Do 

Vậy phương trình vô nghiệm

\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=10x-5x^2-11x-22\)

\(\Leftrightarrow-x^2+2x-1=-5x^2-x-22\)

\(\Leftrightarrow4x^2+3x+21=0\)

Ta có \(\Delta=3^2-4.4.21< 0\)

Vậy pt vô nghiệm

a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)

mn giúp vs ạ , e đang cần gấp . Ai trả lời đúng và nhanh đầu tiên e hứa sẽ k ạ 

e sẽ k ạ

\(\left(x-1\right)^3-x\left(x-1\right)^2=5x\left(2x-x\right)-11\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-2\)

\(\Leftrightarrow2x-1=-x-22\)

\(\Leftrightarrow3x=-21\)

\(\Leftrightarrow x=7\)

\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=15-8=7\)

\(\Leftrightarrow x=\frac{-7}{2}\)

Vậy \(x=\frac{-7}{2}\)

mn ơi giúp mk vs

mk sửa lại đề ạ :v

\(\hept{\begin{cases}\left|x+1\right|-2\left|y-1\right|=3\\2\left|x+1\right|-\left|1-y\right|=11\end{cases}}\)

\(\hept{\begin{cases}\left|x+1\right|-2\left|y-1\right|=3\\2\left|x+1\right|-\left|y-1\right|=11\end{cases}\Leftrightarrow}\hept{\begin{cases}2\left|x+1\right|-4\left|y-1\right|=6\\2\left|x+1\right|-\left|y-1\right|=11\end{cases}}\)

\(\Leftrightarrow3\left|y-1\right|=5\Leftrightarrow\left|y-1\right|=\frac{5}{3}\Leftrightarrow\orbr{\begin{cases}y=\frac{8}{3}\\y=-\frac{2}{3}\end{cases}}\)

Từ đó tìm được x

\(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x^3-8x\left(x+2\right)=6\\ \Leftrightarrow\left(x^2+3x+2\right).\left(x+3\right)-x^3-8x^2-16x=6\\ \Leftrightarrow x^3+6x^2+11x+6-x^3-8x^2-16x-6=0\\ \Leftrightarrow-2x^2-5x=0\\ \Leftrightarrow x.\left(-2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\-2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)