Tìm x: |x+4|.|x|=0 x^2=4 1+2+3+4+…+100+x=5050 |2x+4|=16 (2^4-4^2).100!+2020=x 3x+24=|85%-0,85|

Đáp án:

+`x=0;x=-4`

+ `x=4;x=-4`

+`x=100`

+$\left \{ {{x=6} \atop {x=-10}} \right.$ 

+`x=2020`

+`x=-8`

Giải thích các bước giải:

*`|x+4|.|x|=0`
`=>|x+4|=0;|x|=0`
`**|x+4|=0`
`=>x+4=0`
`x=0-4`
`x=-4`
`**|x|=0`
`=>x=0`
Vậy `x=0;x=-4`

*`x^2=4`
`=>x^2=4^2` hoặc `x^2=(-4)^2`(mũ chẵn có 2 đáp số)

`=>x=4` hoặc `x=-4`
Vậy `x=4;x=-4`

`1+2+3+4+…+100+x=5050`
`{(x+1).[(x-1):1+1]}/2=5050`
`{(x+1).[x-1+1]}/2=5050`
`[(x+1).x]/2=5050`
`(x+1).x=5050.2`
`(x+1).x=10100`
`(x+1).x=101.100`
`=>`\(\left[ \begin{array}{l}x+1=101\\x=100\end{array} \right.\) `=>x=100`
Vậy `x=100`
*`|2x+4|=16`
`=>`\(\left[ \begin{array}{l}2x+4=16\\2x+4=-16\end{array} \right.\)`=>` \(\left[ \begin{array}{l}2x=16-4\\2x=(-16)-4\end{array} \right.\) `=>`\(\left[ \begin{array}{l}2x=12\\2x=-20\end{array} \right.\)`=>` \(\left[ \begin{array}{l}x=12:2\\x=(-20):2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=6\\x=-10\end{array} \right.\) 
Vậy \(\left[ \begin{array}{l}x=6\\x=-10\end{array} \right.\) 

*`(2^4-4^2).100!+2020=x`
`(16-16).100!+2020=x`
`0.100!+2020=x`
`0+2020=x`
`=>x=2020`
Vậy `x=2020`

*`3x+24=|85%-0,85|`
`3x+24=|85/100-0,85|`
`3x+24=|0,85-0,85|`
`3x+24=|0|`
`3x+24=0`
`3x=0-24`
`3x=(-24)`
`x=(-24):3`
`x=(-8)`
Vậy `x=-8`