a: Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

b: Ta có: \(x^2-5x-6=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

Ta có : x(x - 3) - 2x + 6 = 0

<=> x(x - 3) - (2x - 6) = 0

=> x(x - 3) - 2(x - 3) = 0

=> (x - 2)(x - 3) = 0

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

1) x (x-2016) + 2015 (2016-x) = 0

 x (x-2016) - 2015 (x- 2016) = 0

(x-2015)(x-2016) =0

\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)

Vậy x= 2015; 2016

2) -5x (x-15) + (15-x) = 0

-5x (x-15) - (x-15) =0

(-5x -1) (x-15) =0

\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)

Vậy x= -1/5; 15

3) 3x (3x-7) - (7-3x) =0

3x(3x-7) + (3x -7) =0

(3x+1) (3x-7) =0

\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)

Vậy x= -1/3 ; 7/3

a) \(\Rightarrow x\left(x+3\right)=0\)

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

b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

vậy giỏi zữ vậy

a)\(x^2+5x=0\Rightarrow x\left(x+5\right)\Rightarrow\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b)\(3x\left(x-1\right)=1-x\Leftrightarrow3x^2-3x-1+x=0\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

5x^3-3x^2+5x-3=0

5x^3+5x-3x^2-3=0

5x(x^2+1)-3(x^2+1)=0

(x^2+1)(5x-3)=0

x^2+1=0 hoặc 5x-3=0

x(vô nghiệm) hoặc x=3/5

\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)

\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)

a) 3

b) 0

a) \(x^2-3x=0\)

\(\Rightarrow x\left(x-3\right)=0\) \(\Rightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=3\end{cases}}\)

Vậy x = 0 hoặc x = 3

b) \(x+5.x^2=0\)

\(\Rightarrow x\left(1+5x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\5x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}}\)

Vậy x = 0 hoặc x =\(-\frac{1}{5}\)

a,\(x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

vậy x=0;x=3

b,\(x+5x^2=0\Leftrightarrow x\left(1+5x\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\1+5x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\5x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=\frac{-1}{5}\end{cases}}}\)

vậy x=0;x\(=\frac{-1}{5}\)