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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 \(C< \dfrac{1}{3}\)
a) \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+............+\dfrac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+.........+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{100}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{100}}\)
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
Bài 2 :
\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)
\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)
\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)
Đặt :
\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)
\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)
\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)
\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)
\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)
\(\Leftrightarrow3S< \dfrac{4}{3}\)
\(\Leftrightarrow S< \dfrac{4}{9}\)
\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)
\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)
\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)
\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)
\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)
Đặt:
\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)
\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)
\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)
\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)
\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)
Thay M vào A ta có:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
\(\dfrac{A}{B}=\dfrac{\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}}{\dfrac{1}{4}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\right)}=1:\dfrac{1}{4}=4\)
a, \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\\ 3B=3+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\\ 3B-B=\left(3+\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2003}}+\dfrac{1}{3^{2004}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2004}}+\dfrac{1}{3^{2005}}\right)\\2B=3-\dfrac{1}{3^{2005}}\\ B=\dfrac{3-\dfrac{1}{3^{2005}}}{2}\)
b,
\(A=1+5+5^2+5^3+...+5^{49}+5^{50}\\ 5A=5+5^2+5^3+5^4+...+5^{50}+5^{51}\\ 5A-A=\left(5+5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(1+5+5^2+5^3+...+5^{49}+5^{50}\right)\\ 4A=5^{51}-1\\ A=\dfrac{5^{51}-1}{4}\)
c,
\(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2-1}\right)......\left(\dfrac{1}{100^2-1}\right)\\ A=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)......\left(\dfrac{1}{10000}-1\right)\\ A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\cdot\dfrac{9999}{10000}\\ A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot\cdot\cdot\cdot\dfrac{99\cdot101}{100\cdot100}\\ A=\dfrac{1\cdot2\cdot3\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\cdot\dfrac{3\cdot4\cdot5\cdot\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\\ A=\dfrac{1}{100}\cdot\dfrac{101}{2}\\ A=\dfrac{101}{200}\)
d,
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\\ A=\left(2^{100}+2^{98}+...+2^2\right)-\left(2^{99}+2^{97}+...+2^1\right)\)
Đặt \(A=B-C\)
\(\Rightarrow B=\left(2^{100}+2^{98}+...+2^2\right)vàC=\left(2^{99}+2^{97}+...+2^1\right)\)
\(B=2^{100}+2^{98}+...+2^2\\ 4B=2^{102}+2^{100}+...+2^4\\ 4B-B=\left(2^{102}+2^{100}+...+2^4\right)-\left(2^{100}+2^{98}+...+2^2\right)\\ 3B=2^{102}-2^2\\ B=\dfrac{2^{102}-2^2}{3}\left(1\right)\)
\(C=2^{99}+2^{97}+...+2^1\\ 4C=2^{101}+2^{99}+...+2^3\\ 4C-C=\left(2^{101}+2^{99}+...+2^3\right)-\left(2^{99}+2^{97}+...+2\right)\\ 3C=2^{101}-2\\ C=\dfrac{2^{101}-2}{3}\left(2\right)\)
Từ (1) và (2) ta có :
\(A=\dfrac{2^{102}-2^2}{3}-\dfrac{2^{101}-2}{3}\\ A=\dfrac{2^{102}-2^2-2^{101}+2}{3}\\ A=\dfrac{2^{102}-2^{101}+2}{3}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(A=2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)\(\Rightarrow A=1-\dfrac{1}{2^{100}}\)
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