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ta có : 1/10 = 1/2.5
1/40 = 1/5.8
suy ra 1/10 + 1/40 + 1/88 + ... + 1/(3x+2).(3x+5) = 3/20 = 1/2.5 + 1/5.8 + 1/8.11 +....+ 1/(3x+2).(3x+5)
= 1/3(1/2 - 1/5) + 1/3(1/5 - 1/8) + ...+ 1/3(1/8 - 1/11) + ....+ 1/3 [(3x+2) (3x+5)] = 3/20
= 1/3(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +.... + 1/3x+2 + 1/3x+5) = 3/20
1/3 ( 1/2 - 1/3x+5) = 3/20
suy ra 1/2 - 1/3x+5 = 3/20 : 1/3 = 9/20
suy ra 1/3x+5 = 1/2 - 9/20 = 1/20
suy ra 3x+5 = 20
suy ra 3x = 20 - 5= 15
suy ra x = 15 : 3 = 5
vay x = 5
nhớ k nha!!!!
a) \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{2007}{2009}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2007}{2009}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2009}\)
=> x + 1 = 2009
=> x = 2009 - 1
=> x = 2008
b) \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)
=> \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(3x+2\right).\left(3x+5\right)}=\frac{4}{25}\)
=> \(\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{\left(3x+2\right)\left(3x+5\right)}\right)=\frac{4}{25}\)
=> \(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{3x+2}-\frac{1}{3x+5}=\frac{4}{25}:\frac{1}{3}\)
=> \(\frac{1}{2}-\frac{1}{3x+5}=\frac{12}{25}\)
=> \(\frac{1}{3x+5}=\frac{1}{2}-\frac{12}{45}\)
=> \(\frac{1}{3x+5}=\frac{1}{50}\)
=> 3x + 5 = 50
=> 3x = 50 - 5
=> 3x = 45
=> x = 45 : 3
=> x = 15
1)
x - 18 = 3x + 4
=> x - 3x = 4 + 18
=> -2x = 22
=> x = 22 : (-2)
=> x = -11
Vậy x = -11
Sửa đề lại :
Tìm x thuộc Z
\(5\times\left[\frac{1}{4}-\frac{1}{5}-\frac{1}{10}\right]\le\frac{x}{20}\le-3\times\left[\frac{1}{2}-\frac{1}{4}-\frac{1}{5}\right]\)
\(\Leftrightarrow5\times\left[\frac{5}{20}-\frac{4}{20}-\frac{2}{20}\right]\le\frac{x}{20}\le-3\times\left[\frac{10}{20}-\frac{5}{20}-\frac{4}{20}\right]\)
\(\Leftrightarrow5\times\frac{-1}{20}\le\frac{x}{20}\le-3\times\frac{1}{20}\)
\(\Leftrightarrow\frac{-5}{20}\le\frac{x}{20}\le\frac{-3}{20}\)
\(\Leftrightarrow-5\le x\le-3\)
\(\Leftrightarrow x\in\left\{-5;-4;-3\right\}\)
Study well >_<