Đáp án + Giải thích các bước giải:
`a)`
`4x^{2}-1=0`
`<=>(2x-1)(2x+1)=0`
`<=>` \(\left[ \begin{array}{l}2x-1=0\\2x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `S={+-(1)/(2)}`
`b)`
`(2x-1)^{2}-(3x+2)^{2}=0`
`<=>(2x-1+3x+2)(2x-1-3x-2)=0`
`<=>(5x+1)(-x-3)=0`
`<=>` \(\left[ \begin{array}{l}5x+1=0\\-x-3=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac{1}{5}\\x=-3\end{array} \right.\)
Vậy `S={-(1)/(5);-3}`
`c)`
`15x(2x-1)+20(1-2x)=0`
`<=>15x(2x-1)-20(2x-1)=0`
`<=>(2x-1)(15x-20)=0`
`<=>` \(\left[ \begin{array}{l}2x-1=0\\15x-20=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{array} \right.\)
Vậy `S={(1)/(2);(4)/(3)}`
`d)`
`(x+2)=(x+2)^{2}`
`<=>(x+2)^{2}-(x+2)=0`
`<=>(x+2)(x+2-1)=0`
`<=>(x+2)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x+2=0\\x+1=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-2\\x=-1\end{array} \right.\)
Vậy `S={-2;-1}`
`e)`
`5-9x^{2}=0`
`<=>9x^{2}=5`
`<=>x^{2}=(5)/(9)`
`<=>x=+-(\sqrt{5})/(3)`
Vậy `S={+-(\sqrt{5})/(3)}`
`f)`
`(4-3x)^{2}=(2x-1)^{2}`
`<=>(4-3x)^{2}-(2x-1)^{2}=0`
`<=>(4-3x+2x-1)(4-3x-2x+1)=0`
`<=>(-x+3)(-5x+5)=0`
`<=>` \(\left[ \begin{array}{l}-x+3=0\\-5x+5=0\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=1\end{array} \right.\)
Vậy `S={3;1}`
