Giúp mình với 

Nếu chưa quen giải toán căn thức, em tìm ĐKXĐ cho x, rồi đặt \(\sqrt{x}=t\ge0\Rightarrow x=t^2\) rồi thế vào giải là nó ra 1 pt bình thường theo biến t thôi

a) Ta có: \(\sqrt{9x}+\sqrt{x}=12\)

\(\Leftrightarrow4\sqrt{x}=12\)

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

hay x=9

b) Ta có: \(\dfrac{\sqrt{x}+3}{4}=\dfrac{\sqrt{x}}{3}\)

\(\Leftrightarrow4\sqrt{x}=3\sqrt{x}+9\)

\(\Leftrightarrow\sqrt{x}=9\)

hay x=81

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

\(\Leftrightarrow5\sqrt{x}-x=2\sqrt{x}\)

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

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

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

hay x=9

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)

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

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

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

=>\(\sqrt{x+3}=2\)

=>x+3=4

=>x=1(nhận)

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

\(hcmuop\underrightarrow{jjjjjjjjj}me\)

\(a,\Leftrightarrow\left(x-9\right)^2-2\left(x-9\right)+1=0\\ \Leftrightarrow\left(x-9-1\right)^2=0\Leftrightarrow x=10\\ b,Sửa:49x^2-14x\sqrt{5}+5=0\\ \Leftrightarrow\left(7x-\sqrt{5}\right)^2=0\Leftrightarrow x=\dfrac{\sqrt{5}}{7}\)

\(\sqrt{X^2}\) =I X I

=>I X I +I X I =12 =>2IxI=12

=>I x I=6

=>x=6 hoặc x=^-6