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Ta có \(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2021}\right)\left(1-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}.\dfrac{2021}{2022}\)
\(B=\dfrac{1}{2022}\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2020}{2021}\cdot\dfrac{2021}{2022}=\dfrac{1}{2022}\)
\(B=\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot\cdot\cdot\left(1-\dfrac{1}{2021}\right)\cdot\left(1-\dfrac{1}{2022}\right)\)
\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\cdot\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\cdot\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\cdot\cdot\cdot\left(\dfrac{2021}{2021}-\dfrac{1}{2021}\right)\cdot\left(\dfrac{2022}{2022}-\dfrac{1}{2022}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\cdot\cdot\dfrac{2020}{2021}\cdot\dfrac{2021}{2022}\)
\(B=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot2020\cdot2021}{2\cdot3\cdot4\cdot\cdot\cdot2021\cdot2022}\)
\(B=\dfrac{1}{2022}\)
\(B=\dfrac{2^{24}\cdot3^5-2^{24}\cdot3^4}{2^{24}\cdot3^5}+1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{301}-\dfrac{1}{303}\)
\(=\dfrac{2^{24}\cdot3^4\left(3-1\right)}{2^{24}\cdot3^5}+\dfrac{302}{303}\)
\(=\dfrac{2}{3}+\dfrac{302}{303}=\dfrac{202+302}{303}=\dfrac{504}{303}\)
=168/101
Bài 1:
Ta có |x-8| > 0 với mọi x
=>A=37-|x-8| > 37 với mọi x
Vậy GTLN của A=37 với x-8=0 =>x=8
Bài 2 tương tự nhé
Học tốt :))
Giải:
\(B=1+2\cdot\left(1+1\right)+3\cdot\left(2+1\right)+...+99\cdot\left(98+1\right)+100\cdot\left(99+1\right)\)
\(B=1+1\cdot2+2\cdot3\cdot3+...+98\cdot99+99+99\cdot100+100\)
\(B=\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)+\left(1+2+3+...+99+100\right)\)
\(B=333300+5050\)
\(B=3338050\)
Giúp tớ với tớ tick cho