Ta có:
\(\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+2xy+y^2=4\)
\(\Leftrightarrow xy=\dfrac{1}{2}\)
Ta có: \(\left(x^2+y^2\right)^2=9\)
\(\Leftrightarrow x^4+2x^2y^2+y^4=9\)
\(\Rightarrow x^4+y^4=\dfrac{17}{2}\)
a/ Ta có: \(\left(x+y\right)\left(x^4+y^4\right)=17\)
\(\Leftrightarrow x^5+y^5+xy\left(x+y\right)\left(x^2-xy+y^2\right)=17\)
\(\Leftrightarrow x^5+y^5+\dfrac{1}{2}.4.\left(3-\dfrac{1}{2}\right)=17\)
\(\Leftrightarrow x^5+y^5=12\)
b/ Ta có: \(\left(x^2+y^2\right)^3=27\)
\(\Leftrightarrow x^6+3x^2y^2\left(x^2+y^2\right)+y^6=27\)
\(\Leftrightarrow x^6+y^6=27-3.\dfrac{1}{4}.3=\dfrac{99}{4}\)
PS: Thật ra cái này không tìm được x,y thực đâu.
Vì \(\left(x+y\right)^2=x^2+2xy+y^2\)
\(\Rightarrow2xy=\left(x+y\right)^2-\left(x^2+y^2\right)=2-3=-1\Rightarrow xy=\dfrac{-1}{2}\)\(\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3\Rightarrow x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=2^3-3\left(\dfrac{-1}{2}\right).2=5\)a,\(\left(x^2+y^2\right)\left(x^3+y^3\right)\) =3.5=15
\(\Leftrightarrow x^5+x^2y^3+x^3y^2+x^5=15\Rightarrow x^5+y^5=15-xy\left(x+y\right)=15-\left(\dfrac{-1}{2}\right).2=16\)
