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.......

Đk: x >/ 5

pt đã cho \(\Leftrightarrow7\sqrt{x-2}-14\cdot\dfrac{\sqrt{x-2}}{7}-3\sqrt{x-5}=4\)

\(\Leftrightarrow7\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-5}=4\)

\(\Leftrightarrow5\sqrt{x-2}-3\sqrt{x-5}=4\)

\(\Leftrightarrow5\sqrt{x-2}=4+3\sqrt{x-5}\)

\(\Leftrightarrow25x-50=16+9x-45+24\sqrt{x-5}\)

\(\Leftrightarrow16x-21=24\sqrt{x-5}\)

\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-672x+441=576x-2880\\x\ge\dfrac{21}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-1248x+3321=0\\x\ge\dfrac{21}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}256x^2-1248x+3321=0\left(vn\right)\\x\ge\dfrac{21}{16}\end{matrix}\right.\)

Kl: ptvn

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

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)

b. ĐKXĐ: $x^2\geq 5$

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

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

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

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

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

ĐKXĐ: \(x\ge2\)

Từ pt đã cho suy ra:

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

⇒ \(2\sqrt{x-2}=8\) ⇒ \(x=18\)

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

Sao ko gửi đc ảnh nhỉ

a) Ta có: \(\sqrt{25x+75}+2\sqrt{9x+27}=5\sqrt{x+3}+18\)

\(\Leftrightarrow5\sqrt{x+3}+6\sqrt{x+3}-5\sqrt{x+3}=18\)

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

\(\Leftrightarrow x+3=9\)

hay x=6

b) Ta có: \(\sqrt{4x-8}-14\sqrt{\dfrac{x-2}{49}}=\sqrt{9x-18}+8\)

\(\Leftrightarrow2\sqrt{x-2}-2\sqrt{x-2}-3\sqrt{x-2}=8\)

\(\Leftrightarrow-3\sqrt{x-2}=8\)(Vô lý)

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

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

  \(\Leftrightarrow\) \(\sqrt{x-2}=24\)

\(\Leftrightarrow\)\(x-2=576\)\(\Leftrightarrow x=578\)

c)\(\Leftrightarrow GTTĐ\left(x-1\right)=\sqrt{2}-1\)\(TH1:x-1>0\)

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

\(TH2:x-1< 0\)

\(\Rightarrow1-x=\sqrt{2}-1\)

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

d)\(TH1:x-10=0\Rightarrow x=10\)

\(TH2:\sqrt{x-4}=0\Rightarrow x=4\)

câu b) thì mik cần thêm time