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Do A = 98 99 + 1 98 89 + 1 > 1 nên A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 98 98 + 1 98 98 88 + 1 = 98 98 + 1 98 88 + 1 = B
Vậy A > B
a) Do A = 98 99 + 1 98 89 + 1 > 1 nên
A = 98 99 + 1 98 89 + 1 > 98 99 + 1 + 97 98 89 + 1 + 97 = 98 ( 98 98 + 1 ) 98 ( 98 88 + 1 ) = 98 98 + 1 98 88 + 1 = B
Vậy A > B
b) Do C = 100 2008 + 1 100 2018 + 1 < 1 nên
C= 100 2008 + 1 100 2018 + 1 > 100 2008 + 1 + 99 100 2018 + 1 + 99 = 100 ( 100 2007 + 1 ) 100 ( 100 2017 + 1 ) = 100 2007 + 1 100 2017 + 1 = D
Vậy C > D.
Sửa đề: \(C=\dfrac{17^{99}+1}{17^{99}-1}\)
\(C=\dfrac{17^{99}-1+2}{17^{99}-1}=1+\dfrac{2}{17^{99}-1}\)
\(D=\dfrac{17^{98}-1+2}{17^{98}-1}=1+\dfrac{2}{17^{98}-1}\)
17^99>17^98
=>17^99-1>17^98-1
=>C<D
D=\(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+........+1-\frac{1}{9900}\)
\(=1-\frac{1}{1.2}+1-\frac{1}{2.3}+........+1-\frac{1}{99.100}\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=98+\frac{1}{100}=\frac{9801}{100}\)
d=1/1.2+5/2.3+11/3.4+...+9899/99.100
=>d=1-1/2+1/2-1/3+...+1/99-1/100
=>d=1-1/100
=>d=99/100
Vậy d=99/100
\(A=100\cdot\left(1+\dfrac{5}{6}+\dfrac{11}{12}+\dfrac{19}{20}+...+\dfrac{9899}{9900}\right)\\ =100\cdot\left(1+1-\dfrac{1}{6}+1-\dfrac{1}{12}+1-\dfrac{1}{20}+...+1-\dfrac{1}{9900}\right)\\ =100\cdot\left[\left(1+1+1+...+1\right)-\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{9900}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{99\cdot100}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\right]\\ =100\cdot\left[99-\left(\dfrac{1}{2}-\dfrac{1}{100}\right)\right]\\ =100\cdot\left[99-\dfrac{49}{100}\right]\\ =100\cdot\dfrac{9851}{100}\\ =9851\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+....+\frac{9899}{9900}\)
\(=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+.....+\left(1-\frac{1}{9900}\right)\)
\(=\left(1-\frac{1}{1.2}\right)+\left(1-\frac{1}{2.3}\right)+\left(1-\frac{1}{3.4}\right)+...+\left(1-\frac{1}{99.100}\right)\)
\(=99-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(=99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)
\(=99-\left(1-\frac{1}{100}\right)=99-1+\frac{1}{100}=98+\frac{1}{100}=\frac{9801}{100}\)