Đáp án:
Giải thích các bước giải:
\(a,x^2+y^2=\left(x+y\right)^2-2xy=a^2-2b\)
\(b,x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-2xy-xy\right]\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]\)
\(=a.\left(a^2-3b\right)\)
\(=a^3-3ab\)
Theo phần a,
\(x^2+y^2=a^2-b\)
Ta có:
\(c) x^4+y^4=\left(x^2\right)^2+\left(y^2\right)^2=\left(x^2+y^2\right)^2-2x^2y^2=\left(a^2-b\right)^2-2b^2\)
d) $x^5+y^5=\left ( x^4+y^4 \right )\left ( x+y \right )-xy^4-x^4y=\left [ \left ( x+y^2 \right )-2xy \right ]\left ( x+y \right )-xy\left ( x^3+y^3 \right )=\left [ \left ( x+y^2 \right )-2xy \right ]\left ( x+y \right )-xy\left [ \left ( x+y \right )^3-3xy\left ( x+y \right ) \right ]$
