c) (2x-5)2000+(3y+4)2002 \(\le0\)

Có (2x-5)2000\(\ge\)0 với mọi x

(3y+4)2002\(\ge\)0 với mọi y

=> (2x-5)2000+(3y+4)2002 \(\ge\) 0

=> Để (2x-5)2000+(3y+4)2002 \(\le0\) thì (2x-5)2000+(3y+4)2002 =0

=> \(\left\{{}\begin{matrix}\left(2x-5\right)^{2000}=0\\\left(3y+4\right)^{2002}=0\end{matrix}\right.=>\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.=>\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{-4}{3}\end{matrix}\right.\)

Vậy x=\(\dfrac{5}{2}\);y=\(\dfrac{-4}{3}\)

Bài 2:

Có A=2100-299+298-...+22-2

=> 2A=2(2100-299+298-...+22-2)

=> 2A= 2101-2100+299-...+23-22

=> 2A= 2101-2100+299-...+23-22

+A= 2100-299+298-...+22-2

=> 3A= 2101-2

=> A=\(\dfrac{2^{101}-2}{3}\)