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

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

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

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

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

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

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

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

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

\(\left(\sqrt{x+2}-\sqrt{x-1}\right)\left(\sqrt{2-x}+1\right)-1=0\) (ĐKXĐ : \(1\le x\le2\) )

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(2+x\right)}+\sqrt{x+2}-\sqrt{\left(2-x\right)\left(x-1\right)}-\sqrt{x-1}-1=0\)

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(2+x\right)}-\left(2-\sqrt{x+2}\right)-\sqrt{\left(2-x\right)\left(x-1\right)}+\left(1-\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(2+x\right)}-\frac{2-x}{\sqrt{x+2}+2}-\sqrt{\left(2-x\right)\left(x-1\right)}+\frac{2-x}{\sqrt{x-1}+1}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{x+2}-\frac{\sqrt{2-x}}{\sqrt{x+2}+2}-\sqrt{x-1}+\frac{\sqrt{2-x}}{\sqrt{x-1}+1}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x-2}=0\\\sqrt{x+2}-\frac{\sqrt{2-x}}{\sqrt{x+2}+2}-\sqrt{x-1}+\frac{\sqrt{2-x}}{\sqrt{x-1}+1}=0\end{array}\right.\)

Với \(\sqrt{x-2}=0\) => x = 2 (TMĐK)

Với \(\sqrt{x+2}-\frac{\sqrt{2-x}}{\sqrt{x+2}+2}-\sqrt{x-1}+\frac{\sqrt{2-x}}{\sqrt{x-1}+1}=0\) , từ điều kiện \(1\le x\le2\) ta luôn có : \(\sqrt{x+2}-\frac{\sqrt{2-x}}{\sqrt{x+2}+2}-\sqrt{x-1}+\frac{\sqrt{2-x}}{\sqrt{x-1}+1}>0\)

Vậy phương trình có nghiệm : x = 2

\(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3\)(ĐKXĐ : \(x\le-1\)hoặc \(x\ge-\frac{1}{4}\))

\(\Leftrightarrow\left(\sqrt{4x^2+5x+1}-2\sqrt{7}x\right)-\left(\sqrt{4x^2-4x+4}-2\sqrt{7}x\right)-\left(9x-3\right)=0\)

\(\Leftrightarrow\frac{\left(4x^2+5x+1\right)-28x^2}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-\frac{\left(4x^2-4x+4\right)-28x^2}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-3\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{-24x^2+5x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}+\frac{24x^2+4x-4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-3\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{-\left(3x-1\right)\left(8x+1\right)}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}+\frac{4\left(3x-1\right)\left(2x+1\right)}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-3\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}3x-1=0\\\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3=0\end{array}\right.\)

Với 3x - 1 = 0 => x = \(\frac{1}{3}\) (TMĐK)

Với \(\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3=0\) , Từ điều kiện \(\left[\begin{array}{nghiempt}x\le-1\\x\ge-\frac{1}{4}\end{array}\right.\) ta luôn có : \(\frac{8x+4}{\sqrt{4x^2-4x+4}+2\sqrt{7}x}-\frac{8x+1}{\sqrt{4x^2+5x+1}+2\sqrt{7}x}-3>0\)

Vậy phương trình có nghiệm : \(x=\frac{1}{3}\)

Điều kiện xác định: x ≥ \(\dfrac{1}{3}\) 

<=> \(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=9x-3\)  

<=>  \(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}=\left(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}\right).\left(\sqrt{4x^2+5x+1}+\sqrt{4x^2-4x+4}\right)\)= \(\left(\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}\right).\left(1-\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}\right)=0\)<=>\(\left[{}\begin{matrix}\sqrt{4x^2+5x+1}=\sqrt{4x^2-4x+4}\left(1\right)\\1=\sqrt{4x^2+5x+1}-\sqrt{4x^2-4x+4}\left(2\right)\end{matrix}\right.\)

từ (1) ta có \(\sqrt{4x^2+5x+1}=\sqrt{4x^2-4x+4}\)

<=> 4x² + 5x + 1 = 4x² - 4x + 4 

<=> 9x = 3 => x = \(\dfrac{1}{3}\)

từ (2) ta có: 1 = 8x² + x + 5 - \(2\sqrt{16x^4+4x^3+16x+4}\)

<=> 8x² + x + 4 = 2\(\sqrt{16x^4+4x^3+16x+4}\) 

ta có xét delta VT thấy pt vô nghiệm 

VP dễ thấy phương trình có nghiệm x = \(\dfrac{-1}{4}\);-1 

ta suy ra 2 vế phương trình không bằng nhau nên pt (2) vô nghiệm.

vậy S={\(\dfrac{1}{3}\)} 

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nếu bạn xem rồi thì cho mình 1 like nha ghi bài giải hơi mệt nên mong bạn cho mình một like 

b: Sửa đề: \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)(1)

ĐKXĐ: \(x>=5\)

\(\left(1\right)\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

c: ĐKXĐ: \(\dfrac{3x-2}{x+1}>=0\)

=>\(\left[{}\begin{matrix}x>=\dfrac{2}{3}\\x< -1\end{matrix}\right.\)

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

=>\(\dfrac{3x-2}{x+1}=9\)

=>9(x+1)=3x-2

=>9x+9=3x-2

=>6x=-11

=>\(x=-\dfrac{11}{6}\left(nhận\right)\)

d: ĐKXĐ: \(\left\{{}\begin{matrix}5x-4>=0\\x+2>0\end{matrix}\right.\Leftrightarrow x>=\dfrac{4}{5}\)

\(\dfrac{\sqrt{5x-4}}{\sqrt{x+2}}=2\)

=>\(\sqrt{\dfrac{5x-4}{x+2}}=2\)

=>\(\dfrac{5x-4}{x+2}=4\)

=>5x-4=4x+8

=>x=12(nhận)

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290