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Đặt D=\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
=>3D=\(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
=>3D-D=(\(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\))-(\(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\))
=>2D=\(1-\dfrac{1}{3^{99}}\)
=>D=\(\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)
C=D+\(\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{3}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{8.3^{98}}=\dfrac{4.3^{98}-1}{8.3^{98}}\)
a) Ta có
S = \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{n.\left(n+1\right).\left(n+2\right)}\)
2S = \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{n\left(n+1\right)}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)2S = \(\dfrac{1}{1.2}-\dfrac{1}{\left(n+1\right).\left(n+2\right)}\)
S = \(\dfrac{1}{4}-\dfrac{1}{\left(n+1\right).\left(n+2\right):2}\)
b) A = \(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{99}\)
A = \(2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
A = \(2-\dfrac{1}{99}\)
A = \(\dfrac{197}{99}\)
c) Ta có
B = \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\)
B = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
B = \(1-\dfrac{1}{100}\)
B = \(\dfrac{99}{100}\)
d) Ta có
C = \(\dfrac{99}{1}+\dfrac{98}{2}+\dfrac{97}{3}+...+\dfrac{1}{99}\)
C = \(1+\left(1+\dfrac{98}{2}\right)+\left(1+\dfrac{97}{3}\right)+...+\left(1+\dfrac{1}{99}\right)\)
C = \(1+50+\dfrac{100}{3}+...+\dfrac{100}{99}\)
C = 51 + 100(\(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\))
Đặt D = \(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{98}-\dfrac{1}{99}\)
D = \(\dfrac{1}{2}-\dfrac{1}{99}\)
D = \(\dfrac{97}{198}\)
=> C = 51 + 100.\(\dfrac{97}{198}\)
C = 51 + \(\dfrac{4850}{99}\)
C = \(\dfrac{9899}{99}\)
Đây là bài làm của mình sai thì nx nha
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
3) \(\dfrac{3}{4}.x-\dfrac{5}{3}.x=\dfrac{7}{12}\)
\(\left(\dfrac{3}{4}-\dfrac{5}{3}\right).x=\dfrac{7}{12}\)
\(-\dfrac{11}{12}.x=\dfrac{7}{12}\)
\(x=\dfrac{7}{12}:\left(-\dfrac{11}{12}\right)\)
\(x=-\dfrac{7}{11}\)
Ta có: \(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{3}{6}+\dfrac{-2}{6}+\dfrac{-1}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)\left(\dfrac{3+\left(-2\right)+\left(-1\right)}{6}\right)\)
\(B=\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right).0=0\)
Tick mk vs !
B = (3\(\dfrac{10}{99}\)+4\(\dfrac{11}{99}\)-5\(\dfrac{8}{299}\)).0
B = 0
Ta có :
\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)\)
\(=100-1-\dfrac{1}{2}-\dfrac{1}{3}-..................-\dfrac{1}{100}\)
\(=99-\dfrac{1}{2}-\dfrac{1}{3}-................-\dfrac{1}{100}\)
\(=\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+..................+\left(1-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{2}+\dfrac{2}{3}+.................+\dfrac{99}{100}\)
Vậy :\(100-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{100}\right)=\dfrac{1}{2}+\dfrac{2}{3}+....................+\dfrac{99}{100}\)
\(\Rightarrowđpcm\)
Đặt :
\(A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+.......................+\dfrac{1}{3^{99}}+\dfrac{1}{3^{99}}\)
\(\Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...................+\dfrac{1}{3^{98}}\)
\(\Rightarrow3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+..............+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^3}+..............+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)\)\(\Rightarrow2A=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}\)
\(\Rightarrow C=A+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}+\dfrac{1}{8.3^{99}}=\dfrac{1}{2}-\dfrac{3}{8.3^{99}}=\dfrac{1}{2}-\dfrac{1}{8.3^{98}}=\dfrac{4.3^{98}-1}{8.3^{98}}\)