Question

Cho x+y = -1  ; x.y = -2 . Tính   B= x³ +y³; C = x³ - y³; D= x⁴ + y⁴; E= x⁴-y⁴

Answer 1

Ta có: \(x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2.\left(-2\right)=5\)

\(\Rightarrow B=x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=\left(-1\right)\left(5+2\right)=-7\)

\(\left(x-y\right)^2=x^2+y^2-2xy=5-2.\left(-2\right)=9\)

\(\Rightarrow x-y=-3;3\)

\(\Rightarrow C=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=3.\left(5-2\right)=9\) 

hoặc \(C=\left(x-y\right)\left(x^2+xy+y^2\right)=-3.\left(5-2\right)=-9\)

\(D=x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=25-8=17\)

\(E=x^4-y^4=\left(x^2\right)^2-\left(y^2\right)^2=\left(x^2-y^2\right)\left(x^2+y^2\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)=\left(-3\right)\left(-1\right).5=15\)

hoặc \(E=3.\left(-1\right).5=-15\)

 

Answer 2

1)B= x³ +y³

=x³+3x²y+3xy²+y³-3x²y-3xy²

=(x+y)³-3xy.(x+y)

=(-1)³-3.(-2).(-1)

=-1-6

=-7

2) ta có:

x²+y²=x²+2xy+y²-2xy

=(x+y)²-2xy

=(-1)²-2.(-2)

=1+4

=5

=>(x-y)²=x²-2xy+y²

=5-2.(-2)

=5+4

=9

=>x-y=3 hoặc x-y=-3

Với x-y=3 thì:

x³-y³=(x-y)(x²+xy+y²)

=3.(5-2)

=3.3=9

Với x-y=-3 thì :

x³-y³=(x-y)(x²+xy+y²)

=-3.(5-2)

=-3.2=-9