`4)x²+4x=4`
`⇔x²+4x-4=0`
`⇔x²+4x+4-8=0`
`⇔x²+4x+4=8`
`⇔x²+2.x.2+2²=8`
`⇔(x+2)²=(\sqrt{8})^2`
`⇔`$\left[\begin{matrix} x+2=\sqrt{8}\\ x+2=-\sqrt{8}\end{matrix}\right.$
`⇔`$\left[\begin{matrix} x=\sqrt{8}-2\\ x=-\sqrt{8}-2\end{matrix}\right.$
Vậy `x\in{\sqrt{8}-2;-\sqrt{8}-2}`
`---------------`
Sửa đề:`x²+4x=4`
`→x²+4x=-4`
`x²+4x=-4`
`⇔x²+4x+4=0`
`⇔x²+2.x.2+2²=0`
`⇔(x+2)²=0`
`⇔x+2=0`
`⇔x=-2`
Vậy `x=-2`
`--------------`
`5)x³-3x²+3x-1=0`
`⇔x³-3.x².1+3.x.1²-1³=0`
`⇔(x-1)³=0`
`⇔x-1=0`
`⇔x=1`
Vậy `x=1`
`6)x²-2x+5+y²-4y=0`
`⇔x²-2x+1+4+y²-4y=0`
`⇔(x²-2x+1)+(y²-4y+4)=0`
`⇔(x²-2.x.1+1²)+(y²-2.y.2+2²)=0`
`⇔(x-1)²+(y-2)²=0`
Ta có:`(x-1)²≥0∀x`
`(y-2)²≥0∀y`
`⇒(x-1)²+(y-2)²≥0∀x,y`
`⇔`$\begin{cases} x-1=0\\y-2=0\end{cases}$
`⇔`$\begin{cases} x=1\\y=2\end{cases}$
Vậy `x=1` và `y=2`
