Question

Tìm x 

a, (x+7)\(^2\)-x(x-3)=12

b, x\(^2\)-3x+2=0

Answer 1

\(a,\Leftrightarrow x^2+14x+49-x^2+3x=12\\ \Leftrightarrow17x=-37\Leftrightarrow x=-\dfrac{37}{17}\\ b,\Leftrightarrow x^2-x-2x+2=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Answer 2

a) \(x^2+2x7+49-x^2+3x=12\Leftrightarrow17x=-37\Leftrightarrow x=\dfrac{-37}{17}\)

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

Answer 3

`a,`

`(x+7)^2-x(x-3)=12`

`<=>x^2+14x+49 - x^2 +3x=12`

`<=> 17x =-37`

`<=>x=(-37)/17`

Vậy `x=(-37)/17`

`b,`

`x^2-3x+2=0`

`<=>x^2-2x-x+2=0`

`<=>x(x-2)-(x-2)=0`

`<=>(x-2)(x-1)=0`

giải ra `x=2,x=1`

Vậy `x=2,x=1`