Question

a) Tìm tất cả các cặp số (x;y) thoả mãn :

\(\left(5x-y\right)^{2018}+|x^2-4|^{2019}\le0\)

b) Tìm x, y \(\in Z^+\) thoả mãn: x.y + x - y = 0

Answer 1

Chỉ biết làm câu b) thôi, thông cảm nha :D

\(xy+x-y=0\\ x\left(y+1\right)-y-1=0-1\\x\left(y+1\right)-\left(y+1\right)=-1\\ \left(y+1\right)\left(x-1\right)=-1=1\cdot\left(-1\right)=\left(-1\right)\cdot1 \)

Ta xét các TH:

\(\circledast\left\{{}\begin{matrix}y+1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=1-1\\x=-1+1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\) (ktm)

\(\circledast\left\{{}\begin{matrix}y+1=-1\\x-1=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-1-1\\x=1+1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}y=-2\\x=2\end{matrix}\right.\) (tm)

Vậy \(\left(x;y\right)=\left(2;-2\right)\)

Answer 2

a) Ta có: (5x - y)²⁰¹⁸ \(\ge\) 0

|x² - 4|²⁰¹⁹ \(\ge\)0

=> (5x - y)²⁰¹⁸ + |x² - 4|²⁰¹⁹ \(\ge\) 0

Mà: (5x - y)²⁰¹⁸ + |x² - 4|²⁰¹⁹ \(\le\)0

=> (5x - y)²⁰¹⁸ + |x² - 4|²⁰¹⁹ = 0

=> \(\left\{{}\begin{matrix}\left(5x-y\right)^2=0\\\left|x^2-4\right|^{2019}=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}5x-y=0\\\left|\left(x-2\right)\left(x+2\right)\right|=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=5x\\\left(x+2\right)\left(x-2\right)=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=5x\\\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2\\y=10\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2\\y=-10\end{matrix}\right.\end{matrix}\right.\)

Vậy ...