a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

\(\Leftrightarrow\sqrt{x+3}=3\)

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

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)

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

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

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

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

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

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

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

f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)

\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)

\(\Leftrightarrow\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)

\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3x-1\\t=-3x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

\(\Delta=1+4\left(x^2-3x+2\right)=\left(2x-3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+2x-3}{-2}=1-x\\t=\dfrac{1-2x+3}{-2}=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{8}{3}\)

\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)

Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)

\(\left(t^2-6\right)^2-6-t=0\)

\(\Leftrightarrow t^4-12t^2-t+30=0\)

\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\sqrt{4x+8}+\sqrt{9x+18}=5\left(dkxd:x\ge-2\right)\)

\(\Leftrightarrow\sqrt{4\left(x+2\right)}+\sqrt{9\left(x+2\right)}=5\)

\(\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=5\)

\(\Leftrightarrow5\sqrt{x+2}=5\)

\(\Leftrightarrow\sqrt{x+2}=1\)

\(\Leftrightarrow x+2=1\)

\(\Leftrightarrow x=-1\left(tmdk\right)\)

ĐKXĐ: \(x\ge2\)

Ta có: \(\sqrt{4x-8}-\sqrt{9x-18}+2\sqrt{x-2}=1\)

\(\Leftrightarrow2\sqrt{x-2}-3\sqrt{x-2}+2\sqrt{x-2}=1\)

\(\Leftrightarrow\sqrt{x-2}=1\)

\(\Leftrightarrow x-2=1\)

hay x=3(nhận)

Vậy: S={3}

Câu 1 :

Xét điều kiện:\(\hept{\begin{cases}x\ge5\\x\le1\end{cases}}\)(Vô lý) 

Vậy pt vô nghiệm

Câu 2 : 

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

Vậy x=-1

Câu 3 : 

\(\sqrt{3x^2-4x+3}=1-2x\)\(\Leftrightarrow3x^2-4x+3=1+4x^2-4x\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt{2}\)

Câu 4 : 

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

\(\Leftrightarrow x=15\)

\(3\sqrt{x-2}-\sqrt{4x-8}+4\sqrt{\dfrac{9x-18}{4}}=14\left(x\ge0;x\ne2\right)\\ \Leftrightarrow3\sqrt{x-2}-\sqrt{4\left(x-2\right)}+4\cdot\dfrac{1}{2}\sqrt{9\left(x-2\right)}=14\\ \Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+6\sqrt{x-2}=14\\ \Leftrightarrow7\sqrt{x-2}=14\\ \Leftrightarrow\sqrt{x-2}=2\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\)