Về học lại hằng đẳng thức nha .-.

\(\Leftrightarrow\dfrac{\left(x+2\right)+5}{2-x}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow-\left(x+2\right)+5\left(x+2\right)=2x-3\\ \Leftrightarrow6x+12-2x+3=0\\ \Leftrightarrow4x+15=0\\ \Leftrightarrow x=\dfrac{-15}{4}\)

\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\left(đk:x\ne\pm2\right)\)

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

\(\Leftrightarrow x-2-5x-10-2x-3=0\)

\(\Leftrightarrow-6x-15=0\)

\(\Leftrightarrow-6x=15\)

\(\Leftrightarrow x=-\dfrac{15}{6}\left(n\right)\)

Vậy \(S=\left\{-\dfrac{15}{6}\right\}\)

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

=>x=0

2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5=7\)

=>6x-12=7

=>6x=19

hay x=19/6

1: \(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)-3=-6\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=-3\)

\(\Leftrightarrow14x=-3\)

hay x=-3/14

2: \(\Leftrightarrow3x-2x-7-x+6x-5=x+2-x+5\)

=>4x-12=7

=>4x=19

hay x=19/4

\(5\left(x+2\right)-x^2-2x=0\)

\(\Rightarrow5\left(x+2\right)-\left(x^2+2x\right)=0\)

\(\Rightarrow5\left(x+2\right)-x\left(x+2\right)=0\)

\(\Rightarrow\left(5-x\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5-x=0\\x+2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)

khó phết

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

1) 2x – (3 – 5x) = 4( x +3)

<=>2x-3+5x=4x+12

<=>2x-3+5x-4x-12=0

<=>3x-15=0

<=>x=5

2) 5(2x-3) - 4(5x-7) =19 - 2(x+11)

<=>10x-15-20x+28=19-2x-22

<=>10x-15-20x+28-19+2x+22=0

<=>-8x+16=0

<=>x=2