\(x^3-3x^2=0\)

\(\Leftrightarrow x^2\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

khỉ nghĩ như này..

x³-3x²=0

(=)x² (x-3)=0

(=)x²=0,hoac x-3=0

(=)x=3

Giải:

a) \(x\left(x-2\right)-\left(x+3\right).x+7+9x=6\)

\(\Leftrightarrow x^2-2x-\left(x^2+3x\right)+7+9x=6\)

\(\Leftrightarrow x^2-2x-x^2-3x+7+9x=6\)

\(\Leftrightarrow4x=-1\)

\(\Leftrightarrow x=-\dfrac{1}{4}\)

Vậy ...

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-\left(10x+2-15x^2+6x\right)=4\)

\(\Leftrightarrow21x-35-15x^2+25x-10x-2+15x^2-6x=4\)

\(\Leftrightarrow30x-37=4\)

\(\Leftrightarrow30x=41\)

\(\Leftrightarrow x=\dfrac{41}{30}\)

Vậy ...

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14x\) (Sửa đề)

\(\Leftrightarrow x^3+8-x^3-3=14x\)

\(\Leftrightarrow5=14x\)

\(\Leftrightarrow x=\dfrac{5}{14}\)

Vậy ...

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

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

\(\Leftrightarrow1-3x=2\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

Vậy ...

a) \(x\left(x-2\right)-\left(x+3\right)x+7+9x=6\)

=> \(x^2-2x-x-3x+7+9x=6\)

=> \(x^2-2x-x^2-3x+7+9x=6\)

=> \(\left(x^2-x^2\right)+\left(-2x-3x+9x\right)=6-7\)

=> \(4x=-1\)

Vậy \(x=\dfrac{-1}{4}\)

b) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)

=>\(21x-15x^2-35+25x-10x+15x^2-4+6x=4\)

=> \(\left(21x+25x-10x+6x\right)\)\(+\left(-15x^2+15x^2\right)\)\(=4+35+4\)

=> \(42x=43\)

Vậy \(x=\dfrac{43}{42}\)

c) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3+3\right)=14\)

=> \(x^3-2x^2+4x+2x^2-4x+8-x^3-3\)\(=14x\)

=>\(\left(x^3-x^3\right)+\left(-2x^2+2x^x\right)+\left(4x-4x\right)+\left(8-3\right)\)\(=14x\)

=> \(5=14x\)

Vậy \(x=\dfrac{5}{14}\)

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

=> \(x^3+x^2+x+x^2-x+1-x^3-3x=2\)

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

=> \(-3x=1\)

Vậy \(x=\dfrac{-1}{3}\)

1000 mình làm violympic roi

1. (-2x - 1)(x² - x - 3) - (x + 2)(x + 1)²

= -2x³ + 2x² + 6x - x² + x + 3 - (x + 2)(x² + 2x + 1)

= -2x³ + x² + 7x + 3 - x³ - 2x² - x - 2x² - 2x - 2

= -3x³ - 3x² + 4x + 1

2. (x + 2)(x - 1) - (x - 3)(x + 2) = 3

=> (x + 2)(x - 1 - x + 3) = 3

=> (x + 2).0 = 3

...(xem lại đề)

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

\(\Leftrightarrow\left(x+2\right)\left(x-1-x+3\right)=3\)

\(\Leftrightarrow2\left(x+2\right)=3\)

\(\Leftrightarrow x+2=\frac{3}{2}\)

\(\Leftrightarrow x=\frac{3}{2}-2\)

\(\Leftrightarrow x=-\frac{1}{2}\)

\(5x\left(x-3\right)=x-3\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)

\(A=\left(5m^2-8m^2-9m^2\right)\left(-n^3+4n^3\right)=-12m^2.3n^3=-36m^2n^3\)

Để A\(\ge0\) thì \(m^2n^3\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\in Q\\n\le0\end{matrix}\right.\)

Để A thì 

Phương trình bậc nhất một ẩn duy nhất là câu a phương trình 2x+3=7.