Đáp án:

Giải thích các bước giải:

Bài `3:`

`a)(2x+1)(x-1)=0`

`↔` \(\left[ \begin{array}{l}2x+1=0\\x-1=0\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=-\dfrac{1}{2}\\x=1\end{array} \right.\) 

Vậy `S={-1/2;1}`

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

`↔` \(\left[ \begin{array}{l}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{array} \right.\) 

Vậy `S={-2/3;1/2}`

`c)(3x-1)(2x-3)(x+5)=0`

`↔` \(\left[ \begin{array}{l}3x-1=0\\2x-3=0\\x+5=0\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5\end{array} \right.\) 

Vậy `S={1/3;3/2;-5}`

`d)3x-15=2x(x-5)`

`↔3(x-5)=2x(x-5)`

`↔3(x-5)-2x(x-5)=0`

`↔(3-2x)(x-5)=0`

`↔` \(\left[ \begin{array}{l}3-2x=0\\x-5=0\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=\dfrac{3}{2}\\x=5\end{array} \right.\) 

Vậy `S={3/2;5}`

`e)x^2-x=0`

`↔x(x-1)=0`

`↔` \(\left[ \begin{array}{l}x=0\\x-1=0↔x=1\end{array} \right.\)

Vậy `S={0;1}`

`f)x^2-2x=0`

`↔x(x-2)=0`

`↔` \(\left[ \begin{array}{l}x=0\\x-2=0↔x=2\end{array} \right.\) 

Vậy `S={0;2}`

`g)x^2-3x=0`

`↔x(x-3)=0`

`↔` \(\left[ \begin{array}{l}x=0\\x-3=0↔x=3\end{array} \right.\) 

Vậy `S={0;3}`

`h)(x+1)(x+2)=(2-x)(x+2)`

`↔(x+1)(x+2)-(2-x)(x+2)=0`

`↔(x+2)(x+1-2+x)=0`

`↔(x+2)(2x-1)=0`

`↔` \(\left[ \begin{array}{l}x+2=0\\2x-1=0\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=-2\\x=\dfrac{1}{2}\end{array} \right.\)

Vậy `S={-2;1/2}`

`~rai~`

\(\text{Bài 3:}\\a)\left(2x+1\right)\left(x-1\right)=0\\\Leftrightarrow \left[\begin{array}{I}2x+1=0\\x-1=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-0,5\\x=1.\end{array}\right.\\\text{Vậy S={-0,5;1}.}\\b)\left(x+\dfrac{2}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\\Leftrightarrow \left[\begin{array}{I}x+\dfrac{2}{3}=0\\x-\dfrac{1}{2}=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}.\end{array}\right.\\\text{Vậy }S=\{-\dfrac{2}{3};\dfrac{1}{2}\}.\\c)\left(3x-1\right)\left(2x-3\right)\left(x+5\right)=0\\\Leftrightarrow \left[\begin{array}{I}3x-1=0\\2x-3=0\\x+5=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=\dfrac{1}{3}\\x=\dfrac{3}{2}\\x=-5.\end{array}\right.\\\text{Vậy }S=\{\dfrac{1}{3};\dfrac{3}{2};-5\}.\\d)3x-15=2x(x-5)\\\Leftrightarrow 3\left(x-5\right)=2x\left(x-5\right)\\\Leftrightarrow 3\left(x-5\right)-2x\left(x-5\right)=0\\\Leftrightarrow \left(3-2x\right)\left(x-5\right)=0\\\Leftrightarrow \left[\begin{array}{I}3-2x=0\\x-5=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=1,5\\x=5.\end{array}\right.\\\text{Vậy S={1,5;5}.}\\e)x^2-x=0\\\Leftrightarrow x\left(x-1\right)=0\\\Leftrightarrow \left[\begin{array}{I}x=0\\x-1=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=0\\x=1.\end{array}\right.\\\text{Vậy S={0;1}.}\\f)x^2-2x=0\\\Leftrightarrow x\left(x-2\right)=0\\\Leftrightarrow \left[\begin{array}{I}x=0\\x-2=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=0\\x=2.\end{array}\right.\\\text{Vậy S={0;2}.}\\g)x^2-3x=0\\\Leftrightarrow x\left(x-3\right)=0\\\Leftrightarrow \left[\begin{array}{I}x=0\\x-3=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=0\\x=3.\end{array}\right.\\\text{Vậy S={0;3}.}\\h)\left(x+1\right)\left(x+2\right)=\left(2-x\right)\left(x+2\right)\\\Leftrightarrow \left(x+1\right)\left(x+2\right)-\left(2-x\right)\left(x+2\right)=0\\\Leftrightarrow [\left(x+1\right)-\left(2-x\right)].\left(x+2\right)=0\\\Leftrightarrow \left(2x-1\right)\left(x+2\right)=0\\\Leftrightarrow \left[\begin{array}{I}2x-1=0\\x+2=0\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=0,5\\x=-2.\end{array}\right.\\\text{Vậy S={0,5;-2}.}\)