Đáp án `+` Giải thích các bước giải `!`
`a)`
`4x(x-1)-x+1 = 0`
`<=> 4x(x-1)-(x-1) = 0`
`<=> (4x-1)(x-1) = 0`
`⇔` \(\left[ \begin{array}{l}4x-1=0\\x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}4x=1\\x=1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{1}{4}\\x=1\end{array} \right.\)
Vậy `S= {1/4; 1}`
`b)`
`x^3-11x = 0`
`<=> x(x^2-11) = 0`
`⇔` \(\left[ \begin{array}{l}x=0\\x^2-11=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x^2=11\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\x=\sqrt{11}\\x=-\sqrt{11}\end{array} \right.\)
Vậy `S= {0; \sqrt{11}; -\sqrt{11}}`
`c)`
`5-36x^2 = 0`
`<=> (\sqrt{5}-6x)(\sqrt{5}+6x) = 0`
`<=>` \(\left[ \begin{array}{l}6x=\sqrt{5}\\6x=-\sqrt{5}\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac{\sqrt{5}}{6}\\x=-\dfrac{\sqrt{5}}{6}\end{array} \right.\)
Vậy `S= {\frac{\sqrt{5}}{6}; -\frac{\sqrt{5}}{6}}`
`d)`
`x^2+x+1/4 = 0`
`<=> x^2+2. x. 1/2+(1/2)^2 = 0`
`<=> (x+1/2)^2 = 0`
`<=> x+1/2 = 0`
`<=> x = -(1)/2`
Vậy `S= {-(1)/2}`
`e)`
`4x^2-4 = 0`
`<=> 4x^2 = 4`
`<=> x^2 = 1`
`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)
Vậy `S= {1; -1}`
