Question

1) x^2-5|x|=0
2) 2x^2+|x|-3x=0
3) x^2+|x+4|=4
4) x^2-|x-5|=5
Сделайте пожалуйста

Answer 1

$1) \;\; x^2-5\mid x \mid=0 \\ \\ x^2-5x=0=\ \textgreater \ x(x-5)=0=\ \textgreater \ x=0 \;\; x-5=0=\ \textgreater \ x=0 \;\; x=5 \\ x^2-5*(-x)=0=\ \textgreater \ x(x+5)=0=\ \textgreater \ x=0 \;\; x+5=0=\ \textgreater \ x=-5 \\ \\ \boxed{x=0 \;\; x=5 \;\; x=-5} \\ \\ \\ 2) \;\; 2x^2+\mid x \mid -3x=0 \\ \\ 2x^2+x-3x=0=\ \textgreater \ 2x^2-2x=0 =\ \textgreater \ 2x(x-1)=0=\ \textgreater \ x=0 \;x=1\\ 2x^2-x-3x=0=\ \textgreater \ 2x^2-4x=0=\ \textgreater \ 2x(x-2)=0-\ \textgreater \ x=0 \; x=2 \\ \\ No \;\; tak \;\; kak \;\; vo \;\; vtorom \;\; x\ \textless \ 0 \;\; znachit \;\; x\notin R \\ \\ \boxed {x=0 \;\; x=1} \\ \\ \\ 3)\;\; x^2+\mid x+4 \mid =4 \\ \\ x^2+x+4=4,x \geq -4 \\ x^2-(x+4)=4, x+4 \leq -4 \\ \\ x^2+x=0=\ \textgreater \ x(x+1)=0=\ \textgreater \ x=0 \;\; x=-1 \\ \\ vo \;\; vtorom \;\; ne \;\; budet \;\; kornei\;\; tak \;\; kak \;\; x \leq -4 \\ \\ \boxed {x=-1 \;\; x=0} \\ \\ \\ 4)\;\; x^2-\mid x-5 \mid =5 \\ \\ x^2-(x-5)=5 ,x \geq 5 \\ x^2-(-(x-5))=5,x\ \textless \ 5 \\ \\ x^2-x+5=5=\ \textgreater \ x^2-x=0=\ \textgreater \ x=0 \;\; x=1 \;\; NO \;\; x \geq 5 \\ \\ x^2+(x-5)=5=\ \textgreater \ x^2+x-5=5=\ \textgreater \ x^2+x-10=0 \\ \\ D=41; \sqrt{D}=\sqrt{41} \\ \\ \boxed {x= \frac{1\pm \sqrt{41} }{2}}$