Question

Cho x>y>0 và x-y=7, xy=60 

a) \(x^2-y^2\)

b) \(x^4+y^4\)

Answer 1

a) \(\left(x-y\right)^2=x^2-2xy+y^2=x^2+y^2-2xy\)

\(\Rightarrow x^2+y^2=\left(x-y\right)^2+2xy=7^2+2.60\)

\(\Rightarrow x^2+y^2=169\)

\(\left(x+y\right)^2=x^2+y^2+2xy=169+2.60\)

\(\Rightarrow\left(x+y\right)^2=289=17^2\)

\(\Rightarrow x+y=17\)

\(x^2-y^2=\left(x+y\right)\left(x-y\right)=17.7=119\)

b) \(\left(x^2+y^2\right)^2=\left(x^2\right)^2+\left(y^2\right)^2+2x^2y^2=x^4+y^4+2\left(xy\right)^2\)

\(\Rightarrow x^4+y^4=\left(x^2+y^2\right)^2-2\left(xy\right)^2=169^2-2.60^2\)

\(\Rightarrow x^4+y^4=28561-7200=21361\)