Ta có : 

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

\(\Leftrightarrow\)\(4x^2-4x-3x^2+15=x-3-x-4\)

\(\Leftrightarrow\)\(x^2-4x+15=-7\)

\(\Leftrightarrow\)\(\left(x^2-2.x.2+2^2\right)+11=-7\)

\(\Leftrightarrow\)\(\left(x-2\right)^2=-18\)

Mà \(\left(x-2\right)^2\ge0\) \(\left(\forall x\inℝ\right)\)

\(\Rightarrow\)\(x\in\left\{\varnothing\right\}\)

Vậy không có giá trị nào của x thoã mãn đề bài 

Chúc bạn học tốt ~ 

\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

`@` `\text {Ans}`

`\downarrow`

`2+(x+3)=7`

`\Rightarrow x+3=7-2`

`\Rightarrow x+3=5`

`\Rightarrow x=5-3`

`\Rightarrow x=2`

`5+(3+x)=10`

`\Rightarrow 3+x=10-5`

`\Rightarrow 3+x=5`

`\Rightarrow x=5-3`

`\Rightarrow x=2`

`(4+x)+1=7`

`\Rightarrow 4+x=7-1`

`\Rightarrow 4+x=6`

`\Rightarrow x=6-4`

`\Rightarrow x=2`

`(x+5)+3=9`

`\Rightarrow x+5=9-3`

`\Rightarrow x+5=6`

`\Rightarrow x=6-5`

`\Rightarrow x=1`

`(x-1)-4=7`

`\Rightarrow x-1=7+4`

`\Rightarrow x-1=11`

`\Rightarrow x=11+1`

`\Rightarrow x=12`

`4-(6-x)=1`

`\Rightarrow 6-x=4-1`

`\Rightarrow 6-x=3`

`\Rightarrow x=6-3`

`\Rightarrow x=3`

\(2+\left(x+3\right)=7\)

\(\Rightarrow2+x+3=7\)

\(\Rightarrow x+5=7\)

\(\Rightarrow x=2\)

\(5+\left(3+x\right)=10\)

\(\Rightarrow5+3+x=10\)

\(\Rightarrow x+8=10\)

\(\Rightarrow x=2\)

\(\left(4+x\right)+1=7\)

\(\Rightarrow4+x+1=7\)

\(\Rightarrow x+5=7\)

\(\Rightarrow x=2\)

\(\left(x+5\right)+3=9\)

\(=x+5+3=9\)

\(\Rightarrow x+8=9\)

\(\Rightarrow x=1\)

\(\left(x-1\right)-4=7\)

\(\Rightarrow x-1-4=7\)

\(\Rightarrow x-5=7\)

\(\Rightarrow x=12\)

\(4-\left(6-x\right)=1\)

\(\Rightarrow4-6-x=1\)

\(\Rightarrow-2-x=1\)

\(\Rightarrow x=-3\)

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

\(\Rightarrow x+2-2=0\)

\(\Rightarrow x=0\)

\(\left(x+3\right)+1=7\)

\(\Rightarrow x+3+1=7\)

\(\Rightarrow x+4=7\)

\(\Rightarrow x=3\)

\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)

\(\Rightarrow3x=12\)

\(\Rightarrow x=4\)

\(\left(5x+4\right)-1=13\)

\(\Rightarrow5x+4-1=13\)

\(\Rightarrow5x+3=13\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

\(\left(4x-8\right)-3=5\)

\(\Rightarrow4x-8-3=5\)

\(\Rightarrow4x-11=5\)

\(\Rightarrow4x=16\)

\(\Rightarrow x=4\)

\(8-\left(2x+4\right)=2\)

\(\Rightarrow8-2x-4=2\)

\(\Rightarrow4-2x=2\)

\(\Rightarrow2x=2\)

\(\Rightarrow x=1\)

\(7+\left(5x+2\right)=14\)

\(\Rightarrow7+5x+2=14\)

\(\Rightarrow9+5x=14\)

\(\Rightarrow5x=5\)

\(\Rightarrow x=1\)

\(5-\left(3x-11\right)=1\)

\(\Rightarrow5-3x+11=1\)

\(\Rightarrow16-3x=1\)

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

a: Ta có: \(\dfrac{x+2}{5}=\dfrac{1}{x-2}\)

\(\Leftrightarrow x^2-4=5\)

hay \(x\in\left\{3;-3\right\}\)

d: Ta có: \(\dfrac{x-3}{x+1}=\dfrac{x+2}{x-5}\)

\(\Leftrightarrow x^2-8x+15=x^2+3x+2\)

\(\Leftrightarrow-11x=-13\)

hay \(x=\dfrac{13}{11}\)

a) ĐKXĐ: \(x\ne2\)

\(\Rightarrow\left(x+2\right)\left(x-2\right)=5.1\)

\(\Rightarrow x^2-4=5\Rightarrow x^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)

b) ĐKXĐ: \(x\ne-1\)

\(\Rightarrow\left(x+1\right)^2=2.8=16\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

c) giống câu a

d) ĐKXĐ: \(x\ne5,x\ne-1\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)=\left(x-3\right)\left(x-5\right)\)

\(\Rightarrow x^2+3x+2=x^2-8x+15\)

\(\Rightarrow11x=13\)

\(\Rightarrow x=\dfrac{13}{11}\left(tm\right)\)

chiu

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

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

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