Cho M = 1/2.3 + 1/3.4 + 1/4.5 +...+ 1/299.300
Giúp mk nha!!! Cám ơn trc
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a)=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}\)
\(=\frac{49}{100}\)
b)\(=\frac{201.204+1}{\left(201+2\right).204-407}\)
\(=\frac{201.204+1}{201.204+2.204-407}\)
\(=\frac{201.204+1}{201.204+1}\)
=1
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
A=1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9
A=1/3-1/9
A=2/9
các câu 2;3 còn lại giống câu 1 bạn nhé
bạn thay số vào rồi làm tương tự
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)
=1-1/20
=19/20
đặt A=1/1.2+1/2.3+1/3.4+..........1/49.50
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}<1\)
vậy A<1
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/49.50
1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/49 - 1/50
1 - 1/50 < 1
A= 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6
A= 1/2 - 1/3+ 1/3-1/4 + 1/4-1/5+ 1/5-1/6
A= 1/2- 1/6
A= 1/3
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{6}\)
\(\Rightarrow A=\frac{1}{3}\)
Xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\)
Khi đó:
\(1-\frac{2}{2.3}=\frac{1.4}{2.3}\) ; \(1-\frac{2}{3.4}=\frac{2.5}{3.4}\) ; ... ; \(1-\frac{2}{101.102}=\frac{100.103}{101.102}\)
\(\Rightarrow M=\frac{1.4}{2.3}\cdot\frac{2.5}{3.4}\cdot\cdot\cdot\frac{100.103}{101.102}\)
\(M=\frac{\left(1.2...100\right).\left(4.5...103\right)}{\left(2.3...101\right).\left(3.4...102\right)}=\frac{103}{101.3}=\frac{103}{303}\)
Vậy \(M=\frac{103}{303}\)
Cho A=1/1.2 + 1/2.3 + + 1/ 3.4+...+1/49.50 ; B = 1.2+2.3+3.4+4.5+5.6+...+49.50
Tính 50 mủ 2 A – B/17
\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{89\cdot90}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{89}-\dfrac{1}{90}\\ =\dfrac{1}{2}-\dfrac{1}{90}=\dfrac{22}{45}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{89.90}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{89}-\dfrac{1}{90}\\ =\dfrac{1}{2}-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-...-\left(\dfrac{1}{89}-\dfrac{1}{89}\right)-\dfrac{1}{90}\\ =\dfrac{1}{2}-0-0-...-0-\dfrac{1}{90}\\ =\dfrac{1}{2}-\dfrac{1}{90}\\ =\dfrac{45}{90}-\dfrac{1}{90}\\ =\dfrac{44}{90}\\ =\dfrac{22}{45}\)
\(M=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{299\cdot300}\)
\(M=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{300}\)
\(M=\frac{1}{2}-\frac{1}{300}\)
\(M=\frac{149}{300}\)
M = \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ \(\frac{1}{4.5}\)+ ......+\(\frac{1}{299.300}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+......+ \(\frac{1}{299}\)- \(\frac{1}{300}\)
= \(\frac{1}{2}\)- \(\frac{1}{300}\)
= \(\frac{149}{300}\)
Lưu ý: Nếu đúng, bạn kết bạn với mình và chọn đúng cho mình nha. Được không?