Question

Giải hộ mình vs

Answer 1

Bài 6.

Ta có: \(x+y=5\) ; \(xy=3\) 

\(\Leftrightarrow\left(x+y\right)^2=5^2\)

\(\Leftrightarrow x^2+2xy+y^2=25\)

\(\Leftrightarrow x^2+y^2=25-2\cdot3=19\) ( vì \(xy=3\))

Mặt khác: \(x+y=5\Leftrightarrow\left(x+y\right)^3=5^3\)

\(\Leftrightarrow x^3+3x^2y+3xy^2+y^3=125\)

\(\Leftrightarrow x^3+y^3+3xy\left(x+y\right)=125\)

\(\Leftrightarrow x^3+y^3=125-3\cdot3\cdot5=80\) (vì \(x+y=5;xy=3\))

Khi đó: \(x^2+y^2-2xy=19-2\cdot3\) 

\(\Leftrightarrow\left(x-y\right)^2=13\)

\(\Leftrightarrow x-y=\sqrt{13}\)

#\(Ayumu\)

Answer 2

7: x^2+2xy+y^2+4x+4y=21

=>(x+y)^2+4(x+y)-21=0

=>(x+y+7)(x+y-3)=0(2)

x,y dương

=>x+y>0

=>x+y+7>7>0(1)

Từ (1), (2) suy ra x+y-3=0

=>x+y=3