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A= 1/1-1/2+1/2-1/3+1/4-1/5+...+1/101-1/102
A=1-1/102=102/102-1/102=101/102
ý b thì chờ mình tí tìm cách lập luận đã nhé
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}+\frac{1}{101.102}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{101}-\frac{1}{102}\)
\(A=1-\frac{1}{102}\)
\(A=\frac{101}{102}\)
\(X=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{99\cdot100}\)
\(X=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(X=1-\frac{1}{100}=\frac{99}{100}\).
x = 1/1*2 + 1/2*3 +1/3*4 + 1/4*5 + ... + 1/99*100
x = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
x = 1- 1/100
x = 99/100
Bài 1
a) \(P=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}=\frac{9}{10}\)
b) \(S=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)
c)\(Q=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{10}{20}-\frac{1}{20}\)
\(=\frac{9}{20}\)
Tk mình nha!!
Câu 2:
\(P=\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{99}\right)\)
\(=\left(\frac{2}{2}+\frac{1}{2}\right).\left(\frac{3}{3}+\frac{1}{3}\right).\left(\frac{4}{4}+\frac{1}{4}\right)...\left(\frac{99}{99}+\frac{1}{99}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{100}{99}\)
\(=\frac{3\cdot4\cdot5...100}{2.3.4...99}\)
\(=\frac{3\cdot100}{2}\)
\(=\frac{300}{2}=150\)
\(A=\frac{1\cdot1}{1\cdot2}\cdot\frac{2\cdot2}{2\cdot3}\cdot\frac{3\cdot3}{3\cdot4}\cdot\frac{4\cdot4}{4\cdot5}=\frac{1\cdot2\cdot3\cdot4}{1\cdot2\cdot3\cdot4}\cdot\frac{1\cdot2\cdot3\cdot4}{2\cdot3\cdot4\cdot5}=\frac{1}{5}\)
`Answer:`
Bài 1:
a. \(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}-\frac{2}{3}x+\frac{1}{3}=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{6}-\frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow-\frac{2}{3}=\frac{2}{3}-\frac{5}{6}\)
\(\Leftrightarrow-\frac{2}{3}x=-\frac{1}{6}\)
\(\Leftrightarrow x=-\frac{1}{6}:-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b. \(\frac{3}{x+5}=15\%\left(ĐKXĐ:x\ne-5\right)\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow\frac{60}{20\left(x+5\right)}=\frac{3\left(x+5\right)}{20\left(x+5\right)}\)
\(\Leftrightarrow60x=3x+15\)
\(\Leftrightarrow-3x=-45\)
\(\Leftrightarrow x=15\)
Bài 2:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
giúp mk nha. mk sẽ k cho bn nào trả lời giúp mk mà đúng
(1/1*2+1/2*3+1/3*4+...+1/8*9+1/9*10)*100-[5/2:(x+206/100)]:1/2=89
Đặt A=1/1*2+1/2*3+1/3*4+...+1/8*9+1/9*10
A=1-1/2+1/2-1/3+1/3-1/4+...+1/8-1/9+1/9-1/10
A=1-1/10
A=9/10
=>(1/1*2+1/2*3+1/3*4+...+1/8*9+1/9*10)*100-[5/2:(x+206/100)]:1/2=89
=9/10*100-[5/2:(x+206/100)]:1/2=89
90-[5/2:(x+206/100)]:1/2=89
5/2:(x+206/100):1/2=90-89
5/2:(x+206/100):1/2=1
x+206/100:1/2=5/2:1
x+206/100:1/2=5/2
x+103/25=5/2
x=5/2-103/25
x=-81/50