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: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)

\(\Leftrightarrow\sqrt{x-2}=4\)

=>x-2=16

hay x=18

b: \(\Leftrightarrow\left|3x+2\right|=4x\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)

c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

\(\Leftrightarrow4\sqrt{x-2}=40\)

=>x-2=100

hay x=102

d: =>5x-6=9

hay x=3

\(a,\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\left(dk:x\ge2\right)\)

\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)

\(\Leftrightarrow\sqrt{x-2}=4\)

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\left(tmdk\right)\)

b,\(\sqrt{9x^2-12x+4=3x\left(dk:x\ge0\right)}\)

\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x\)

\(\Leftrightarrow\left|3x-2\right|=3x\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=3x\\3x-2=-3x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=\dfrac{1}{3}\left(tmdk\right)\end{matrix}\right.\)

Các câu còn lại làm tương tự nhé 

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)

\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)

\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)

\(-\sqrt{x-2}=-4\)

\(\sqrt{x-2}=4\)

\(\left|x-2\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)

d. \(\sqrt{9x^2+12x+4}=4\)

<=> \(\sqrt{\left(3x+2\right)^2}=4\)

<=> \(|3x+2|=4\)

<=> \(\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

c: Ta có: \(\dfrac{5\sqrt{x}-2}{8\sqrt{x}+2.5}=\dfrac{2}{7}\)

\(\Leftrightarrow35\sqrt{x}-14=16\sqrt{x}+5\)

\(\Leftrightarrow x=1\)

a: \(=\dfrac{4x\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}=\dfrac{4x}{3x-1}\)

b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)

d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)

h) \(x-2-\sqrt{4-4x+x^2}\)

\(=x-2-\sqrt{\left(2-x\right)^2}\)

\(=x-2-\left|2-x\right|\)

\(=x-2-2+x\)

\(=2x-4\)

g) \(x-2-\sqrt{4-4x+x^2}\)

\(=x-2-\sqrt{\left(2-x\right)^2}\)

\(=x-2-\left|2-x\right|\)

\(=x-2-\left[-\left(2-x\right)\right]\)

\(=x-2+2-x\)

\(=0\)

i) \(3-x+\sqrt{9+6x+x^2}\)

\(=3-x+\sqrt{\left(3+x\right)^2}\)

\(=3-x+\left|3+x\right|\)

\(=3-x-3-x\)

\(=-2x\)

điều kiện -4<=x<=4x<=4

\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)

\(A=\left|x+4\right|+\left|x-4\right|\)

KẾT HỢP ĐIỀU KIỆN

\(A=x+4+4-x\)

\(A=8\)

\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)

\(B=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(B=\left|3x-1\right|+\left|2x-3\right|\)

\(TH1:x>=\frac{3}{2}\)

\(B=3x-1+2x-3\)

\(B=5x-4\)

\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)

\(B=3x-1-2x+3\)

\(B=x+2\)

\(TH3:x< \frac{1}{3}\)

\(B=-3x+1-2x+3\)

\(B=4-5x\)

câu c và câu d tương tự

câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)

còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

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

bạn tự làm nốt câu c, d nha 

a, Với \(-4\le x\le4\)

 \(A=\sqrt{x^2+8x+16}+\sqrt{x^2-8x+16}\)

\(=\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}=\left|x+4\right|+\left|x-4\right|\)

b, \(B=\sqrt{9x^2-6x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(3x\right)^2-2.3x+1}+\sqrt{\left(2x\right)^2-2.2x.3x+3^2}\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}=\left|3x-1\right|+\left|2x-3\right|\)

+) ta có : \(A=6x+\sqrt{9x^2-12x+4}=6x+\sqrt{\left(3x-2\right)^2}\)

\(=6x+\left|3x-2\right|\) \(\Rightarrow\left[{}\begin{matrix}A=9x-2\left(x\ge\dfrac{2}{3}\right)\\A=3x+2\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

+) ta có : \(B=5x-\sqrt{x^2+4x+4}=5x-\sqrt{\left(x+2\right)^2}\)

\(=5x-\left|x+2\right|\) \(\Rightarrow\left[{}\begin{matrix}A=4x-2\left(x\ge-2\right)\\6x+2\left(x< -2\right)\end{matrix}\right.\)