Question

a) (x^2 +2x+1)-x-1 phần 3=6(x+1)^2-5x-5 phần 6 b)(x-1)-2(x-1) phần 3=1 + (2x-2) phần 2

Answer 1

a: Sửa đề: \(\dfrac{x^2+2x+1-x-1}{3}=\dfrac{6\left(x+1\right)^2-5x-5}{6}\)

=>\(\dfrac{x^2+x}{3}=\dfrac{6\left(x^2+2x+1\right)-5x-5}{6}\)

=>\(2\left(x^2+x\right)=6\left(x^2+2x+1\right)-5x-5\)

=>\(6x^2+12x+6-5x-5=2x^2+2x\)

=>\(6x^2+7x+1-2x^2-2x=0\)

=>\(4x^2+5x+1=0\)

=>(x+1)(4x+1)=0

=>\(\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{4}\end{matrix}\right.\)

b: \(\left(x-1\right)-\dfrac{2\left(x-1\right)}{3}=1+\dfrac{2x-2}{2}\)

=>\(\dfrac{1}{3}\left(x-1\right)=1+\left(x-1\right)\)

=>\(\dfrac{1}{3}\left(x-1\right)-\left(x-1\right)=1\)

=>\(\dfrac{-2}{3}\left(x-1\right)=1\)

=>\(x-1=-\dfrac{3}{2}\)

=>\(x=-\dfrac{3}{2}+1=-\dfrac{1}{2}\)

Answer 2

Giúp nhanh vs ak