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\(\frac{2}{5}.\frac{1}{x}+\frac{1}{x}.2+\frac{2}{5}=0,5\)
\(\Rightarrow\frac{2}{5x}+\frac{2}{x}+\frac{2}{5}=\frac{1}{2}\)
\(\Rightarrow2.\left(\frac{1}{5x}+\frac{1}{x}+\frac{1}{5}\right)=\frac{1}{2}\)
\(\Rightarrow\frac{1}{5x}+\frac{5}{5x}+\frac{x}{5x}=\frac{1}{2}:2=\frac{1}{4}\)
\(\Rightarrow\frac{1+5+x}{5x}=\frac{1}{4}\)
\(\Rightarrow4.\left(1+5+x\right)=5x\)
\(\Rightarrow4+20+4x=5x\)
\(\Rightarrow24+4x=5x\)
\(\Rightarrow5x-4x=24\)
\(\Rightarrow x=24\)
Ta có: \(1-\frac{1}{2}.\frac{2}{3}....\frac{98}{99}\)
= \(1-\frac{1}{99}\)
= \(\frac{98}{99}\)
P=\(1-\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).........\left(1-\frac{1}{99}\right)\)
P=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.........\frac{98}{99}=\frac{1}{99}\)
Bài 1.
b) \(\frac{5+55+555+5555}{9+99+999+9999}\)
= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)
c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)
= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)
= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)
d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)
= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)
Bài 2.
a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)
= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)
= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)
= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)
x.2011-x=2011.2009+2001
2010x= 4042100
x= 4042100:2010 => x= 404210/201
=> \(\frac{56}{3}\) - [x.(1,73+3,27)] = \(\frac{71}{4}\)
=> 5x = \(\frac{56}{3}\) - \(\frac{71}{4}\) =\(\frac{11}{12}\)
=> x= \(\frac{11}{12}\) : 5= \(\frac{11}{60}\) Vậy x= \(\frac{11}{60}\)
\(18\frac{2}{3}-5x=\frac{71}{4}\)
\(5x=\frac{56}{3}-\frac{71}{4}=\frac{11}{12}\)
\(x=\frac{11}{12}\div5=\frac{11}{60}\)
\(\frac{37-2\left(y-3,27\right)}{5}=7,06\)
\(\Rightarrow37-2y+6,54=35,3\)
\(\Rightarrow-2y=35,3-37-6,54\)
\(\Rightarrow-2y=-8,24\)
\(\Rightarrow y=\frac{8,24}{2}=4,12\)
\(\frac{37-2.\left(y-3,27\right)}{5}=7,06\)
\(37-2.\left(y-3,27\right)=7,06.5\)
\(37-2y+6,54=35,3\)
\(2y=37+6,54-35,3\)
\(2y=8,04\)
\(y=4,02\)