Tính : \(\left(\frac{3}{5}\right)^{2003}:\left(\frac{9}{25}\right)^{1000}\)
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$\textbf{a)}$
$A=\dfrac{\left(\dfrac49\right)^2\cdot\left(-\dfrac9{16}\right)\cdot(-1)^{19}}{\left(\dfrac4{25}\right)^2\cdot\left(-\dfrac{25}{144}\right)^2\cdot\left(-\dfrac{49}{144}\right)^2}$
$=\dfrac{\dfrac{16}{81}\cdot\left(-\dfrac9{16}\right)\cdot(-1)}{\dfrac{16}{625}\cdot\dfrac{625}{144^2}\cdot\dfrac{49^2}{144^2}}$
$=\dfrac{\dfrac19}{\dfrac{49^2}{144^4}}$
$=\dfrac19\cdot\dfrac{144^4}{49^2}$
$=\dfrac19\cdot\dfrac{(2^4\cdot3^2)^4}{7^4}$
$=\dfrac{2^{16}\cdot3^6}{7^4}$
$=\dfrac{47775744}{2401}.$
$\textbf{b)}$
$B=\dfrac{\left(\dfrac23\right)^3\cdot\left(-\dfrac34\right)^2\cdot(-1)^{2003}}{\left(\dfrac25\right)^2\cdot\left(-\dfrac5{12}\right)^3}$
$=\dfrac{\dfrac8{27}\cdot\dfrac9{16}\cdot(-1)}{\dfrac4{25}\cdot\left(-\dfrac{125}{1728}\right)}$
$=\dfrac{-\dfrac16}{-\dfrac5{432}}$
$=\dfrac16\cdot\dfrac{432}{5}$
$=\dfrac{72}{5}.$
\(\left(\frac{3}{5}\right)^{2012}:\left(\frac{9}{25}\right)^{1000}=\left(\frac{3}{5}\right)^{2012}:\left(\frac{3}{5}\right)^{2000}=\left(\frac{3}{5}\right)^{12}\)
Gộp nhóm 4 => A = -4 * 500+2001+2002-2003=0
B = X = 2 11 1 x^2 1
a) tạm bỏ số 1 ra => có 2012 số hạng=> có 1006 cặp =(-1)
=> A=1+-(-1).1006=-1005
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2019}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}:\sqrt{\frac{9}{25}}\)\(=\frac{\frac{2^3}{3^3}.\frac{-3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{12^3}}:\frac{3}{5}\)
\(=\frac{\frac{2^3}{5^3}.\frac{-3^2}{2^4}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{-5^3}{2^6.3^3}}:\frac{3}{5}=\frac{\frac{-1}{3.2}}{\frac{-5}{2^4.3^3}}:\frac{3}{5}\)\(=\frac{-1}{3.2}.\frac{-2^4.3^3}{5}.\frac{5}{3}\)
\(=\frac{2^3.3^2}{5}.\frac{5}{3}=24\)
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}\)=\(\frac{\frac{8}{27}.\frac{9}{16}.-1}{\frac{4}{25}.\frac{-125}{1728}}\)=\(\frac{\frac{-1}{6}}{-\frac{5}{432}}\)=\(\frac{-1}{6}:\frac{-5}{432}=\frac{-1}{6}.-\frac{432}{5}=\frac{72}{5}\)
Bài này dễ mà bn
\(\left(\frac{3}{5}\right)^{2003}:\left(\frac{9}{25}\right)^{1000}\)
\(=\left(\frac{3}{5}\right)^{2003}:\left(\left(\frac{3}{5}\right)^2\right)^{1000}\)
\(=\left(\frac{3}{5}\right)^{2003}:\left(\frac{3}{5}\right)^{2000}\)
\(=\left(\frac{3}{5}\right)^3\)
\(=\frac{27}{125}\)
\(\left(\frac{3}{5}\right)^{2003}:\left(\frac{9}{25}\right)^{1000}\)
\(=\frac{3}{5}.\left[\left(\frac{3}{5}\right)^2\right]^{1000}:\left(\frac{9}{25}\right)^{1000}\)
\(=\frac{3}{5}.\left(\frac{9}{25}\right)^{1000}:\left(\frac{9}{25}\right)^{1000}\)
\(=\frac{3}{5}\)