Question

giải phương trình sau 

a)2(x+1)=4-x

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

Answer 1

\(a,2\left(x+1\right)=4-x\\ =>2x+2-4+x=0\\ =>3x-2=0\\ =>x=\dfrac{2}{3}\\ b,x^2-3x+2=0\\ =>x^2-2x-x+2=0\\ =>x\left(x-2\right)-\left(x-2\right)=0\\ =>\left(x-1\right)\left(x-2\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Answer 2

`2(x+1)=4-x`

`<=> 2x+2=4-x`

`<=> 2x+x=4-2`

`<=> 3x=2`

`<=>x=2/3`

`------`

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

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)