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Bài này mình chắc 100%, 1 đúng nha vì ghi cực khổ lắm:
1) Ta có: \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}...+\frac{50-49}{49.50}\)
\(=\frac{2}{1.2}-\frac{1}{1.2}+\frac{3}{2.3}-\frac{2}{2.3}+...+\frac{50}{49.50}-\frac{49}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}
Bài 5:
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}< 1\)
Vậy A<1.
Bài 4: Bn ghi nhầm đề rồi.
Đề đúng: \(A=\frac{4}{1.3}+\frac{4}{3.5}+...+\frac{4}{2011.2013}\)
\(\frac{1}{2}A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\)
\(\frac{1}{2}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\)
\(\frac{1}{2}A=1-\frac{1}{2013}\)
\(A=2.\frac{2012}{2013}=\frac{4024}{2013}\)
\(A=2.3+3.4+4.5+...+49.50\)
\(3A=2.3.3+3.4.3+4.5.3+...+49.50.3\)
\(3A=2.3.\left(4-1\right)+3.4.\left(5-2\right)+4.5.\left(6-3\right)+...+49.50.\left(51-48\right)\)
\(3A=2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+49.50.51-48.49.50\)
\(3A=-1.2.3+49.50.51\)
\(3A=-6+48450\)
\(3A=48444\)
\(A=\frac{48444}{3}\)
\(A=16148\)
Chúc bạn học tốt ~
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{25.26.27}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{25.26.27}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{25.26}-\frac{1}{26.27}\)
\(2B=\frac{1}{1.2}-\frac{1}{26.27}\)
\(2B=\frac{1}{2}-\frac{1}{702}\)
\(2B=\frac{175}{351}\)
\(B=\frac{175}{251}:2\)
\(B=\frac{175}{502}\)
Chúc bạn học tốt ~
Tách: 1/1.2=1-1/2; 1/2.3=1/2-1/3; ....; 1/49.50=1/49-1/50
Và rút gọn các số liền kề thì còn lại kết quả
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
A=1/1.2+1/2.3+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50
=49/50
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+\frac{1}{3.4.5.6}+...+\frac{1}{27.28.29.30}\)
\(A=\frac{1}{4.6}+\frac{1}{10.12}+\frac{1}{18.20}+...+\frac{1}{810.812}\)
.......
~ Chúc học tốt ~
Ai ngang qua xin để lại 1 L - I - K - E
\(A=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+.....+\frac{1}{27.28.29.30}\)
\(3A=3.\left(\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+......+\frac{1}{27.28.29.30}\right)\)
\(3A=\frac{3}{1.2.3.4}+\frac{3}{2.3.4.5}+..........+\frac{3}{27.28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+........+\frac{1}{27.28.29}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{1.2.3}-\frac{1}{28.29.30}\)
\(3A=\frac{1}{6}-\frac{1}{24360}\)
\(3A=\frac{1353}{8120}\)
\(A=\frac{1353}{8120}:3\)
\(A=\frac{451}{8120}\)
\(\Rightarrow\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{1}{1.2}-\frac{1}{99.100}\right).y=\frac{49}{100}\Leftrightarrow\left(\frac{99.50-1}{99.100}\right).y=\frac{49}{100}\)
\(\Leftrightarrow\left(\frac{99.50-1}{99}\right).y=49\Leftrightarrow\left(99.50-1\right).y=99.49\Rightarrow y=\frac{99.49}{99.50-1}\)
Ta có :
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(A=\frac{49}{50}\)
Vậy \(A=\frac{49}{50}\)
Chúc bạn học tốt ~
A= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\)
A= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
A= \(1-\frac{1}{50}\)
A= \(\frac{49}{50}\)