9}}=\frac{\sqrt{6\sqrt3-9}}{3}\)
Bài 2:
a: ĐKXĐ: x>=0; x<>4
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2}{\sqrt{x}+3}-\frac{9\sqrt{x}-3}{x+\sqrt{x}-6}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+2\left(\sqrt{x}-2\right)-9\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{x+4\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-3\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}-1}{\sqrt{x}+3}\)
b: \(x^2-20x+64=0\)
=>(x-4)(x-16)=0
=>x=4(loại) hoặc x=16(nhận)
Thay x=16 vào A, ta được:
\(A=\frac{4-1}{4+3}=\frac37\)
