\(A=2^2+2^4+...+2^{20}\)

\(=2^2\left(1+2^2+...+2^{18}\right)=4\left(1+2^2+...+2^{18}\right)⋮4\)

\(A=2^2+2^4+...+2^{18}+2^{20}\)

\(=2^2\left(1+2^2\right)+...+2^{18}\left(1+2^2\right)\)

\(=5\left(2^2+2^6+...+2^{18}\right)⋮5\)

\(A=1+3+3^2+...+3^{11}\)

\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)

\(=40+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)

\(=40\left(1+3^4+3^8\right)⋮40\)

Để ý thấy rằng \(1+3+3^2+3^3=40\)

\(A=1+3+3^2+3^3+...+3^{11}\)

\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)

\(=40+3^4\times40+3^8\times40\)

\(=40\left(1+3^4+3^8\right)\)

Do đó A chia hết cho 40

Áp dụng BĐT trị tuyệt đối, ta có:

 \(\left|x-9\right|+\left|2-x\right|\ge\left|x-9+2-x\right|=\left|7\right|=7\)

Dấu "=" xảy ra khi: \(\left(x-9\right)\left(2-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-9\ge0\\2-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-9\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge9\\x\le2\end{matrix}\right.\left(\text{vô lí}\right)\\\left\{{}\begin{matrix}x\le9\\x\ge2\end{matrix}\right.\end{matrix}\right.\\ \Rightarrow2\le x\le9\)

\(\left|x-9\right|+\left|2-x\right|=7\)

Ta có : \(\left|x-9\right|+\left|2-x\right|\ge\left|x-9+2-x\right|=7\)

Nên \(x=0\) là nghiệm phương trình đã cho.

Gọi 2 số cần tìm là \(\overline{5ab};\overline{3cd}\) theo đề bài

\(\overline{5ab}+\overline{3cd}=836\)

\(500+\overline{ab}+300+\overline{cd}=836\)

\(\Rightarrow\overline{ab}+\overline{cd}=36\)

Đến đây là dạng toán tìm 2 số khi biết tổng và tỷ ở lớp 5

Chia 2 trường hợp \(\overline{ab}=2x\overline{cd}\) hoặc \(\overline{cd}=2x\overline{ab}\)

Bạn tự làm nốt nhé

\(M=\dfrac{202}{7\cdot1010}+\dfrac{202}{10\cdot1313}+\dfrac{202}{13\cdot1616}+...+\dfrac{202}{91\cdot9494}\\ =\dfrac{202}{7\cdot10\cdot101}+\dfrac{202}{10\cdot13\cdot101}+\dfrac{202}{13\cdot16\cdot101}+...+\dfrac{202}{91\cdot94\cdot101}\\ =\dfrac{202}{101}\cdot\left(\dfrac{1}{7\cdot10}+\dfrac{1}{10\cdot13}+\dfrac{1}{13\cdot16}+...+\dfrac{1}{91\cdot94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{7\cdot10}+\dfrac{3}{10\cdot13}+...+\dfrac{3}{91\cdot94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{94}\right)\\ =\dfrac{2}{3}\cdot\dfrac{87}{658}\\ =\dfrac{29}{329}\)

Ta có:

\(\left|x-\dfrac{1}{3}\right|+\left|x-\dfrac{1}{15}\right|+...+\left|x-\dfrac{1}{399}\right|\ge0\forall x\)

\(\Rightarrow-11x\ge0\forall x\Rightarrow x\le0\)

\(\Rightarrow x-\dfrac{1}{3};x-\dfrac{1}{15};...;x-\dfrac{1}{399}< 0\)

\(\Rightarrow x-\dfrac{1}{3}+x-\dfrac{1}{15}+...+x-\dfrac{1}{399}=11x\)

\(\Rightarrow x+x+...+x-\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right)=11x\)

Vì số lượng \(x\) ở vế trái bằng số lượng số hạng là phân số

\(\Rightarrow\) Số lượng \(x\) ở vế trái là:\(\left(19-1\right):2+1=10\left(số\right)\)

\(\Rightarrow10x-\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)=11x\)

\(\Rightarrow-x=1-\dfrac{1}{21}\)

\(\Rightarrow x=-\dfrac{20}{21}\)

\(a.\dfrac{-3}{7}\cdot\dfrac{15}{13}-\dfrac{3}{7}\cdot\dfrac{11}{13}-\dfrac{3}{7}\\ =-\dfrac{3}{7}\cdot\left(\dfrac{15}{13}+\dfrac{11}{13}+1\right)\\ =-\dfrac{3}{7}\cdot\left(\dfrac{26}{13}+1\right)\\ =\dfrac{-3}{7}\cdot3\\ =\dfrac{-9}{7}\\ b.\dfrac{-1}{9}\cdot\dfrac{-3}{5}+\dfrac{5}{-6}\cdot\dfrac{-3}{5}-\dfrac{7}{2}\cdot\dfrac{3}{5}\\ =-\dfrac{3}{5}\cdot\left(\dfrac{-1}{9}+\dfrac{-5}{6}+\dfrac{7}{2}\right)\\ =-\dfrac{3}{5}\cdot\dfrac{23}{9}\\ =-\dfrac{23}{15}\)

\(Ư\left(240\right)=\left\{\text{1;2,3,4,5,6,8,10,24,30,40,48,60,80,120,240}\right\}\)

240 = 2⁴.3.5 

Ư(240) ={1;2;3;4;5;6;8;10;12;15;16;20;24;30;40;48;60;80;120;240}