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

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

ĐKXĐ:

\(\left(2x+2-2\sqrt{5x-1}\right)+\left(\sqrt{5x^2+x+3}-\left(2x+1\right)\right)+x^2-3x+2=0\)

\(\Leftrightarrow\dfrac{2\left(x^2-3x+2\right)}{x+1+\sqrt{5x-1}}+\dfrac{x^2-3x+2}{\sqrt{5x^2+x+3}+2x+1}+x^2-3x+2=0\)

\(\Leftrightarrow\left(x^2-3x+2\right)\left(\dfrac{2}{x+1+\sqrt{5x-1}}+\dfrac{1}{\sqrt{5x^2+x+3}+2x+1}+1\right)=0\)

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

Đặt \(\sqrt{x^2+5x+10}=a\ge0\)

\(PT\Leftrightarrow a^2+2a-8=0\\ \Leftrightarrow a=2\left(a\ge0\right)\\ \Leftrightarrow x^2+5x+10=4\\ \Leftrightarrow x^2+5x+6=0\\ \Leftrightarrow\left[{}\begin{matrix}x_1=-2\\x_2=-3\end{matrix}\right.\Leftrightarrow x_1^2+x_2^2=4+9=13\)

Vậy ...

Đặt \(\sqrt{x^2+5x+10}=t>0\Rightarrow x^2+5x=t^2-10\)

Phương trình trở thành:

\(t^2-10+2+2t=0\)

\(\Leftrightarrow t^2+2t-8=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-4\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x^2+5x+10}=2\)

\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)

\(PT\Leftrightarrow-5x^2-24x+60=\left(x^2+5x-10\right)^2\\ \Leftrightarrow-5x^2-24x+60=x^4+10x^3+5x^2-100x+100\\ \Leftrightarrow x^4+10x^3+10x^2-76x+40=0\\ \Leftrightarrow x^4+4x^3-10x^2+6x^3+24x^2-60x-4x^2-16x+40=0\\ \Leftrightarrow\left(x^2+4x-10\right)\left(x^2+6x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+4x-10=0\\x^2+6x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt{14}\\x=-2-\sqrt{14}\\x=-3+\sqrt{13}\\x=-3-\sqrt{13}\end{matrix}\right.\)

x-1 + x-3 =1 <=> 2x -4=1 tu giai not

\(\Leftrightarrow\left(x^2+2\right)\sqrt{x^2+x+1}-2\left(x^2+2\right)+x^3-x^2-5x+6=0\)

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

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

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

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

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

Đặt \(\sqrt{x^2+x+1}=t>0\Rightarrow x^2=t^2-x-1\)

\(\Rightarrow t^2+x-3=\left(2-x\right)t\)

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

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

\(\Leftrightarrow\left(t+1\right)\left(t+x-3\right)=0\)

\(\Leftrightarrow t=3-x\)

\(\Leftrightarrow\sqrt{x^2+x+1}=3-x\) (\(x\le3\))

\(\Leftrightarrow x^2+x+1=x^2-6x+9\)

\(\Leftrightarrow x=\dfrac{8}{7}\)

ĐKXĐ: \(x\ge1\)

\(\sqrt{5x-1}=\sqrt{3x-2}+\sqrt{x-1}\)

\(\Leftrightarrow5x-1=3x-2+x-1+2\sqrt{\left(3x-2\right)\left(x-1\right)}\)

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

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

\(\Leftrightarrow11x^2-24x+4=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{11}\left(loại\right)\\x=2\end{matrix}\right.\)

a: \(\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\)

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

hay \(x=\sqrt{5}\)

b: \(\Leftrightarrow4x^4-9x^2+4x^2-9=0\)

\(\Leftrightarrow4x^2-9=0\)

=>x=3/2hoặc x=-3/2