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\(A=1+3+3^2+...+3^{2016}\)
\(3A=3.\left(1+3+3^2+...+3^{2016}\right)\)
\(3A=3+3^2+3^3+...+3^{2017}\)
\(3A-A=\left(3+3^2+3^3+...+3^{2017}\right)-\left(1+3+3^2+...+3^{2016}\right)\)
\(2A=3^{2017}-1\)
\(A=\left(3^{2017}-1\right):2\)
\(B=1+6+6^2+...+6^{200}\)
\(6B=6.\left(1+6+6^2+...+6^{200}\right)\)
\(6B=6+6^2+6^3+...+6^{201}\)
\(6B-B=\left(6+6^2+6^3+...+3^{201}\right)-\left(1+6+6^2+...+6^{200}\right)\)
\(5B=6^{201}-1\)
\(B=\left(6^{201}-1\right):5\)
\(3^{x-2}.4=324\)
\(3^{x-2}=324:4\)
\(3^{x-2}=81\)
\(3^{x-2}=3^4\)
\(x-2=4\)
\(x=4+2\)
\(x=6\)
\(2x< 20\)
\(\Rightarrow x=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
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cau 1:8/45:8/9-2/5.x=8/45*9/8-2/5.x=8.9/45.8-2/5.x=1/5-2/5.x=2/-3. =>2/5.x=1/5-2/-3=3/15+2/3=3/15+10/15=13/15. =>x=13/15:2/5=13/15.5/2=13.5/15.2=13/6. Vậy x = 13/6. k nha co j ket ban
8/45 : 8/9 - 2/5 . X = 2/-3
8/45 . 9/8 - 2/5 . X = -2/3
1/5 - 2/5 . X = -2/3
-1/5 . X = -2/3
X = -2/3 : -1/5
X = -2/3 x 5/-1 = 10/3
=> x = 10/3
tk ủng hộ nhé !
nếu có sai nhờ anh/ chị / bạn sửa giùm em ạ !!!!!
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h: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
m: \(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{2021^2}< \dfrac{1}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)
Do đó: \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2021^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
=>\(B< 1-\dfrac{1}{2021}< 1\)