=>3x-2x^2-3+2x+2x^2+6x-x-3=14

=>10x-6=14

=>x=2

\(\left(x-1\right)\left(3-2x\right)+\left(2x-1\right)\left(x+3\right)=14\)

\(\Leftrightarrow3x-2x^2-3+2x+2x^2+6x-x-3=14\)

\(\Leftrightarrow\left(-2x^2+2x^2\right)+\left(3x+2x+6x-x\right)-\left(3+3\right)=14\)

\(\Leftrightarrow10x-6=14\)

\(\Leftrightarrow10x=14+6\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=20:10\)

\(\Leftrightarrow x=2\)

Vậy: \(x=2\)

`75 xx 1/x + 25 xx 1/x = 10`

`1/x xx (75+25) = 10`

`=> 1/x xx 100 = 10`

`=> 1/x = 10/100`

`=> 1/x = 1/10`

`=> x = 10`.

=>100:x=10

hay x=10

`x : 75 : x + 25 : x = 10`

`x : 100 = 10`

`x = 10xx100`

`x = 1000`

\(\dfrac{3}{x-2}=\dfrac{-2}{x-4}\left(dk:x\ne2;x\ne4\right)\)

\(\Rightarrow3\cdot\left(x-4\right)=-2\cdot\left(x-2\right)\)

\(\Rightarrow3x-12=-2x+4\)

\(\Rightarrow3x+2x=4+12\)

\(\Rightarrow5x=16\)

\(\Rightarrow x=\dfrac{16}{5}\left(tm\right)\)

\(ĐK:x\ne2;x\ne4\\ Có:\dfrac{3}{x-2}=\dfrac{-2}{x-4}\\ \Leftrightarrow3\left(x-4\right)=-2\left(x-2\right)\\ \Leftrightarrow3x-12=-2x+4\\ \Leftrightarrow3x+2x=4+12\\ \Leftrightarrow5x=16\\ \Leftrightarrow x=\dfrac{16}{5}\left(TM\right)\\ Vậy:x=\dfrac{16}{5}\)

\(x+\dfrac{1}{2}+\dfrac{3}{8}=\dfrac{17}{16}\\ x+\dfrac{7}{8}=\dfrac{17}{16}\\ x=\dfrac{17}{16}-\dfrac{7}{8}\\ x=\dfrac{3}{16}\)

\(\dfrac{1}{x}\times\left(75+25\right)=10\)

\(\dfrac{1}{x}=\dfrac{10}{75+15}=\dfrac{1}{10}\)

=> x = 10

`=>100:x=10`

hay `x=10`

 x : 75 : x + 25 : X = 10

 x :(75+25)            =10

 x  :  100                =10

 x                           =10x100

 x                           =1000

\(12-x\left(x-7\right)=31-3\left(x+12\right)\)

\(\Leftrightarrow12-x^2+7x=31-3x-36\)

\(\Leftrightarrow12-x^2+7x=-5-3x\)

\(\Leftrightarrow17-x^2+10x=0\)

phương trình bậc 2 lớp 8 mới học :

Ta có: \(\left|y+3\right|\ge0\forall y\)

\(\left|2x+y\right|\ge0\forall x,y\)

Do đó: \(\left|y+3\right|+\left|2x+y\right|\ge0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}y+3=0\\2x+y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-3\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\y=-3\end{matrix}\right.\)

Có 

\(6x+1⋮2x-1\)

\(3\left(2x-1\right)⋮2x-1\)

\(\Rightarrow\left(\left(6x+1\right)-3\left(2x-1\right)\right)⋮2x-1\)

\(\Rightarrow\left(6x+1-6x+3\right)⋮2x-1\)

\(\Rightarrow4⋮2x-1\)

\(\Rightarrow\left(2x-1\right)\inƯ_{\left(4\right)}\)

mà \(2x-1\)lẻ

\(\Rightarrow2x-1\in\pm1\)

Ta có bảng giá trị

Thử lại : Ta thấy đều thỏa mãn

\(2x+x+x+83=-2\\\Rightarrow3x+x+83=-2\\\Rightarrow4x+83=-2\\\Rightarrow4x=-2-83\\\Rightarrow4x=-85\\\Rightarrow x=\dfrac{-85}{4}\)

2x + x + x + 83 = -2

(2x + 2x) + 83 = -2

4x + 83 = -2

4x = (-2) - 83

4x = -85

x = -21,25