KQ:\(\frac{1}{5}\)

cho tớ xin cách lm

\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)

\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)

\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)

\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)

\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)

~ Hok tốt ~

Ta có:

 \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)

nha

Mk ko biết lm nhưng cứ k thoải mái nha

SORRY

Ta có

\(A=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)...\left(\frac{3^6}{9}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)...\left(\frac{729}{9}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)=0\)

vì 729/9=81

Vậy A=0

k me đi

\(\left(\frac{3}{4}-81\right).\left(\frac{3^2}{5}-81\right).\left(\frac{3^3}{6}-81\right).\left(\frac{3^4}{7}-81\right).\left(\frac{3^5}{8}-81\right).\left(\frac{729}{9}-81\right)....\left(\frac{3^{2013}}{2016}-81\right)\)

=>....................................................................................................................(81-81)..............................................

=>.....................................................................................................................0.....................................................

=>A=0