Bài 35:
a: ĐKXĐ: x>=0; x<>1
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\left(\frac{1-x}{\sqrt2}\right)^2\)
\(=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\frac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)
\(=\frac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{1}\cdot\frac{\left(\sqrt{x}-1\right)}{2}=-\sqrt{x}\left(\sqrt{x}-1\right)\)
b: P>0
=>\(-\sqrt{x}\left(\sqrt{x}-1\right)>0\)
=>\(-\left(\sqrt{x}-1\right)>0\)
=>\(\sqrt{x}-1<0\)
=>\(\sqrt{x}<1\)
=>0<x<1
c: \(P=-\sqrt{x}\left(\sqrt{x}-1\right)\)
\(=-x+\sqrt{x}-\frac14+\frac14=-\left(\sqrt{x}-\frac12\right)^2+\frac14\le\frac14\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}-\frac12=0\)
=>\(\sqrt{x}=\frac12\)
=>x=1/4(nhận)
Bài 36:
a: ĐKXĐ: x>0; x<>1
\(P=\frac{x\cdot\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}=\frac{2\sqrt{x}+x+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: \(P=\frac92\)
=>\(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac92\)
=>\(2\left(x+2\sqrt{x}+1\right)=9\sqrt{x}\)
=>\(2x-5\sqrt{x}+2=0\)
=>\(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x}-2=0\\ 2\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=2\\ \sqrt{x}=\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(nhận\right)\\ x=\frac14\left(nhận\right)\end{array}\right.\)
