Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

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

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

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

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

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

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

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

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

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

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

\(x^2-2x+5=0\)

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

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

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

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

\(a,\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow3x\left(x-1\right)+\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\\ c,\Leftrightarrow\left(x+2\right)\left(2x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{2}\end{matrix}\right.\)

a) \(5x\left(x-1\right)=x-1\)

\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

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

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

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

Vậy \(x=1\) hoặc \(x=\frac{1}{5}\)

b) \(2\left(x+5\right)-x^2-5x\)

\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)\)

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

\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy \(x=-5\)hoặc \(x=2\)

b) \(\Leftrightarrow5x^2-6x+1=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow x=\frac{1}{5}\) hoặc x = 1

c) \(\Leftrightarrow x^2+4x-21-x^2-4x+5=0\Leftrightarrow-16=0\) (vô lí) => PT vô nghiệm

d) \(\Leftrightarrow x^2+3x-10=0\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\Leftrightarrow\)x = 2 hoặc x = -5

e) \(\Leftrightarrow x\left(x-2\right)=0\)<=> x = 0 hoặc x = 2

sửa lại:

a) \(5x\left(x-1\right)=x-1\)

=> \(5x\left(x-1\right)-\left(x-1\right)=0\)

=> \(\left(x-1\right)\left(5x-1\right)=0\)

=> x - 1 = 0   hoặc   5x - 1 = 0

=> x = 1       hoặc    5x = 1  =>  x = 1/5

Vậy x = 1 hoặc x = 1/5

b) \(2\left(5+x\right)-x^2-5x=0\)

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

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

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

=> x + 5 = 0  hoặc   2 - x = 0

=> x = -5   hoặc   x = 2

a, 5x(x-1)=x-1

<=>5x(x-1)-(x-1)=0

<=>(5x-1)(x-1)=0

<=>5x-1=0 hoặc x-1=0

<=>x=1/5 hoặc x=1

b,2(5+x)-x²-5x=0

<=>2(x+5)-x(x+5)=0

<=>(2-x)(x+5)=0

<=>2-x=0 hoặc x+5=0

<=>x=2 hoặc x=-5

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

\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow8x+16-5x^2-10x+4x^2+4x-8x-8+2x^2-8=0\)

\(\Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

                                    Vậy S = { 0, 6}

- Help Me :((

\(25\left(x+3\right)^2+\left(1-5x\right)\left(1+5x\right)=0\)

\(25\left(x^2+6x+9\right)+1-25x^2=0\)

\(25x^2+150x+225+1-25x^2=0\)

\(150x=-226\)

\(x=-\frac{113}{75}\)