\(a,ĐK:x\ge2\\ PT\Leftrightarrow4\sqrt{x-2}-2\sqrt{x-2}-7\sqrt{x-2}=-10\\ \Leftrightarrow-5\sqrt{x-2}=-10\\ \Leftrightarrow\sqrt{x-2}=2\Leftrightarrow x-2=4\\ \Leftrightarrow x=6\left(tm\right)\\ b,ĐK:x\ge1\\ PT\Leftrightarrow x-3=\sqrt{x-1}\\ \Leftrightarrow x^2-6x+9=x-1\\ \Leftrightarrow x^2-7x+10=0\\ \Leftrightarrow\left(x-2\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\left(tm\right)\)

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

d. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{49}.\sqrt{x-2}-14\sqrt{\frac{1}{49}}\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8$

$\Leftrightarrow 2\sqrt{x-2}=8$

$\Leftrightarrow \sqrt{x-2}=4$

$\Leftrightarrow x=4^2+2=18$ (tm)

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

a) \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\) (ĐKXĐ : \(x\ge2\) )

\(\Leftrightarrow5\sqrt{x+3}+3\sqrt{x-2}-4\sqrt{x+3}-3\sqrt{x-2}=2\)

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

\(\Leftrightarrow x+3=4\)

\(\Leftrightarrow x=1\) ( Thỏa mãn ĐKXĐ )

c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐKXĐ : \(x\ge-5\) )

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

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

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

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\) ( Thỏa mãn ĐKXĐ )

Vậy.......

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

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

c: \(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=18\)

=>2 căn x-2=18

=>x-2=81

=>x=83

Giải:

\(\sqrt{49x-98}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

ĐKXĐ: \(x-2\ge0\Leftrightarrow x\ge2\)

\(7\sqrt{x-2}-2\sqrt{x-2}=3\sqrt{x-2}+8\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

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

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

\(\Leftrightarrow x-2=16\)

\(\Leftrightarrow x=18\) (thỏa mãn)

Vậy ...

1/ \(\Leftrightarrow\left|2x-1\right|=7\Leftrightarrow\left[{}\begin{matrix}2x-1=7\\2x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

2/ \(\Leftrightarrow6\sqrt{x+2}-2\sqrt{x+2}=9\sqrt{x+2}-10\)

\(\Leftrightarrow5\sqrt{x+2}=10\)

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

\(\Leftrightarrow x=2\)

a) \(4\sqrt{4x-8}-2\sqrt{9x-18}+\sqrt{16x-32}=5\)

\(\rightarrow4.2\sqrt{x-2}-2.3\sqrt{x-2}+4\sqrt{x-2}=5\)

\(\rightarrow\sqrt{x-2}\left(8-6+4\right)=5\)

\(\rightarrow6\sqrt{x-2}=5\)

\(\rightarrow\sqrt{x-2}=\frac{5}{6}\)

\(\rightarrow x-2=\frac{25}{36}\)

\(\Rightarrow x=\frac{97}{36}\)

b)\(\sqrt{x^2+6x+9}-2=7\)

\(\rightarrow\sqrt{\left(x+3\right)^2}=9\)

\(\rightarrow x+3=9\)

\(\Rightarrow x=6\)

Nhớ tick mik nha

a, \(\Leftrightarrow4\sqrt{4\left(x-2\right)}-2\sqrt{9\left(x-2\right)}+\sqrt{16\left(x-2\right)}=5\)

\(\Leftrightarrow8\sqrt{x-2}-6\sqrt{x-2}+4\sqrt{x-2}=5\)

\(\Leftrightarrow\sqrt{x-2}\left(8-6+4\right)=5\)

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

\(\Leftrightarrow x=3\)

b, ĐKXĐ: \(x^2+6x+9\ge0\)

\(\Leftrightarrow\sqrt{x^2+6x+9}=9\\ \Leftrightarrow x^2+6x+9=81\\ \Leftrightarrow x^2+6x-72=0\\ \Leftrightarrow\left(x-6\right)\left(x+12\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-12\end{matrix}\right.\left(tm\right)\)

Vậy nghiệm của phương trình đã cho là 6, -12