Đáp án + Giải thích các bước giải:
`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` $\\$ `<=> (x-5)(3-2x)=0 <=> `\(\left[ \begin{array}{l}x=5\\x=\dfrac{3}{2}\end{array} \right.\)
Vậy `S={5;3/2}`
`e)` `x^2-x=0` $\\$ `<=>x(x-1)=0`$\\$ `<=> ` \(\left[ \begin{array}{l}x=0\\x=1\end{array} \right.\)
Vậy `S={0;1}`
`f)` `x^2 - 2x = 0` $\\$ `<=>x(x-2)=0` $\\$ `<=> `\(\left[ \begin{array}{l}x=0\\x=2\end{array} \right.\)
Vậy `S={0;2}`
`g)` `x^2 - 3x = 0` $\\$ `<=>x(x-3)=0` $\\$ `<=> `\(\left[ \begin{array}{l}x=0\\x=3\end{array} \right.\) 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)(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}`
