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Ta có: \(M=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)
\(\Rightarrow M\) \(\)\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(\Rightarrow M=1-\dfrac{1}{50}< 1\)
Vậy M < 1.
M=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=1-\dfrac{1}{50}=\dfrac{50}{50}-\dfrac{1}{50}=\dfrac{49}{50}.\)
Vậy M=\(\dfrac{49}{50}\)
*Trước dấu = là 1 chữ M
Sửa lại đề:
\(M=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{49.50}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-..........-\dfrac{1}{49}-\dfrac{1}{50}\)
\(M=1-\dfrac{1}{50}\)
\(M=\dfrac{50}{50}-\dfrac{1}{50}\)
\(M=\dfrac{49}{50}\)
Đề bài là thu gọn / tính giá trị biểu thức nhé chứ không phải là So sánh , thiếu dữ kiện kìa
\(a,\dfrac{3}{4}-1\dfrac{1}{2}+0,5:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}.\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{1}{2}.\dfrac{12}{5}.\)
\(=-\dfrac{3}{4}+\dfrac{12}{10}.\)
\(=-\dfrac{3}{4}+\dfrac{6}{5}.\)
\(=-\dfrac{15}{20}+\dfrac{24}{20}=\dfrac{9}{20}.\)
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.....
a: \(B=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}=1-\dfrac{1}{2008}=\dfrac{2007}{2008}\)
b: \(Q=\dfrac{7}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{2009\cdot2011}\right)\)
\(=\dfrac{7}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{7}{2}\cdot\dfrac{2010}{2011}\simeq3,50\)
Câu 2:
a: =>-11/12x=-1/6-3/4=-2/12-9/12=-11/12
=>x=1
b: =>x-42=57-x-50=7-x
=>2x=49
hay x=49/2
d: =>x+1=3 hoặc x+1=-3
=>x=2 hoặc x=-4
e: =>2x+3=5 hoặc 2x+3=-5
=>2x=2 hoặc 2x=-8
=>x=1 hoặc x=-4
\(A=\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
\(A=\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)-\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100.101}\right)\)
\(A=\left(1-\dfrac{1}{100}\right)-\left(\dfrac{\dfrac{1}{1.2}-\dfrac{1}{100.101}}{2}\right)\)
Bấm máy nha
\(B=\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+\dfrac{5}{3.4.5.6}+...+\dfrac{5}{98.99.100.101}\)
\(B=\dfrac{5}{3}.\left(\dfrac{3}{1.2.3.4}+\dfrac{3}{2.3.4.5}+...+\dfrac{3}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4-1}{1.2.3.4}+\dfrac{5-2}{2.3.4.5}+...+\dfrac{101-98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{4}{1.2.3.4}-\dfrac{1}{1.2.3.4}+\dfrac{5}{2.3.4.5}-\dfrac{2}{2.3.4.5}+...+\dfrac{101}{98.99.100.101}-\dfrac{98}{98.99.100.101}\right)\)
\(B=\dfrac{5}{3}.\left(\dfrac{1}{1.2.3}-\dfrac{1}{99.100.101}\right)\)
\(B=\dfrac{5}{3}.\dfrac{166649}{999900}\approx0,3\)
Giải: Ta có: \(\dfrac{1}{2.3}\)=\(\dfrac{1}{2}-\dfrac{1}{3}\) ; \(\dfrac{1}{2}-\dfrac{1}{3}\)(giữ nguyên)
Vì \(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{2}-\dfrac{1}{3}\) nên \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)
Vậy \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\)