\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

a) \(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\)

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

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

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

Vậy tập nghiệm của phương trình là \(S=\left\{-3;2\right\}\)

c) \(3x\left(x-5\right)-x^2+25=0\)

\(\Leftrightarrow3x\left(x-5\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow3x\left(x-5\right)-\left(x-5\right)\left(x+5\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\2x=5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{2}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{5;\frac{5}{2}\right\}\) 

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)

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

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

Vậy \(x\in\left\{0;\frac{5}{3}\right\}\)

a) \(x\left(3x-5\right)=0\)

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

b) \(3x^2-27=0\)

\(\Leftrightarrow3x^2=27\)

\(\Leftrightarrow x^2=9\)

\(\Leftrightarrow x=\pm3\)

c) \(\left(x-5\right)^2=x-5\)

\(\Leftrightarrow x^2-10x+25-x+5=0\)

\(\Leftrightarrow x^2-11x+30=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}}\)

d) \(2\left(x+7\right)-x^2-7x=0\)

\(\Leftrightarrow2x+14-x^2-7x=0\)

\(\Leftrightarrow-x^2-5x+14=0\)

\(\Leftrightarrow\left(x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=2\end{cases}}}\)

e)\(7x\left(x-3\right)+2.3x=0\)

\(\Leftrightarrow7x^2-21x+6x=0\)

\(\Leftrightarrow7x^2-15x=0\)

\(\Leftrightarrow x\left(7x-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\7x-15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{15}{7}\end{cases}}}\)

#H

Rảnh rỗi thật sự .-.

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

\(2x^2-7x+3=4x^2+4x-3\)

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

\(2x\left(x+6\right)-\left(x+6\right)=0\)

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

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

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

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

a) \(2\chi-3=3\left(\chi+1\right)\)

\(\Leftrightarrow2\chi-3=3\chi+3\)

\(\Leftrightarrow2\chi-3\chi=3+3\)

\(\Leftrightarrow\chi=-6\)

Vậy phương trình có tập nghiệm S= \(\left\{-6\right\}\)

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

\(\Leftrightarrow3\chi-3=2\chi+2\)

\(\Leftrightarrow3\chi-2\chi=2+3\)

\(\Leftrightarrow\chi=5\)

Vậy phương trình có tập nghiệm S= \(\left\{5\right\}\)

b) \(\left(3\chi+2\right)\left(4\chi-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi+2=0\\4\chi-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-2\\4\chi=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-2}{3}\\\chi=\dfrac{5}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-2}{3};\dfrac{5}{4}\right\}\)

\(\left(3\chi+5\right)\left(4\chi-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi+5=0\\4\chi-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3\chi=-5\\4\chi=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\chi=\dfrac{-5}{3}\\\chi=\dfrac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S= \(\left\{\dfrac{-5}{3};\dfrac{1}{2}\right\}\)

c) \(\left|\chi-7\right|=2\chi+3\)

Trường hợp 1: 

Nếu \(\chi-7\ge0\Leftrightarrow\chi\ge7\)

Khi đó:\(\left|\chi-7\right|=2\chi+3\)

 \(\Leftrightarrow\chi-7=2\chi+3\)

\(\Leftrightarrow\chi-2\chi=3+7\)

\(\Leftrightarrow\chi=-10\) (KTMĐK)

Trường hợp 2:

Nếu \(\chi-7\le0\Leftrightarrow\chi\le7\)

Khi đó: \(\left|\chi-7\right|=2\chi+3\)

\(\Leftrightarrow-\chi+7=2\chi+3\)

\(\Leftrightarrow-\chi-2\chi=3-7\)

\(\Leftrightarrow-3\chi=-4\)

\(\Leftrightarrow\chi=\dfrac{4}{3}\)(TMĐK)

Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{4}{3}\right\}\)

\(\left|\chi-4\right|=5-3\chi\)

Trường hợp 1:  

Nếu \(\chi-4\ge0\Leftrightarrow\chi\ge4\)

Khi đó: \(\left|\chi-4\right|=5-3\chi\)

\(\Leftrightarrow\chi-4=5-3\chi\)

\(\Leftrightarrow\chi+3\chi=5+4\)

\(\Leftrightarrow4\chi=9\)

\(\Leftrightarrow\chi=\dfrac{9}{4}\)(KTMĐK)

Trường hợp 2: Nếu \(\chi-4\le0\Leftrightarrow\chi\le4\)

Khi đó: \(\left|\chi-4\right|=5-3\chi\)

\(\Leftrightarrow-\chi+4=5-3\chi\)

\(\Leftrightarrow-\chi+3\chi=5-4\)

\(\Leftrightarrow2\chi=1\)

\(\Leftrightarrow\chi=\dfrac{1}{2}\)(TMĐK)

Vậy phương trình có tập nghiệm S=\(\left\{\dfrac{1}{2}\right\}\)

a,\(\left(x-1\right)^2-\left(2x\right)^2=0< =>\left(x-1-2x\right)\left(x-1+2x\right)=0\)

\(< =>\left(-x-1\right)\left(3x-1\right)=0< =>\orbr{\begin{cases}x=-1\\x=\frac{1}{3}\end{cases}}\)

b,\(\left(3x-5\right)^2-x\left(3x-5\right)=0< =>\left(3x-5\right)\left(3x-5-x\right)=0\)

\(< =>\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{5}{2}\end{cases}}\)

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

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow x=-1;x=\frac{1}{3}\)

b, \(\left(3x-5\right)^2-x\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x-5-x\right)=0\Leftrightarrow\left(3x-5\right)\left(2x-5\right)=0\Leftrightarrow x=\frac{5}{3};x=\frac{5}{2}\)

\(2x\left(x^2-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(9+x\right)\left(3x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)

\(\left(2x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

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