Ta có: \(\left(2x-5\right)^{2000}\ge0\forall x\)
\(\left(3y+4\right)^{2002}\ge0\forall y\)
\(\Rightarrow\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\ge0\forall x,y\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+5\right)^{2000}=0\\\left(3y+4\right)^{2002}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\3y+4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=5\\3y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
a. \(7^6+7^5-7^4\)
\(=7^4.7^2+7^4.7-7^4\)
\(=7^4.\left(7^2+7-1\right)\)
\(=7^4.55\)
Mà \(55⋮11\)
\(\Rightarrow7^4.55⋮11\Rightarrow7^6+7^5-7^4⋮11\left(dpcm\right)\)
b. \(1+2+2^2+2^3+...+2^{59}\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{58}+2^{59}\right)\)
\(=3+2^2.\left(1+2\right)+...+2^{58}.\left(1+2\right)\)
\(=3+2^2.3+...+2^{58}.3\)
\(=3.\left(1+2^2+2^4+2^6+...+2^{58}\right)\)
Mà \(3.\left(1+2^2+2^4+2^6+...+2^{58}\right)⋮3\)
\(\Rightarrow1+2+2^2+...+2^{59}⋮3\)
3.
a. \(3^{34}>3^{30}=3^{3.10}=\left(3^3\right)^{10}=27^{10}\)
\(5^{20}=5^{2.10}=\left(5^2\right)^{10}=25^{10}\)
Mà \(27^{10}>25^{10}\)
\(\Rightarrow3^{34}>5^{20}\)
b. \(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Mà \(8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)