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Ta có : \(32^{17}+16^{21}-2^{82}=\left(2^5\right)^{17}+\left(2^4\right)^{21}-2^{82}=2^{85}+2^{84}-2^{82}\)
\(=2^{80}.2^5+2^{80}.2^4-2^{80}.2^2=2^{80}.\left(2^5+2^4-2^2\right)=2^{80}.44⋮44\)
Vậy \(32^{17}+16^{21}-2^{82}⋮44\left(đpcm\right)\)
Vì \(\left(x+2y-3\right)^{2016}\ge0;\left|2x+3y-5\right|\ge0\forall x;y\)
\(\Rightarrow\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|\ge0\forall x;y\)
Mà \(\left(x+2y-3\right)^{2016}+\left|2x+3y-5\right|=0\) \(\Leftrightarrow\left(x+2y-3\right)^{2016}=0\) ; \(\left|2x+3y-5\right|=0\)
\(\Rightarrow x+2y-3=0;2x+3y-5=0\)
\(\Leftrightarrow x+2y=3;2x+3y=5\)
\(\Rightarrow x=3-2y\)
\(\Rightarrow2\left(3-2y\right)+3y=5\Leftrightarrow6-4y+3y=5\Leftrightarrow6-y=5\Rightarrow y=1\)
\(\Rightarrow x=3-2.1=1\)
Vậy \(x=1;y=1\)
\(\left(1+11+21+31+41+27+37+47\right)^{11}-\left(45+27+17\right)^6⋮2\)
mk ghi lại đề nha
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{9}\right)\left(1+\frac{1}{10}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{10}{9}\cdot\frac{11}{10}\)
\(=\frac{3.4.5.....10.11}{2.3.4....10}=\frac{11}{2}\)