a: \(=\dfrac{\left|x+2\right|}{x-1}\)

b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)

c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)

Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

\(A=1-\left(\dfrac{2}{1+2\sqrt{x}}-\dfrac{5\sqrt{x}}{4x-1}-\dfrac{1}{1-2\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)

\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)

\(A=1-\dfrac{4\sqrt{x}-2-5\sqrt{x}+2\sqrt{x}+1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)

\(A=1-\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}\cdot\dfrac{\left(2\sqrt{x}+1\right)^2}{\sqrt{x}-1}\)

\(A=1-\dfrac{2\sqrt{x}+1}{2\sqrt{x}-1}=\dfrac{2\sqrt{x}-1-2\sqrt{x}-1}{2\sqrt{x}-1}=\dfrac{-2}{2\sqrt{x}-1}\)

Tick hộ nha

Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
    \(3x=16+2=18\)
    \(x=18:3=6\)
    Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
    \(2x+1=5+11=16\)
    \(2x=16-1=15\)
    \(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
    \(2x-1=-5+11=6\)
    \(2x=6+1=7\)
    \(x=7:2=3,5\)
    Vậy \(x=\left\{7,5;3,5\right\}\) 
    (Câu này mình không chắc chắn lắm)   
    (Học sinh lớp 6 đang làm bài này)    

Bài 4:

a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)

b: C-6<0

=>C<6

=>\(2\sqrt{x}< 6\)

=>\(\sqrt{x}< 3\)

=>0<=x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)

Bài 3

a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)

Vậy PT có nghiệm x=6

b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)

+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)

+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)

Bài 4

a)\(C=1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\Leftrightarrow C=\dfrac{x-1}{\sqrt{x}}\left(\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\Leftrightarrow C=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\dfrac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Leftrightarrow C=\dfrac{2x}{\sqrt{x}}\Leftrightarrow C=2\sqrt{x}\)

\(Vậy\) \(C=2\sqrt{x}\)

\(A=1-\left(\frac{2}{1+2\sqrt{x}}-\frac{5\sqrt{x}}{4x-1}-\frac{1}{1-2\sqrt{x}}\right):\frac{\sqrt{x}-1}{4x+4\sqrt{x}+1}\)

\(=1-\left(\frac{2\left(1-2\sqrt{x}\right)+5\sqrt{x}-1-2\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}\right):\frac{\sqrt{x}-1}{\left(1+2\sqrt{x}\right)^2}\)

\(=1-\frac{1-\sqrt{x}}{\left(1+2\sqrt{x}\right)\left(1-2\sqrt{x}\right)}.\frac{\left(1+2\sqrt{x}\right)^2}{\sqrt{x}-1}=1-\frac{1+2\sqrt{x}}{1-2\sqrt{x}}=2-\frac{2}{1-2\sqrt{x}}\)

để A là số nguyên thì \(1-2\sqrt{x}\) là ước của 2 khi đó ta tìm được \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)