a. 

ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow 6\sqrt{2x}-4\sqrt{2x}+5\sqrt{2x}=21$
$\Leftrightarrow 7\sqrt{2x}=21$

$\Leftrightarrow \sqrt{2x}=3$

$\Leftrightarrow 2x=9$

$\Leftrightarrow x=\frac{9}{2}$ (tm)

b.

ĐKXĐ: $x\geq -2$

PT $\Leftrightarrow \sqrt{25(x+2)}+3\sqrt{4(x+2)}-2\sqrt{16(x+2)}=15$

$\Leftrightarrow 5\sqrt{x+2}+6\sqrt{x+2}-8\sqrt{x+2}=15$

$\Leftrightarrow 3\sqrt{x+2}=15$

$\Leftrightarrow \sqrt{x+2}=5$

$\Leftrightarrow x+2=25$

$\Leftrightarrow x=23$ (tm)

c.

$\sqrt{(x-2)^2}=12$

$\Leftrightarrow |x-2|=12$

$\Leftrightarrow x-2=12$ hoặc $x-2=-12$

$\Leftrightarrow x=14$ hoặc $x=-10$

e.

PT $\Leftrightarrow |2x-1|-x=3$

Nếu $x\geq \frac{1}{2}$ thì $2x-1-x=3$

$\Leftrightarrow x=4$ (tm)

Nếu $x< \frac{1}{2}$ thì $1-2x-x=3$

$\Leftrightarrow x=\frac{-2}{3}$ (tm)

f.

ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{3(x-2)}-(x-2)=0$

$\Leftrightarrow \sqrt{x-2}(\sqrt{3}-\sqrt{x-2})=0$

$\Leftrightarrow \sqrt{x-2}=0$ hoặc $\sqrt{3}-\sqrt{x-2}=0$

$\Leftrightarrow x=2$ hoặc $x=5$ (tm)

h. ĐKXĐ: $x\leq \frac{3}{2}$

PT $\Leftrightarrow \sqrt{3-2x}=x+2$

\(\Rightarrow \left\{\begin{matrix} x+2\geq 0\\ 3-2x=(x+2)^2=x^2+4x+4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ x^2+6x+1=0\end{matrix}\right.\)

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

Vậy.......

a,ĐK: x≥4

Ta có: \(2\sqrt{x-4}-\dfrac{1}{3}\sqrt{9x-36}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}-\sqrt{x-4}=4-\sqrt{x-4}\)

      \(\Leftrightarrow2\sqrt{x-4}=4\)

      \(\Leftrightarrow\sqrt{x-4}=2\Leftrightarrow x-4=4\Leftrightarrow x=8\left(tm\right)\)

b, ĐK: x≥2

Ta có: \(3\sqrt{x-2}-\sqrt{x^2-4}=0\)

      \(\Leftrightarrow3\sqrt{x-2}-\sqrt{\left(x-2\right)\left(x+2\right)}=0\)

      \(\Leftrightarrow\sqrt{x-2}\left(3-\sqrt{x+2}\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=0\\3-\sqrt{x+2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\\sqrt{x+2}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x+2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)

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: 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

a) - 3 =0

<=> (√x-3)(√x+3)-3√x-3=0

<=> (√x-3)(√x+3-3)=0

<=> (√x-3)√x=0

<=> √x-3=0

<=>x=9

b)=x - 3

<=> √(2x -3)² =x-3

<=> 2x-3=x-3

<=>2x-x=-3+3

<=>x=0

c)=3x-1

<=> √(x+3)² =3x-1

<=> x+3=3x-1

<=> -2x=-4

<=>  x=2

Nhớ cho mình 1 tim nha bạn

Lời giải:

a. ĐKXĐ: $x\geq 3$

PT $\Leftrightarrow \sqrt{(x-3)(x+3)}-3\sqrt{x-3}=0$

$\Leftrightarrow \sqrt{x-3}(\sqrt{x+3}-3)=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}-3=0$

$\Leftrightarrow \sqrt{x-3}=0$ hoặc $\sqrt{x+3}=3$

$\Leftrightarrow x=3$ hoặc $x=6$ (tm)

b.

PT \(\Rightarrow \left\{\begin{matrix} x-3\geq 0\\ 4x^2-12x+9=(x-3)^2=x^2-6x+9\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x^2-6x=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 3x(x-2)=0\end{matrix}\right.\)

$\Rightarrow$ không có giá trị $x$ nào thỏa mãn 

Vậy pt vô nghiệm.

c.

PT \(\Rightarrow \left\{\begin{matrix} 3x-1\geq 0\\ x^2+6x+9=(3x-1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ x^2+6x+9=9x^2-6x+1\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 8x^2-12x-8=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{1}{3}\\ 4(x-2)(2x+1)=0\end{matrix}\right.\Leftrightarrow x=2\)

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)

a) \(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)

\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )

b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)

\(\Leftrightarrow\left|2x-1\right|=5x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)

\(\Leftrightarrow x+3=4\)

\(\Rightarrow x=1\)

a: =>|2x-1|=3

=>2x-1=3 hoặc 2x-1=-3

=>2x=-2 hoặc 2x=4

=>x=2 hoặc x=-1

c: \(\Leftrightarrow\left|x-3\right|=11-x\)

=>x<=11 và (x-3)^2=(11-x)^2

=>x<=11 và x^2-6x+9=x^2-22x+121

=>x<=11 và 16x=112

=>x=7

d:

ĐKXĐ: 3x+19>=0

=>x>=-19/3

PT =>x>=-3 và (3x+19)=(x+3)^2=x^2+6x+9

=>x>=-3 và x^2+6x+9-3x-19=0 

=>x>=-3 và (x+5)(x-2)=0

=>x=2

e: =>\(\sqrt{x^2+x+5}=x+1\)

=>x>=-1 và x^2+x+5=x^2+2x+1

=>x>=-1 và 2x+1=x+5

=>x=4