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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:
\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)
\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)
\(S=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{100}\left(1+2+3+...+100\right)\)
\(=1+\dfrac{1}{2}.\dfrac{2\left(1+2\right)}{2}+\dfrac{1}{3}.\dfrac{3\left(1+3\right)}{2}+\dfrac{1}{4}.\dfrac{4\left(1+4\right)}{2}+...+\dfrac{1}{100}.\dfrac{100\left(1+100\right)}{2}\)
\(=1+\dfrac{2\left(1+2\right)}{2.2}+\dfrac{3\left(1+3\right)}{2.3}+\dfrac{4\left(1+4\right)}{2.4}+...+\dfrac{100\left(1+100\right)}{2.100}\)
\(=1+\dfrac{1+2}{2}+\dfrac{1+3}{2}+\dfrac{1+4}{2}+...+\dfrac{1+100}{2}\)
\(=1+\dfrac{3+4+5+...+101}{2}\)
\(=1+\dfrac{\dfrac{99\left(101+3\right)}{2}}{2}\)
\(=1+2574=2575\)
\(\)
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:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
\(A=\left[\dfrac{1}{100}-1^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{3}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)\(=\left[\dfrac{1}{100}-1^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right].\left[\dfrac{1}{100}-\left(\dfrac{1}{3}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)Mà \(\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2=\dfrac{1}{100}-\dfrac{1}{100}=0\)
\(\Rightarrow A=0\)
\(\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)
\(=\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{10}\right)^2\right]...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)
\(=\left(\dfrac{1}{100}-1^2\right)\left[\dfrac{1}{100}-\left(\dfrac{1}{2}\right)^2\right]...0...\left[\dfrac{1}{100}-\left(\dfrac{1}{20}\right)^2\right]\)
\(=0\)
Vậy...
a/ \(\dfrac{\left(1+2+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right)\left(6,3.12-21.36\right)}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+2+3+.....+100\right)\left(\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{9}\right).0}{\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}}\)
\(=\dfrac{0}{\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{100}}\)
\(=0\)
\(C=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{1-99^2}{100^2}\right)\left(\dfrac{100^2-1}{99^2}\right)=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{-2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{-98.100}{99^2}\right)\left(\dfrac{99.101}{100^2}\right)=-\dfrac{101}{200}\)
Áp dụng \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)
\(\Rightarrow\dfrac{1}{n}\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2n}=\dfrac{n+1}{2}\)
Vậy:
\(A=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{101}{2}=\dfrac{1+2+3+...+100}{2}-1\)
\(=\dfrac{100.101}{2}-1=5049\)