a) Ta có:
\(a^2+b^2=\left(a+b\right)^2-2ab=23^2-2.132=265\)
b) Ta có:
\(x^3+3xy+y^3=x^3+3xy\left(x+y\right)+y^3=\left(x+y\right)^3=1\)
b,Ta có:
\(x+y=1\Rightarrow x=1-y\)(1)
Thay (1) vào biểu thức cần tìm ta có:
\(\left(1-y\right)^3+3\left(1-y\right)y+y^3\)
\(=1-3y+3y^2-y^3+3\left(y-y^2\right)+y^3\)
\(=1-3y+3y^2-y^3+3y-3y^2+y^3\)
\(=1\)
Vậy.....
Chúc bạn học tốt!!!
a) Ta có: \(a+b=23\Rightarrow\left(a+b\right)^2=23^2=529\)
\(\left(a+b\right)^2=a^2+2ab+b^2\Leftrightarrow a^2+b^2=\left(a+b\right)^2-2ab=529-2.132=265\)
Vậy \(a^2+b^2=265\)
b) Ta có: \(x+y=1\Leftrightarrow\left(x+y\right)^3=1^3=1\Rightarrow x^3+3x^2y+3xy^2+y^3=1\)
\(\Rightarrow x^2+3xy\left(x+y\right)+y^3=1\Rightarrow x^2+3xy.1+y^3=1\)
Vậy \(x^2+3xy+y^3=1\)
