Question

Tìm x :

1, \(\left(x+1\right)^3-\left(x-1\right)^3=6\cdot\left(x+2\right)^2-9\)

2, \(\left(2x-1\right)\cdot\left(4x^2+2x+1\right)+\left(1-2x\right)^3=3\cdot\left(2x+3\right)^2\)

Answer 1

1) \(\left(x+1\right)^3-\left(x-1\right)^3=6.\left(x+2\right)^2-9\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)=6\left(x^2+4x+4\right)-9\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+24x+24-9\)

\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2-24x-24+9=0\)

\(\Leftrightarrow-24x-13=0\Leftrightarrow-24x=13\Leftrightarrow x=\dfrac{-13}{24}\) vậy \(x=\dfrac{-13}{24}\)

Answer 2

2) \(\left(2x-1\right).\left(4x^2+2x+1\right)+\left(1-2x\right)^3=3.\left(2x+3\right)^2\)

\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3=3\left(.4x^2+12x+9\right)\)

\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3=12x^2+36x+27\)

\(\Leftrightarrow8x^3+4x^2+2x-4x^2-2x-1+1-6x+12x^2-8x^3-12x^2-36x-27=0\)

\(\Leftrightarrow-42x-27=0\Leftrightarrow-42x=27\Leftrightarrow x=\dfrac{-27}{42}\) vậy \(x=\dfrac{-27}{42}\)

Answer 3

a, \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x+2\right)^2-9\)

\(\Rightarrow x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)=6\left(x^2+4x+4\right)-9\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+24x+24-9\)

\(\Rightarrow6x^2+2-6x^2-24x=24-9\)

\(\Rightarrow-24x=15-2\Rightarrow-24x=13\Rightarrow x=-\dfrac{13}{24}\)

Câu b tương tự

Chúc bạn học tốt!!!