a)

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

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

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

b)

\(\Rightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(3x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\3x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=\frac{2}{3}\end{array}\right.\)

c)

\(\Rightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)

\(\Rightarrow\left(3x-2\right)^2.2=0\)

\(\Rightarrow3x-2=0\)

\(\Rightarrow x=\frac{2}{3}\)

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

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

vt rõ đề câu a đi lm 1 thể

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow9x^2-3x-9x^2+1=0\)

\(\Leftrightarrow3x=1\)

hay \(x=\dfrac{1}{3}\)

b: Ta có: \(x^2-5x+25-5x=0\)

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

\(\Leftrightarrow x-5=0\)

hay x=5

a)\(3x\left(x-2\right)+2\left(2-x\right)=0\)

\(\Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\)

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

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

b)\(5x\left(3x-1\right)+x\left(3x-1\right)-2\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(5x+x-2\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(6x-2\right)=0\)

\(\Leftrightarrow2\left(3x-1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)^2=0\Rightarrow3x-1=0\Rightarrow x=\frac{1}{3}\)

a/3x(x-2)+2(2-x)=0

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

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

b/5x(3x-1)+x(3x-1)-2(3x-1)=0

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

=>(6x-2)(3x-1)=0

=>\(\orbr{\begin{cases}6x-2=0\\3x-1=0\end{cases}}\)=>\(\orbr{\begin{cases}6x=2\\3x=1\end{cases}}\)=>x=\(\frac{1}{3}\)

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)<=> 3x-5-x=0

   <=>    2x-5=0

   <=>        x=5/2

c) x.(1+2+3+4+...+100)=0

    x.5050=0

     x=0:5050=0

Vậy x=0

d) x.(1+2+3+4+5+...+100)=5050

    x.5050=5050

    x=1

Vậy x=1

e) x+1+x+2+x+3+x+4+...+x+100=5050

    (x+x+x+x+...+x)+(1+2+3+4+...+100)=5050

     100 số hạng x

    x.100+5050=5050

    x.100=0

    x=0

Vậy x=0

a , x = \(\frac{5}{2}\)