Question

[doraemon]

Cho 2 số thực x, y thỏa mãn: x + y = 5; xy = - 2

Tính \(x^3+y^3\); \(x^5+y^5\)

Answer 1

[Lê Song Phương]

Ta có: \(\left(x+y\right)^2=x^2+2xy+y^2\) \(\Rightarrow5^2=x^2+y^2-4\)(vì \(\hept{\begin{cases}x+y=5\\xy=-2\end{cases}}\))   \(\Rightarrow x^2+y^2=29\)

Mặt khác \(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=5\left(29+4\right)=165\)(vì \(\hept{\begin{cases}x+y=5\left(đề\right)\\xy=-2\left(đề\right)\\x^2+y^2=29\left(cmt\right)\end{cases}}\))

\(\Rightarrow x^3+y^3=165\)(ý thứ nhất)

Ta có \(xy=-2\Rightarrow x^2y^2=4\); \(\hept{\begin{cases}x+y=5\\xy=-2\end{cases}}\Rightarrow xy\left(x+y\right)=5.\left(-2\right)\Rightarrow x^3y+xy^3=-10\Rightarrow-\left(x^3y+xy^3\right)=10\)

Lại có \(\left(x+y\right)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\)\(\Rightarrow5^4=x^4+y^4+4\left(x^3y+xy^3\right)+6.4\)( bởi lẽ \(\hept{\begin{cases}x+y=5\left(đề\right)\\x^2y^2=4\left(cmt\right)\end{cases}}\))

\(\Rightarrow625=x^4+y^4+4.\left(-10\right)+24\)(vì \(x^3y+xy^3=-10\left(cmt\right)\))\(\Rightarrow x^4+y^4=625-24+40=641\)

Mặt khác nữa, ta có \(x^5+y^5=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)

\(\Rightarrow x^5+y^5=5.\left[\left(x^4+y^4\right)-\left(x^3y+xy^3\right)+x^2y^2\right]\)(vì \(x+y=5\left(đề\right)\))

\(\Rightarrow x^5+y^5=5\left(641+10+4\right)=3275\)(vì \(\hept{\begin{cases}x^4+y^4=641\left(cmt\right)\\-\left(x^3y+xy^3\right)=10\left(cmt\right)\\x^2y^2=4\left(cmt\right)\end{cases}}\)

Vậy \(x^5+y^5=3275\)(ý thứ hai)