Question

[Minh Anh Doan]

giúp với em sắp phải nộp deadline r ạ:((

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)

\(\Leftrightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)

\(\Leftrightarrow-9x=-18\)

hay x=2

b: ta có: \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3+x-6x^2+5x-3=0\)

\(\Leftrightarrow17x=-5\)

hay \(x=-\dfrac{5}{17}\)

c: Ta có: \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)

\(\Leftrightarrow8x^3-12x^2+6x-1+12x^2-12-14x+13=0\)

\(\Leftrightarrow8x^3-8x=0\)

\(\Leftrightarrow8x\left(x-1\right)\left(x+1\right)=0\)

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

Answer 2

[Lấp La Lấp Lánh]

a) \(3\left(x-2\right)^2+\left(x-1\right)^3-x^3=-7\)

\(\Rightarrow3x^2-12x+12+x^3-3x^2+3x-1-x^3=-7\)

\(\Rightarrow-9x=-18\)

\(\Rightarrow x=2\)

b) \(\left(x+2\right)^3-x\left(x-1\right)\left(x+1\right)=6x^2-5x+3\)

\(\Rightarrow x^3+6x^2+12x+8-x^3+x=6x^2-5x+3\)

\(\Rightarrow18x=-5\)

\(\Rightarrow x=-\dfrac{5}{18}\)

c) \(\left(2x-1\right)^3+12\left(x-1\right)\left(x+1\right)=14x-13\)

\(\Rightarrow8x^3-12x^2+6x-1+12x^2-12=14x-13\)

\(\Rightarrow8x^3-8x=0\)

\(\Rightarrow8x\left(x-1\right)\left(x+1\right)=0\)

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