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a, 2/3+1/3:x=-7

1/3:x=-7-2/3

1/3:x=-23/3

x=1/3:-23/3

x=-1/23

Vậy x=-1/23

c, (2x-3)(6-2x)=0

*TH1: 2x-3=0

2x=3

x=3/2

*TH2: 6-2x=0

2x=6

x=6/2

x=3

Vậy x=3/2 hoặc x=3

d,x:3/4+1/4=-2/3

x:3/4=-2/3-1/4

x:3/4=-11/12

x=-11/12*3/4

x=-11/16

\(2x^2=x\)

\(\Rightarrow2x^2-x=0\)

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

\(\Rightarrow\orbr{\begin{cases}x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

Vậy \(x=0\)hoặc \(x=\frac{1}{2}\)

\(x^3=x^5\)

\(\Rightarrow x^5-x^3=0\)

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

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

Vậy \(x=0\)hoặc \(x=1\)

\(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)

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

\(x.\left(x+1\right)\left(x+2\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)hoặc \(x=-2\)

Vậy \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\) hoặc \(x=-2\)

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

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

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

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

Vậy \(x=\frac{3}{2}\)hoặc \(x=-2\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

\(S\left\{0;\frac{1}{2}\right\}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{1}\end{cases}}\)

\(S=\left\{0;\pm\sqrt{1}\right\}\)

các câu sau tương tự nha bn

d) x^3-x^5=0

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

x=0 hoặc 1-x^2=0

x^2=1

x=1hoặc =-1

e)(x+1)(x+2)x=0

x+1=0 =>x=-1

x+2= 0=>x=-2

x=0

g)(2x-3)(x+2)

2x-3=0 =>x=3/2

x+2=0 =>x=-2

a: Ta có: \(x^2-2x=0\)

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

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

b: Ta có: \(\left(x-1\right)\cdot x-2\left(1-x\right)=0\)

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

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

c: Ta có: \(x^3+2x^2+x=0\)

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

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

d: Ta có: \(x^3-3x^2=0\)

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

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

a) Ta có: \(\left(5x-15\right)\left(4+6x\right)=0\)

\(\Leftrightarrow5\left(x-3\right)\cdot2\cdot\left(2+3x\right)=0\)

\(\Leftrightarrow10\left(x-3\right)\left(2+3x\right)=0\)

Vì 10\(\ne\)0 nên

\(\left[{}\begin{matrix}x-3=0\\2+3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-2}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{3;\frac{-2}{3}\right\}\)

b) Ta có: \(\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\5x=6\\\frac{1}{2}x=\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=\frac{3}{4}:\frac{1}{2}=\frac{3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{2};\frac{6}{5};\frac{3}{2}\right\}\)

c) Ta có: \(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\)

\(\Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=3\\x=\frac{25}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{12}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{3}{4};\frac{25}{12}\right\}\)

d) Ta có: \(\left(\frac{2}{3}x-\frac{1}{6}\right)\left[5\left(x-1\right)-\frac{3}{2}-\frac{\left(2-3\right)\left(x-1\right)}{3}\right]=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left[5x-5-\frac{3}{2}-\frac{-1\left(x-1\right)}{3}\right]=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(5x-5-\frac{3}{2}-\frac{1-x}{3}\right)=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(5x-\frac{13}{2}-\frac{1}{3}+\frac{x}{3}\right)=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(\frac{15x}{3}-\frac{41}{6}+\frac{x}{3}\right)=0\)

\(\Leftrightarrow\left(\frac{2}{3}x-\frac{1}{6}\right)\left(\frac{16x}{3}-\frac{41}{6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\frac{2}{3}x-\frac{1}{6}=0\\\frac{16x}{3}-\frac{41}{6}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{2}{3}x=\frac{1}{6}\\\frac{16}{3}\cdot x=\frac{41}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{6}:\frac{2}{3}\\x=\frac{41}{6}:\frac{16}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{41}{32}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{1}{4};\frac{41}{32}\right\}\)

\(a.\left(5x-15\right)\left(4+6x\right)=0\\ \left[{}\begin{matrix}5x-15=0\\4+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-2}{3}\end{matrix}\right.\)

\(b.\left(2x-1\right)\left(5x-6\right)\left(\frac{1}{2}x-\frac{3}{4}=0\right)\\ \left[{}\begin{matrix}2x-1=0\\5x-6=0\\\frac{1}{2}x-\frac{3}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{6}{5}\\x=-\frac{3}{2}\end{matrix}\right.\)

c.

\(\left(3-4x\right)\left(2x-\frac{3}{4}-x-\frac{4}{3}\right)=0\\ \Leftrightarrow\left(3-4x\right)\left(x-\frac{25}{12}\right)=0\\ \left[{}\begin{matrix}3-4x=0\\x-\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{25}{2}\end{matrix}\right.\)

2.a) \(8x^2-4x=0\Rightarrow4x\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)

b) \(5x\left(x-3\right)+7\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(5x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\5x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1.4\end{matrix}\right.\)

c) \(2x^2=x\Rightarrow2x^2-x=0\Rightarrow x\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0.5\end{matrix}\right.\)

d) \(x^3=x^5\Rightarrow x^3-x^5=0\Rightarrow x^3\left(1-x^2\right)=0\\ \Rightarrow x^3\left(1-x\right)\left(1+x\right)=0\Rightarrow\left[{}\begin{matrix}x^3=0\\1-x=0\\1+x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

e) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)

\(\Rightarrow\left(x+1\right)\left(x^2+2x\right)=0\Rightarrow\left(x+1\right)x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)

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

\(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\\ \Rightarrow\left(2x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1.5\\x=-2\end{matrix}\right.\)

1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42