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a/ Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+.........+\dfrac{1}{3^{50}}\)
\(\Leftrightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{49}}\)
\(\Leftrightarrow3A-A=\left(1+\dfrac{1}{3}+....+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+....+\dfrac{1}{3^{50}}\right)\)
\(\Leftrightarrow2A=1-\dfrac{1}{3^{50}}\)
còn sao nx thì mk chịu =.=
\(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \Rightarrow P+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\\ \Rightarrow P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\\ =\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=2018.\dfrac{2021}{4034}=1011.000992\\ \Rightarrow P=1008.000992\)
Giả sử trong 2021 số nguyên dương đã cho không có số nào bằng nhau.
Và a1 < a2 < a3 < ... < a2021 . Ta có :
\(\dfrac{1}{a_1}+\dfrac{1}{a_2}+...+\dfrac{1}{a_{2021}}\le\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2021}\)
\(\Rightarrow\dfrac{1}{a_1}+\dfrac{1}{a_2}+...+\dfrac{1}{a_{2021}}< \dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2}=1+1010=1011\)
(mâu thuẫn)
⇒Điều giả sử sai. ⇒ Ít nhất 2 trong số 2021 số nguyên dương đã cho bằng nhau.
\(a)3\dfrac{1}{2}.\dfrac{4}{49}-\left[2,\left(4\right):2\dfrac{5}{11}\right]:\left(\dfrac{-42}{5}\right)\)
\(=\dfrac{7}{2}.\dfrac{4}{49}-\dfrac{88}{27}:\left(\dfrac{-42}{7}\right)\)
\(=\dfrac{2}{7}-\dfrac{-220}{567}\)
\(=\dfrac{382}{567}\)
các phần con lại dễ nên bn tự lm đi nhé mk bn lắm
Chúc bạn học tốt!
a) \(\frac{1}{8}>0>\frac{-3}{8}=>\frac{1}{8}>\frac{-3}{8}\)
b) \(\frac{-3}{7}< 0< 2\frac{1}{2}=>\frac{-3}{7}< 2\frac{1}{2}\)
c) \(-3.9< 0< 0.1=>-3.9< 0.1\)
d) \(-2.3< 0< 3.2=>-2.3< 3.2\)
\(\Rightarrow-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{100^2}\right)\)
\(-A=\dfrac{3}{2^2}.\dfrac{8}{3^2}...\dfrac{9999}{100^2}\)
\(-A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(-A=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(-A=\dfrac{1}{100}.\dfrac{101}{2}=\dfrac{101}{200}\)
\(\Rightarrow A=\dfrac{-101}{200}\)
Mà \(\dfrac{-101}{200}< \dfrac{-100}{200}=\dfrac{-1}{2}\)
\(\Rightarrow A< \dfrac{-1}{2}\)
a, Ta có :\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}+\dfrac{1}{2^{50}}\\ \Rightarrow2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\\ \Rightarrow2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{49}}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{50}}\right)\\ \Rightarrow A=1-\dfrac{1}{2^{50}}< 1\\ \Rightarrow A< 1\) Vậy \(A< 1\)
b, Ta có :
\(B=\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\\ \Rightarrow3B-B=\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\\ \Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}\)Vậy \(B< \dfrac{1}{2}\)
c, Ta có :
\(C=\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\\ \Rightarrow4C=1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\\\Rightarrow4C-C=\left(1+\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{999}}\right)-\left(\dfrac{1}{4^1}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{1000}}\right)\\ \Rightarrow3C=1-\dfrac{1}{4^{1000}}< 1\\ \Rightarrow C< \dfrac{1}{3}\)Vậy \(C< \dfrac{1}{3}\)