a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(x+3\right)^2}=12\)

=>\(\left|x+3\right|=12\)

=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)

b: ĐKXĐ: x>=1

\(\sqrt{25x-25}-\sqrt{9x-9}=10\)

=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)

=>\(2\sqrt{x-1}=10\)

=>x-1=25

=>x=26(nhận)

ĐK: \(x+2\ge0\Leftrightarrow x\ge-2\)

\(3\sqrt{x+2}-\sqrt{x+2}-4\sqrt{x+2}=-10\)

\(-2\sqrt{x+2}=-10\)

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

\(\left\{{}\begin{matrix}5\ge0\left(ld\right)\\x+2=25\end{matrix}\right.\)\(\Leftrightarrow x=23\left(n\right)\)

a)

\(\sqrt{9x^2}=12\\ < =>\left(\sqrt{9x^2}\right)^2=12^2\\ < =>9x^2=144\\ < =>x^2=16\\ < =>\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

b)

\(\sqrt{25x^2}=\left|-50\right|\\ < =>\sqrt{25x^2}=50\left(vì-50< 0\right)\)

\(< =>\left(\sqrt{25x^2}\right)^2=50^2\\ =>25x^2=2500\\ < =>x^2=100\\ < =>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

a, x = 4

b, x = 10 

\(a)\sqrt{9x^2}=12\\ \Leftrightarrow\sqrt{\left(3x\right)^2}=12\\ \Leftrightarrow\left|3x\right|=12\\ \Leftrightarrow\left|x\right|=\dfrac{12}{3}\\ \Leftrightarrow x=4\)

Vậy x=4

\(b)\sqrt{25x^2}=\left|-50\right|\\ \Leftrightarrow\sqrt{\left(5x\right)^2}=\left|-50\right|\\ \Leftrightarrow\left|5x\right|=50\\ \Leftrightarrow\left|x\right|=\dfrac{50}{5}\\ \Leftrightarrow x=10\)

Vậy x=10

Suy ra :

d)

làm sao ?? cho e hỏi cái

<=>   \(\sqrt{64\left(x+1\right)}-\sqrt{25\left(x+1\right)}+\sqrt{4\left(x+1\right)}=20\)

<=> \(8\sqrt{\left(x+1\right)}-5\sqrt{\left(x+1\right)}+2\sqrt{\left(x+1\right)}=20\)

<=>   . \(5\sqrt{\left(x+1\right)}=20\)

<=>  \(\sqrt{\left(x+1\right)}=4\)

=> x+1=16

=> x=15

1) ĐKXĐ: \(x\ge-2\)

\(pt\Leftrightarrow\dfrac{\sqrt{x+2}}{2}+5\sqrt{x+2}-2\sqrt{x+2}=14\)

\(\Leftrightarrow\dfrac{\sqrt{x+2}+6\sqrt{x+2}}{2}=14\Leftrightarrow7\sqrt{x+2}=28\)

\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)

2) ĐKXĐ: \(x\ge0\)

\(pt\Leftrightarrow2x+3=x^2\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

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

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

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

4) ĐKXĐ: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1\ge0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1\le0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{2}\\x\le-1\end{matrix}\right.\)

\(pt\Leftrightarrow\dfrac{x+1}{2x-1}=4\Leftrightarrow x+1=8x-4\)

\(\Leftrightarrow7x=5\Leftrightarrow x=\dfrac{5}{7}\left(tm\right)\)

5) ĐKXĐ: \(x\ge2\)

\(pt\Leftrightarrow\dfrac{x-2}{3x+1}=36\)

\(\Leftrightarrow x-2=108x+36\Leftrightarrow107x=-38\Leftrightarrow x=-\dfrac{38}{107}\left(ktm\right)\)

Vậy \(S=\varnothing\)

Ta có: \(x^{10}=25\cdot x^8\)

- Với x=0 => x thỏa man đề bài.

- Với x khác 0 ta có: 

\(x^{10}=25\cdot x^8\)

<=> \(\frac{x^{10}}{x^8}=25\)

<=> \(x^2=25\)

<=> \(x=\pm5\)

Vậy x\(\in\left\{-5;0;5\right\}\)

Ta có:

     x¹⁰ = 25 . x⁸

=> x¹⁰ : x⁸ = 25

=>      x²    = 25

=>    x²    =   5²      hoặc  x² = (-5)²

=> x = 5 hoặc x = -5 

x^10= 25x^8 suy ra x^10-25x^8=0 suy ra x^8.(x^2-25)=0

suy ra x^8=0 hoặc x^2-25=0

Do đó x bằng 0 hoặc -5 hoặc 5

2) \(\frac{1}{5}\sqrt{25x+50}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}\sqrt{25\left(x+2\right)}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.\sqrt{25}.\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9x+18}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9\left(x+2\right)}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+\sqrt{9}.\sqrt{x+2}+9=0\)

\(\frac{1}{5}.5\sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0\)

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

\(-\sqrt{x+2}=-9\)

\(x+2=81\)

\(\Rightarrow x=79\)

3) \(\sqrt{x^2-4x+4}=7x-1\)

\(\sqrt{x^2-2.x.2+2^2}=7x-1\)

\(\sqrt{\left(x-2\right)^2}=7x-1\)

\(x-2=7x-1\)

\(-2=7x-1-x\)

\(-2+1=7x-x\)

\(-1=6x\)

\(-\frac{1}{6}=x\)

\(\Rightarrow x=-\frac{1}{6}\)