Question

[Hannnaa]

Cho x-y=2 và xy=+1. Tính (x^2+y^2); (x^3-y^3); (x^2-y^2)^2; x^6-y^6

Answer 1

[Nguyễn Ngọc Anh Minh]

1/

\(x^2+y^2=\left(x-y\right)^2+2xy=2^2+2.1=6\)

2/

\(x^3-y^3=\left(x-y\right)\left(x^2+y^2+xy\right)=2\left(6+1\right)=14\)

3/

\(x^2-y^2=\left(x-y\right)\left(x+y\right)=2\left(x+y\right)\) (3)

Ta có

\(x^2+y^2=\left(x+y\right)^2-2xy=\left(x+y\right)^2-2=6\)

\(\Rightarrow\left(x+y\right)^2=8\Rightarrow\left(x+y\right)=\pm2\sqrt{2}\) Thay vào (3)

\(\Rightarrow x^2-y^2=2.\pm2\sqrt{2}=\pm4\sqrt{2}\)

4/

\(x^6-y^6=\left(x^3-y^3\right)\left(x^3+y^3\right)\) (4)

Ta có

\(x^3-y^3=14\) (cmt)

Ta có

\(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right).5=\pm2\sqrt{2}.5=\pm10\sqrt{2}\)

\(\Rightarrow x^6-y^6=\pm10\sqrt{2}.14=\pm140\sqrt{2}\)

Answer 2

[tranvanoi]

9(a-b)^2 - 4(x-y)^2