Đáp án+Giải thích các bước giải:
Bài `7:`
`1) x^2-4=0`
`<=> x^2-2^2=0`
`<=> (x-2).(x+2)=0`
`<=>`\(\left[ \begin{array}{l}x-2=0\\x+2=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=2\\x=-2\end{array} \right.\)
Vậy `x∈{±2}`
`2) 16x^2-1=0`
`<=> (4x)^2-1^2=0`
`<=> (4x-1).(4x+1)=0`
`<=>`\(\left[ \begin{array}{l}4x-1=0\\4x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}4x=1\\4x=-1\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=\dfrac{1}{4}\\x=\dfrac{-1}{4}\end{array} \right.\)
Vậy `x∈{±1/4}`
`3) x^3-x=0`
`<=> x.(x^2-1)=0`
`<=>x.(x-1).(x+1)=0`
`<=>`\(\left[ \begin{array}{l}x=0\\x-1=0\\x+1=0\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x=0\\x=1\\x=-1\end{array} \right.\)
Vậy `x∈{0; ±1}`
