Bài 3:
a: \(P=\left(\frac{1-a\cdot\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\frac{1-\sqrt{a}}{1-a}\)
\(=\left(\frac{\left(1-\sqrt{a}\right)}{1-\sqrt{a}}\left(1+\sqrt{a}+a\right)+\sqrt{a}\right)\cdot\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\)
\(=\frac{\left(1+a+2\sqrt{a}\right)}{1+\sqrt{a}}=\sqrt{a}+1\)
b: \(a=\sqrt{\frac52-\sqrt6}=\sqrt{\frac{5-2\sqrt6}{2}}\)
\(=\frac{\sqrt{5-2\sqrt6}}{\sqrt2}=\frac{\left(\sqrt3-\sqrt2\right)}{\sqrt2}=\frac{\sqrt6-2}{2}\)
\(P=\sqrt{a}+1\)
\(=\sqrt{\frac{\sqrt6-2}{2}}+1=\sqrt{\frac{2\sqrt6-4}{4}}+1=\frac{\sqrt{2\sqrt6-4}}{2}+1=\frac{\sqrt{2\sqrt6-4}+2}{2}\)
Bài 4:
a: ĐKXĐ: x>0; x<>4
b: \(P=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}+2}{x-2\sqrt{x}}+\frac{3}{\sqrt{x}}\right)\)
\(=\frac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{\sqrt{x}+2+3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}+2+3\sqrt{x}-6}\)
\(=\frac{-4x-8\sqrt{x}}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{4\sqrt{x}-4}=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{4\left(\sqrt{x}-1\right)}=\frac{-x}{\sqrt{x}-1}\)
c: P=-4
=>\(\frac{-x}{\sqrt{x}-1}=-4\)
=>\(-x=-4\left(\sqrt{x}-1\right)=-4\sqrt{x}+4\)
=>\(x-4\sqrt{x}+4=0\)
=>\(\left(\sqrt{x}-2\right)^2=0\)
=>\(\sqrt{x}-2=0\)
=>\(\sqrt{x}=2\)
=>x=4(loại)
