Bài 4:
a: ĐKXĐ: x>0; x<>1
\(D=\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\left(\sqrt{x}-1\right)\cdot\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1+\sqrt{x}+2-x}=\frac{x\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x}{\sqrt{x}-1}\)
b: D>1
=>D-1>0
=>\(\frac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\)
=>\(\sqrt{x}-1>0\)
=>\(\sqrt{x}>1\)
=>x>1
d: Để D là số nguyên thì x⋮\(\sqrt{x}-1\)
=>\(x-1+1\) ⋮\(\sqrt{x}-1\)
=>1⋮\(\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\lbrace1;-1\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace2;0\right\rbrace\)
=>x∈{0;4}
e: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>2x=4 hoặc 2x=-6
=>x=2(nhận) hoặc x=-3(loại)
Thay x=2 vào D, ta được:
\(D=\frac{2}{\sqrt2-1}=2\left(\sqrt2+1\right)=2\sqrt2+2\)
f: \(D=\frac92\)
=>\(\frac{x}{\sqrt{x}-1}=\frac92\)
=>\(2x=9\sqrt{x}-9\)
=>\(2x-9\sqrt{x}+9=0\)
=>\(\left(\sqrt{x}-3\right)\left(2\sqrt{x}-3\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x}=3\\ \sqrt{x}=\frac32\end{array}\right.\Rightarrow\left[\begin{array}{l}x=9\left(nhận\right)\\ x=\frac94\left(nhận\right)\end{array}\right.\)
Bài 5:
a: ĐKXĐ: x>=0; x<>1
\(E=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-x}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{1-\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\frac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-\sqrt{x}-\sqrt{x}}{x-1}\cdot\frac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}=\frac{2x+1}{4\sqrt{x}}\)
c: \(x=17-4\sqrt{13}=\left(\sqrt{13}-2\right)^2\)
\(E=\frac{2x+1}{4\sqrt{x}}\)
\(=\frac{2\left(17-4\sqrt{13}\right)+1}{4\cdot\sqrt{\left(\sqrt{13}-2\right)^2}}=\frac{34-8\sqrt{13}+1}{4\left(\sqrt{13}-2\right)}=\frac{35-8\sqrt{13}}{4\left(\sqrt{13}-2\right)}\)
d: \(E=\frac98\)
=>\(\frac{2x+1}{4\sqrt{x}}=\frac98=\frac{9\sqrt{x}}{8\sqrt{x}}\)
=>\(2\left(2x+1\right)=9\sqrt{x}\)
=>\(4x-9\sqrt{x}+2=0\)
=>\(4x-8\sqrt{x}-\sqrt{x}+2=0\)
=>\(\left(\sqrt{x}-2\right)\left(4\sqrt{x}-1\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x}-2=0\\ 4\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\left(nhận\right)\\ x=\frac{1}{16}\left(nhận\right)\end{array}\right.\)
