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Lời giải:
Ta có:
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{100-99}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
Vậy ta có đpcm.
\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)
\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)
\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)
Vậy.....
\(b,\left(-2\right)^2-1\dfrac{5}{27}.\left(-\dfrac{3}{2}\right)^3.\)
\(=4-1\dfrac{5}{27}.\left(-\dfrac{27}{8}\right).\)
\(=4-\dfrac{32}{27}.\left(-\dfrac{27}{8}\right).\)
\(=4-\left(-4\right).\)
\(=4+4=8.\)
Vậy.....
\(c,\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}.\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{99}\right)-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}+0+0+...+0-\dfrac{1}{100}.\)
\(=\dfrac{1}{2}-\dfrac{1}{100}.\)
\(=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{100}.\)
Vậy.....
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.............+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3
2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4
...
Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)
N=\(98+\dfrac{1}{100}-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-....-\dfrac{1}{99.100}\)
Xét P=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)
P=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
P=\(\dfrac{1}{2}-\dfrac{1}{100}\)
Vậy N=98-1+\(\dfrac{1}{50}\)
N=\(97+\dfrac{1}{50}\)
Vậy 97<N<98(ĐPCM)
A = \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{99.100}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{99}\) - \(\dfrac{1}{100}\)
A = \(\dfrac{1}{2}\) - \(\dfrac{1}{100}\)
Vậy: A = \(\dfrac{49}{100}\)
A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
A=\(\dfrac{1}{2}-\dfrac{1}{100}\)
A=\(\dfrac{49}{100}\)
\(B=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{99.100}\)
\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)
(do \(\dfrac{1}{a.\left(a+1\right)}=\dfrac{1}{a}-\dfrac{1}{a+n}\) với \(a\in N\)*)
\(B=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Chúc bạn học tốt!!!
\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..............+\dfrac{1}{99.100}\)
\(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+................+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B=1-\dfrac{1}{100}\)
\(B=\dfrac{99}{100}\)
a)
\(\dfrac{1}{2\cdot3}x+\dfrac{1}{3\cdot4}x+...+\dfrac{1}{49\cdot50}x=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=1\\ x\cdot\dfrac{12}{25}=1\\ x=1:\dfrac{12}{25}=1\cdot\dfrac{25}{12}=\dfrac{25}{12}\)
\(\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)..............\left(1-\dfrac{2}{99.100}\right)\)
\(=\left(\dfrac{6}{2.3}-\dfrac{2}{2.3}\right).\left(\dfrac{12}{3.4}-\dfrac{2}{3.4}\right)..............\left(\dfrac{9900}{99.100}-\dfrac{2}{99.100}\right)\)
\(=\dfrac{4}{2.3}.\dfrac{10}{3.4}..........................\dfrac{9898}{99.100}\)
\(=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}..............\dfrac{98.101}{99.100}\)
\(=\dfrac{1.2.3.....98}{2.3......99}.\dfrac{4.5.6.....101}{3.4.......100}\)
\(=\dfrac{1}{99}.\dfrac{101}{3}=\dfrac{101}{297}\)
a) A = \(\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}.\dfrac{4^2}{4.5}\)
A = \(\dfrac{1.1}{1.2}.\dfrac{2.2}{2.3}.\dfrac{3.3}{3.4}.\dfrac{4.4}{4.5}\)
A = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)= \(\dfrac{1}{5}\)
b) B = \(\dfrac{2^2}{1.3}.\dfrac{3^2}{2.4}.\dfrac{4^2}{3.5}.\dfrac{5^2}{4.6}\)
B = \(\dfrac{2.3.4.5}{1.2.3.4}.\dfrac{2.3.4.5}{3.4.5.6}\)= \(\dfrac{5}{3}\)
\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\\ =2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\right)\\ =2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)
Đặt A = \(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
\(\Rightarrow A=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\right)\)
\(=2\left(\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+...+\dfrac{100-99}{99.100}\right)\)
\(=2\left(\dfrac{3}{2.3}-\dfrac{2}{2.3}+\dfrac{4}{3.4}-\dfrac{3}{3.4}+...+\dfrac{100}{99.100}-\dfrac{99}{99.100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=1-\dfrac{1}{50}=\dfrac{49}{50}\)