Bài 27:
1: \(M=\left(\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}+1}\right)\cdot\frac{\sqrt{x}-1}{2}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\frac{\sqrt{x}-1}{2}=\frac{2\sqrt{x}}{2\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
2: \(M\le\frac12\)
=>\(M-\frac12\le0\)
=>\(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac12\le0\)
=>\(\frac{2\sqrt{x}-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}\le0\)
=>\(\sqrt{x}-1\le0\)
=>\(\sqrt{x}\le1\)
=>0<=x<1
3: \(\frac{1}{M}=\frac{\sqrt{x}+1}{\sqrt{x}}=1+\frac{1}{\sqrt{x}}\)
Để 1/M là số nguyên thì 1⋮\(\sqrt{x}\)
=>\(\sqrt{x}=1\)
=>x=1(loại)
Bài 28:
1: \(M=\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right):\frac{2}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}-2}{2}\)
\(=\frac{2\sqrt{x}+2}{2\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+2}\)
2: \(M=\frac45\)
=>\(\frac{\sqrt{x}+1}{\sqrt{x}+2}=\frac45\)
=>\(5\left(\sqrt{x}+1\right)=4\left(\sqrt{x}+2\right)\)
=>\(5\sqrt{x}+5=4\sqrt{x}+8\)
=>\(\sqrt{x}=3\)
=>x=9(nhận)
3: \(M^2-M=M\left(M-1\right)\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}-1\right)=\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{-1}{\sqrt{x}+2}=\frac{-\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)^2}<0\) ∀x thỏa mãn ĐKXĐ
=>\(M^2
