Question

(3x+2).(x-1)+(x+3).(x-7)+2x+23=0

tìm x

 

Answer 1

\(\left(3x+2\right)\left(x-1\right)+\left(x+3\right)\left(x-7\right)+2x+23=0\\ \Leftrightarrow3x^2+2x-3x-2+x^2+3x-7x-21+2x+23=0\\ \Leftrightarrow3x^2-x^2+2x-3x+3x-7x+2x-2-21+23=0\\ \Leftrightarrow x^2-3x=0\\ \Leftrightarrow x.\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)

<=> x=0 hoặc x=3

Answer 2

(3x+2)(x-1)+(x+3)(x-7)+2x+23=0

=>3x²+2x-3x-2+x²+3x-7x-21+2x=-23

=>(3x²+x²)+(2x-3x+3x-7x+2x) -(2+21)=-23

=>4x²-3x-23=-23

=>4x²-3x=-23+23=0

=>x(4x-3)=0

=>x=0 hoặc 4x-3=0

=>x=0 hoặc x=3/4.