giải pt $\frac{x^2-4-|x-2|}{2}$ = x( x + 1) 28/08/2021 Bởi Bella giải pt $\frac{x^2-4-|x-2|}{2}$ = x( x + 1)
`\text{~~Holi~~}` #Moon `(x^2-4-|x-2|)/2=x(x+1)` `->(x^2-4-|x-2|)/2=x^2+x` `->x^2-4-|x-2|=2x^2+2x` `->x^2-|x-2|-2x^2-2x=4` `->-x^2-|x-2|-2x=4` `->`\(\left[ \begin{array}{l}-x^2-(x-2)-2x=4.(Đk:x-2\ge0)\\-x^2-[-(x-2)]-2x=4.(Đk:x-2<0)\end{array} \right.\) `->`\(\left[ \begin{array}{l}\begin{cases}x=-2\\x=-1\end{cases}\\x\not{\in}\mathbb{R}\end{array} \right.\) `->`\(\left[ \begin{array}{l}x\in∅\\x\not{\in}\mathbb{R}\end{array} \right.\) `->`$x\not{\in}\mathbb{R}$ Vậy $S\not{\in}\mathbb{R}$ Bình luận
`\text{~~Holi~~}`
#Moon
`(x^2-4-|x-2|)/2=x(x+1)`
`->(x^2-4-|x-2|)/2=x^2+x`
`->x^2-4-|x-2|=2x^2+2x`
`->x^2-|x-2|-2x^2-2x=4`
`->-x^2-|x-2|-2x=4`
`->`\(\left[ \begin{array}{l}-x^2-(x-2)-2x=4.(Đk:x-2\ge0)\\-x^2-[-(x-2)]-2x=4.(Đk:x-2<0)\end{array} \right.\)
`->`\(\left[ \begin{array}{l}\begin{cases}x=-2\\x=-1\end{cases}\\x\not{\in}\mathbb{R}\end{array} \right.\)
`->`\(\left[ \begin{array}{l}x\in∅\\x\not{\in}\mathbb{R}\end{array} \right.\)
`->`$x\not{\in}\mathbb{R}$
Vậy $S\not{\in}\mathbb{R}$