lx

lỗi r bn

lx

a: =>(x^2+x)^2-2(x^2+x)+(x^2+x)-2=0

=>(x^2+x-2)(x^2+x+1)=0

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

b: ĐKXĐ: x<>4; x<>1

PT =>\(\dfrac{x+3+3x-12}{x-4}=\dfrac{6}{1-x}\)

=>(4x-9)(1-x)=6(x-4)

=>4x-4x^2-9+9x=6x-24

=>-4x^2+13x-9-6x+24=0

=>-4x^2+7x+15=0

=>x=3(nhận) hoặc x=-5/4(nhận)

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

⇔\(\left(x^5+x^4\right)+\left(x^3+x^2\right)+\left(x+1\right)=0\)

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

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

⇔ \(\left[{}\begin{matrix}x+1=0\\x^4+x^2+1=0\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-1\\x\in\varnothing\end{matrix}\right.\)

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

ĐẶT \(\sqrt{x^2+1}=y\left(y>0\right)\)

\(\Rightarrow x^4=\left(y^2-1\right)^2\)

Từ Đó Ta Có pt mới : \(\left(y^2-1\right)^2+y^3-1=0\left(y>0\right)\)

\(\Rightarrow y^4+y^3-2y^2=0\)

\(\Rightarrow y^2\left(y^2+y-2\right)=0\)

\(\Rightarrow y^2\left(y-1\right)\left(y+2\right)=0\)

\(\Rightarrow y=1\left(y>0\Rightarrow y\notin\left(-2;0\right)\right)\)

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

                                                           VẬY PT trên có nghiệm duy nhất X = 0

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

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

đề có sai k đấy 

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

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