Giải thích các bước giải:
Bài 1:
a) Ta có:
`x+y=2=>(x+y)^2=4`
`=>x^2+2xy+y^2=4`
`=>2xy+20=4`
`=>2xy=-16`
`=>xy=-8`
Ta có: `x^3+y^3=(x+y)(x^2-xy+y^2)`
`=2.[20-(-8)]`
`=2.28`
`=56`
Vậy `x^3+y^3=56.`
b) `A = x^2 + 2xy + y^2 - 6x - 6y - 5 `
`= (x^2 + 2xy + y^2) - (6x + 6y) - 5 `
`= (x+y)^2 - 6(x + y) - 5 `
`= (-9)^2 - 6.(-9) - 5 `
`=135-5`
`=130`
Vậy `A=130.`
Bài 2:
a) `x^3 + y^3 = 152`
`=>(x+y)(x^2-xy+y^2)=152`
`=>10(x+y)=152`
`=>x+y=15,2`
Mà `x-y=2`
`=>x=43/5;y=33/5`
Vậy `x=43/5;y=33/5`
b) `x^2 - 6x + y^2 + 10y + 34 = -(z - 4)^2`
`=>(x^2 - 6x +9)+( y^2 + 10y + 25) = -(z - 4)^2`
`=>(x-3)^2+(y+5)^2 = -(z - 4)^2`
`=>(x-3)^2+(y+5)^2 +(z - 4)^2=0`
Mà `(x-3)^2+(y+5)^2 +(z - 4)^2>=0AAx;y;z`
Dấu "=" xảy ra `<=>`$\left\{\begin{matrix}x-3=0\\y+5=0\\z-4=0\end{matrix}\right.$`=>`$\left\{\begin{matrix}x=3\\y=-5\\z=4\end{matrix}\right.$
Vậy `x=3;y=-5;z=4.`