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: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

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

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

=>\(x^3-x^2-4x^2+4x+4x-4=0\)

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

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

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2: \(x^3+3x^2=x+6\)

=>\(x^3+3x^2-x-6=0\)

=>\(x^3+2x^2+x^2+2x-3x-6=0\)

=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

\(2x+3\sqrt{x}=1\)

=>\(2x+3\sqrt{x}-1=0\)

=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

=>\(x^4-3x^3+4x^2-3x+1=0\)

=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

\(\Leftrightarrow\left(x^3-4x^2+4x\right)-\left(x^2-4x+4\right)=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

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

b.

\(x^3+3x^2-x-6=0\)

\(\Leftrightarrow\left(x^3+x^2-3x\right)+\left(2x^2+2x-6\right)=0\)

\(\Leftrightarrow x\left(x^2+x-3\right)+2\left(x^2+x-3\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

c.

\(2x+3\sqrt{x}+1=0\)

ĐKXĐ: \(x\ge0\)

Do \(x\ge0\Rightarrow\left\{{}\begin{matrix}2x\ge0\\3\sqrt{x}\ge0\end{matrix}\right.\)

\(\Rightarrow2x+3\sqrt{x}+1>0\)

Pt đã cho vô nghiệm

d.

\(x^4+4x^2+1=3x^3+3x\)

\(\Leftrightarrow x^4-3x^3+4x^2-3x+1=0\)

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\) chia cả 2 vế của pt cho \(x^2\)

\(\Rightarrow x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

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

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

Đặt \(x+\dfrac{1}{x}=t\)

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

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=1\end{matrix}\right.\)

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

\(\Rightarrow x=1\)

\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)

giải pt 

c: \(\Leftrightarrow x-3=0\)

hay x=3

c: 

hay x=3

2x-x^2 =t

<=>t^2 -6t+7 =0

t=-1 loại

t=7 => \(\left[{}\begin{matrix}x_1=1-2\sqrt{2}\\x_2=1+2\sqrt{2}\end{matrix}\right.\)

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