\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)

\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)

\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)

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

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

\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)

\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)

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

\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)

Chắc tới đây bạn làm đc rồi nhỉ

À câu a mình tự làm được rồi nhé! Các bạn chỉ cần làm câu b cho mình là được.

b, \(\frac{2\sqrt{x}}{\sqrt{x+1}}+\sqrt{x}=\sqrt{x+9}\)

ĐK \(x\ge0\)

Pt 

<=> \(2\sqrt{x}+\sqrt{x\left(x+1\right)}=\sqrt{\left(x+1\right)\left(x+9\right)}\)

<=> \(4x+x^2+x+4\sqrt{x^2\left(x+1\right)}=x^2+10x+9\)

 <=> \(4x\sqrt{x+1}=5x+9\)

<=> \(16x^2\left(x+1\right)=25x^2+90x+81\)với mọi \(x\ge0\)

<=> \(16x^3-9x^2-90x-81=0\)

<=> \(x=3\)(tm ĐK)

Vậy x=3

a, ĐKXĐ : \(D=R\)

BPT \(\Leftrightarrow x^2+5x+4< 5\sqrt{x^2+5x+4+24}\)

Đặt \(x^2+5x+4=a\left(a\ge-\dfrac{9}{4}\right)\)

BPTTT : \(5\sqrt{a+24}>a\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a+24\ge0\\a< 0\end{matrix}\right.\\\left\{{}\begin{matrix}a\ge0\\25\left(a+24\right)>a^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\\left\{{}\begin{matrix}a^2-25a-600< 0\\a\ge0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-24\le a< 0\\0\le a< 40\end{matrix}\right.\)

\(\Leftrightarrow-24\le a< 40\)

- Thay lại a vào ta được : \(\left\{{}\begin{matrix}x^2+5x-36< 0\\x^2+5x+28\ge0\end{matrix}\right.\)

\(\Leftrightarrow-9< x< 4\)

Vậy ....

b, ĐKXĐ : \(x>0\)

BĐT \(\Leftrightarrow2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< x+\dfrac{1}{4x}+1\)

- Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a\left(a\ge\sqrt{2}\right)\)

\(\Leftrightarrow a^2=x+\dfrac{1}{4x}+1\)

BPTTT : \(2a\le a^2\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le0\\a\ge2\end{matrix}\right.\)

\(\Leftrightarrow a\ge2\)

\(\Leftrightarrow a^2\ge4\)

- Thay a vào lại BPT ta được : \(x+\dfrac{1}{4x}-3\ge0\)

\(\Leftrightarrow4x^2-12x+1\ge0\)

\(\Leftrightarrow x=(0;\dfrac{3-2\sqrt{2}}{2}]\cup[\dfrac{3+2\sqrt{2}}{2};+\infty)\)

Vậy ...

\(a,PT\Leftrightarrow x^2-3x+2+x^2-x\sqrt{3x-2}=0\left(x\ge\dfrac{2}{3}\right)\\ \Leftrightarrow\left(x^2-3x+2\right)+\dfrac{x\left(x^2-3x+2\right)}{x+\sqrt{3x-2}}=0\\ \Leftrightarrow\left(x^2-3x+2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)\left(1+\dfrac{x}{x+\sqrt{3x-2}}\right)=0\)

Vì \(x\ge\dfrac{2}{3}>0\Leftrightarrow1+\dfrac{x}{x+\sqrt{3x-2}}>0\)

Do đó \(x\in\left\{1;2\right\}\)

\(b,ĐK:0\le x\le4\\ PT\Leftrightarrow x+2\sqrt{x}+1=6\sqrt{x}-3-\sqrt{4-x}\\ \Leftrightarrow x-4\sqrt{x}+4=-\sqrt{4-x}\\ \Leftrightarrow\left(\sqrt{x}-2\right)^2=-\sqrt{4-x}\)

Vì \(VT\ge0\ge VP\Leftrightarrow VT=VP=0\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{4-x}=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Vậy PT có nghiệm \(x=4\)

a.

ĐKXĐ: \(x\ne-1\)

\(x^2+5x+2=\left(2x+2\right)\sqrt{x^2+x+2}\)

\(\Leftrightarrow\left(x^2+x+2\right)-2\left(x+1\right)\sqrt{x^2+x+2}+4x=0\)

Đặt \(\sqrt{x^2+x+2}=t>0\)

\(\Rightarrow t^2-2\left(x+1\right)t+4x=0\)

\(\Leftrightarrow t\left(t-2x\right)-2\left(t-2x\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=2x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+2}=2\\\sqrt{x^2+x+2}=2x\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\\x^2+x+2=4x^2\left(x\ge0\right)\end{matrix}\right.\)

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

b.

ĐKXĐ: \(x\ge-1\)

\(x^2-5x+14-4\sqrt{x+1}=0\)

\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\)