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

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

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

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

(3x-4-x-1)(3x-4+x+1)=0
(2x-5)(4x-3)=0
2x-5 = 0 hoặc 4x-3=0
2x=5      hoặc 4x=3
x=5/2     hoặc   x=3/4

(3x - 4 - x - 1)(3x - 4 + x + 1) = 0

(2x - 5)(4x - 3) = 0

2x - 5 = 0           hoặc               4x - 3 = 0

x = 5/2               hoặc               x = 3/4

a)\(2x\left(x-2016\right)-2x+4032=0\)

\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)

\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)

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

\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)

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

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

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

\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)

thank you very much !

\(=\frac{3x^2+9x-3}{x^2+x-2}-\frac{x+1}{x+2}-\frac{x-2}{x-1}\)

\(=\frac{3x^2+9x-3}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x+1\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}-\frac{\left(x-2\right)\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{3x^2+9x-3-\left(x^2-1\right)-\left(x^2-4\right)}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{3x^2+9x-3-x^2+1-x^2+4}{\left(x-1\right)\left(x+2\right)}\)

\(=\frac{x^2+9x+2}{\left(x-1\right)\left(x+2\right)}\)

hi bn 

bn ghi sai đề

Vd1: 

d) Ta có: \(\sqrt{2}\left(x-1\right)-\sqrt{50}=0\)

\(\Leftrightarrow\sqrt{2}\left(x-1-5\right)=0\)

\(\Leftrightarrow x=6\)

1/ (2x+3)(x-4)+(x+5)(x-2)=(3x-5)(x-4)

<=> 2x² - 8x + 3x - 12 + x² - 2x + 5x  - 10 - 3x² + 12x + 5x - 20 = 0

<=> 15x - 20 = 0

<=> 15x = 20

<=> x = 4/3

1: \(\frac{2x+6}{3x^2-x}:\frac{x^2+3x}{1-3x}\)

\(=\frac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\frac{-3x+1}{x\left(x+3\right)}\)

\(=\frac{2}{x}\cdot\frac{-\left(3x-1\right)}{x\left(3x-1\right)}=\frac{-2}{x^2}\)

2: \(\frac{x}{x-2y}+\frac{x}{x+2y}+\frac{4xy}{4y^2-x^2}\)

\(=\frac{x}{x-2y}+\frac{x}{x+2y}-\frac{4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{x\left(x+2y\right)+x\left(x-2y\right)-4xy}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}\)

\(=\frac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\frac{2x}{x+2y}\)

3: \(\frac{1}{3x-2}-\frac{1}{3x+2}-\frac{3x-6}{4-9x^2}\)

\(=\frac{1}{3x-2}-\frac{1}{3x+2}+\frac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\frac{3x+2-\left(3x-2\right)+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\frac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)

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

4: \(\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{x^2-1}\)

\(=\frac{x+3}{x+1}+\frac{2x-1}{x-1}+\frac{x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(x+3\right)\left(x-1\right)+\left(2x-1\right)\left(x+1\right)+x+5}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2+2x-3+2x^2+2x-x-1+x+5}{\left(x-1\right)\left(x+1\right)}=\frac{3x^2+4x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{\left(3x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{3x+1}{x-1}\)

3, 2x(x^2-8x+16)-(x+5)(x^2-4)+2(x^2+10x+25)-x+1

=2x^3-16x^2+32x-(x^3-4x+5x^2-20)+2x^2+20x+50-x+1

=2x^3-16x^2+32x-x^3+4x-5x^2+20+2x^2+20x+50-x+1

=x^3-19x^2+55x+71