Bài 4:
a: \(D=\left(\frac{x-1}{\sqrt{x}-2}-1\right)\cdot\left(\frac{1}{\sqrt{x}+1}-\frac{2-\sqrt{x}}{x\cdot\sqrt{x}+1}\right)\)
\(=\frac{x-1-\sqrt{x}+2}{\sqrt{x}-2}\cdot\left(\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\)
\(=\frac{x-\sqrt{x}+1}{\sqrt{x}-2}\cdot\frac{x-\sqrt{x}+1+\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{x-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
b: Thay x=1/4 vào D, ta được:
\(D=\left(\sqrt{\frac14}-1\right):\left(\sqrt{\frac14}-2\right)=\left(\frac12-1\right):\left(\frac12-2\right)=\frac{-1}{2}:\frac{-3}{2}=\frac13\)
c: \(D=\frac23\sqrt{x}\)
=>\(\frac{\sqrt{x}-1}{\sqrt{x}-2}=\frac{2\sqrt{x}}{3}\)
=>\(2x-4\sqrt{x}=3\sqrt{x}-3\)
=>\(2x-7\sqrt{x}+3=0\)
=>\(\left(\sqrt{x}-3\right)\left(2\sqrt{x}-1\right)=0\)
=>\(\left[\begin{array}{l}x=9\left(nhận\right)\\ x=\frac14\left(nhận\right)\end{array}\right.\)
d: D>=0
=>\(\frac{\sqrt{x}-1}{\sqrt{x}-2}\ge0\)
=>\(\left[\begin{array}{l}\sqrt{x}>2\\ \sqrt{x}\le1\end{array}\right.\Rightarrow\left[\begin{array}{l}x>4\\ 0\le x\le1\end{array}\right.\)
Bài 3:
a: \(C=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b: Thay \(x=7+4\sqrt3=\left(2+\sqrt3\right)^2\) vào C, ta được:
\(C=\frac{3\cdot\sqrt{\left(2-\sqrt3\right)^2}}{\sqrt{\left(2-\sqrt3\right)^2}-3}=\frac{3\left(2-\sqrt3\right)}{2-\sqrt3-3}=\frac{3\left(2-\sqrt3\right)}{-\sqrt3-1}=\frac{-3\left(2-\sqrt3\right)}{\sqrt3+1}=\frac{-3\left(2-\sqrt3\right)\left(\sqrt3-1\right)}{2}=\frac{-3\left(2\sqrt3-2-3+\sqrt3\right)}{2}=\frac{-3\left(3\sqrt3-5\right)}{2}\)
c: C=-5
=>\(3\sqrt{x}=-5\left(\sqrt{x}-3\right)=-5\sqrt{x}+15\)
=>\(8\sqrt{x}=15\)
=>\(\sqrt{x}=\frac{15}{8}\)
=>x=225/64(nhận)
