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mình cần gấp trong hôm nay mong mọi người giúp mình !!!!!!!
\(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{x\times\left(x+2\right)}=\frac{19}{40}\)
\(\frac{1}{2}\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{x\times\left(x+2\right)}\right)=\frac{19}{40}\)
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{19}{40}\)
\(1-\frac{1}{x+2}=\frac{19}{40}\times2\)
\(\frac{x+1}{x+2}=\frac{19}{20}\Rightarrow x=18\)
\(a,x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x-\frac{61}{8}=\frac{5}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)
\(b,x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x+\frac{43}{5}=\frac{37}{4}\)
=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)
\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)
=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)
=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)
d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)
=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)
=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)
=> \(\frac{98x}{198}=99\)
=> 98x = 99 . 198
=> 98x = 19602
=> x = 19602 : 98 = 9801/49
a) \(x-7\frac{5}{8}=1\frac{1}{4}\)
=> \(x=\frac{5}{4}+\frac{61}{8}\)
=> \(x=\frac{71}{8}\)
b) \(x+7\frac{5}{8}=9\frac{1}{4}\)
=> \(x=\frac{37}{4}-\frac{61}{8}\)
=> \(x=\frac{13}{8}\)
c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)
=> \(x-\frac{61}{8}=3.\frac{1}{2}\)
=> \(x-\frac{61}{8}=\frac{3}{2}\)
=> \(x=\frac{3}{2}+\frac{61}{8}\)
=> \(x=\frac{73}{8}\)
d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)
=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)
=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)
=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)
=> \(x.\frac{98}{99}=198\)
=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)
6B=1.3.(5+1)+3.5.(7-1)+...+19.21(23-17)
6B=1.3.5+1.3+3.5.7-1.3.5+...+19.21.23-17.19.21
6B=3+19.21.23
6B=9180
B=9180/6
B=1530
tick nhé ko tick mai đến lớp tao phang
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
= \(1-\dfrac{1}{101}\)
=\(\dfrac{100}{101}\)
\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)
=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)
=\(\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
=\(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)
= \(\dfrac{5}{2}-\dfrac{100}{101}\)
= \(\dfrac{305}{202}\)
M = 5 + 53 + 55 + ... + 547 + 549
52M = 52(5 + 53 + 55 + ... + 547 + 549)
25M = 53 + 55 + 57 + ... + 549 + 551
25M - M = ( 53 + 55 + 57 + ... + 549 + 551) - (5 + 53 + 55 + ... + 547 + 549)
24M = 551 - 5
M = \(\frac{5^{51}-5}{24}\)
\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}\left(1-\frac{1}{11}\right)y=\frac{2}{3}\)
=> \(\frac{1}{2}.\frac{10}{11}y=\frac{2}{3}\)
=> \(\frac{5}{11}y=\frac{2}{3}\)
=>y = \(\frac{2}{3}:\frac{5}{11}\)
=> y = \(\frac{22}{15}\)
cho mk cái lời giải thích chỗ nhân 1/2 ý mk ko hiểu mong bn thông cảm
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)
cậu đang khen hay đang đe doạ vậy