Đáp án + Giải thích các bước giải:
a)
`x^3 - 25%x = 0`
`to x^3 - 1/4x = 0`
`to x . ( x^2 - 1/4 ) = 0`
`to` \(\left[ \begin{array}{l}x=0\\x^2-\dfrac{1}{4}=0\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=0\\x^2=\dfrac{1}{4}\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.\)
Vậy `x in {0;1/2;-1/2}`
b)
`9/2 - | x - 3/4 | = 1/2`
`to | x - 3/4 | = 9/2 - 1/2`
`to | x - 3/4 | = 4`
`to` \(\left[ \begin{array}{l}x-\dfrac{3}{4}=4\\x-\dfrac{3}{4}=-4\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=4+\dfrac{3}{4}\\x=-4+\dfrac{3}{4}\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=\dfrac{19}{4}\\x=-\dfrac{13}{4}\end{array} \right.\)
Vậy `x in {19/4;-13/4}`
