Giả giùm mình cái này nha :
\(\left(\frac{980\div2}{1000}-0,25\right)\times\frac{125}{423}=?\)
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14 tháng 10 2016
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Leftrightarrow x=2010\)
14 tháng 10 2016
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(1-\frac{2}{x+1}=\frac{2009}{2011}\)
\(\frac{2}{x+1}=1-\frac{2009}{2011}\)
\(\frac{2}{x+1}=\frac{2}{2011}\)
\(x+1=2011\)
\(x=2011-1\)
\(\Rightarrow x=2010\)
8 tháng 6 2019
Kết quả là: 1649/151x1826679x330890x15263. Mình không chắc lắm.
VA
12 tháng 7 2016
\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=-\frac{16}{10}:\frac{5}{3}\)
\(A=-\frac{16.3}{10.5}=-\frac{48}{50}=-\frac{24}{25}\)
\(B=1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(B=\frac{14}{10}\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)
\(B=\frac{2.7.3.5}{2.5.7.7}-\left(\frac{12+10}{15}\right):\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}\times\frac{5}{11}=\frac{3}{7}-\frac{2.11.5}{3.5.11}\)
\(B=\frac{3}{7}-\frac{2}{3}=\frac{9-14}{21}=-\frac{5}{21}\)
Ủng hộ mk nha !!! ^_^
NN
21 tháng 8 2020
a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
T
13 tháng 7 2019
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(\left(\frac{980:2}{1000}-0,25\right)\times\frac{125}{423}=\left(\frac{490}{1000}-0,25\right)\times\frac{125}{423}\)
\(=\left(0,49-0,25\right)\times\frac{125}{423}=0,24\times\frac{125}{423}=\frac{6}{25}\times\frac{125}{423}=\frac{10}{141}\)
\(\left(\frac{980:2}{1000}-0,25\right).\frac{125}{423}\)
=\(\left(\frac{49}{100}-\frac{1}{4}\right).\frac{125}{423}\)
=\(\left(\frac{49-25}{100}\right).\frac{125}{423}\)
=\(\frac{6}{25}.\frac{125}{423}\)
=\(\frac{6.125}{25.423}\)
= \(\frac{10}{141}\)