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

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

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

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

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

ĐKXĐ: \(x\ge-4\)

\(\Leftrightarrow\sqrt{x+8}+2=\sqrt{5x+20}\)

\(\Leftrightarrow x+12+4\sqrt{x+8}=5x+20\)

\(\Leftrightarrow\sqrt{x+8}=x+2\left(x\ge-2\right)\)

\(\Leftrightarrow x+8=x^2+4x+4\)

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

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

\(a,ĐK:x\ge-7\\ PT\Leftrightarrow\sqrt{\left(\sqrt{x+7}+1\right)^2}+\sqrt{x+7-\sqrt{x+7}-6}=4\)

Đạt \(\sqrt{x+7}=a\ge0\)

\(PT\Leftrightarrow\sqrt{\left(a+1\right)^2}+\sqrt{a^2-a-6}=4\\ \Leftrightarrow a+1+\sqrt{a^2-a-6}=4\\ \Leftrightarrow\sqrt{a^2-a-6}=3-a\\ \Leftrightarrow a^2-a-6=a^2-6a+9\\ \Leftrightarrow5a=15\Leftrightarrow a=3\\ \Leftrightarrow\sqrt{x+7}=3\\ \Leftrightarrow x+7=9\\ \Leftrightarrow x=2\left(tm\right)\)

Tử số của phân số đầu phải là \(\sqrt{x}+2\) chứ không phải \(\sqrt{x+2}\), vì cái \(\sqrt{x}+2\) nó mới logic để rút gọn: )

\(Q=\left(\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}^3+8}-\dfrac{x-\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\left(\dfrac{x+4\sqrt{x}+4-x+\sqrt{x}}{\sqrt{x}^3+8}\right)\left(\dfrac{5x-10\sqrt{x}+20}{5\sqrt{x}+4}\right)\\ =\dfrac{\left(5\sqrt{x}+4\right).5.\left(x-2\sqrt{x}+4\right)}{\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)\left(5\sqrt{x}+4\right)}\\ =\dfrac{5}{\sqrt{x}+2}\)

https://olm.vn/hoi-dap/question/1102059.html

a)  ĐK:  \(x\ge5\)

 \(\sqrt{4x-20}+\frac{1}{3}\sqrt{9x-45}-\frac{1}{5}\sqrt{16x-80}=0\)

\(\Leftrightarrow\)\(\sqrt{4\left(x-5\right)}+\frac{1}{3}\sqrt{9\left(x-5\right)}-\frac{1}{5}\sqrt{16\left(x-5\right)}=0\)

\(\Leftrightarrow\)\(2\sqrt{x-5}+\sqrt{x-5}-\frac{4}{5}\sqrt{x-5}=0\)

\(\Leftrightarrow\)\(\frac{11}{5}\sqrt{x-5}=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\) (t/m)

Vậy

b)  \(-5x+7\sqrt{x}=-12\)

\(\Leftrightarrow\)\(5x-7\sqrt{x}-12=0\)

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

đến đây tự làm

c) d) e) bạn bình phương lên

f)  \(VT=\sqrt{3\left(x^2+2x+1\right)+9}+\sqrt{5\left(x^4-2x^2+1\right)+25}\)

             \(=\sqrt{3\left(x+1\right)^2+9}+\sqrt{5\left(x^2-1\right)^2}\)

           \(\ge\sqrt{9}+\sqrt{25}=8\)

Dấu "=" xảy ra  \(\Leftrightarrow\)\(\hept{\begin{cases}x+1=0\\x^2-1=0\end{cases}}\)\(\Leftrightarrow\)\(x=-1\)

Vậy...

\(\sqrt{x+8}-\sqrt{5x+2}+2=0\)             

<=> \(\sqrt{x+8}=\sqrt{5x+2}-2\)

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

<=> x + 8 = \(\left(\sqrt{5x+2}\right)^2-4\sqrt{5x+2}+4\)

<=> x + 8 = 5x + 2 - \(4\sqrt{5x+2}+4\)

<=> \(4\sqrt{5x+2}=5x+2+4-x-8\)

<=> \(4\sqrt{5x+2}=4x-2\)

<=> \(4\sqrt{5x+2}=2\left(2x-1\right)\)

<=> \(\sqrt{5x+2}=\dfrac{2\left(2x-1\right)}{4}\)

<=> \(\sqrt{5x+2}=\dfrac{2x-1}{2}\)

<=> 5x + 2 = \(\dfrac{\left(2x-1\right)^2}{4}\)

<=> x = \(\dfrac{\dfrac{\left(2x-1\right)^2}{4}-2}{5}\)

<=> x = -0,278.....