BT5: Tính giá trị biểu thức:
a) \(\dfrac{25^2\cdot4^2}{5^5\cdot\left(-2\right)^5}=\)
b) \(\dfrac{\left(0,125\right)^5\cdot\left(2,4\right)^5}{\left(-0,3\right)^5\cdot\left(0,01\right)^3}=\)
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\(D=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right)...\left(1+\dfrac{1}{2019.2021}\right)=\dfrac{4}{1.3}.\dfrac{9}{2.4}...\dfrac{2019.2021+1}{2019.2021}=\dfrac{2.2}{1.3}.\dfrac{3.3}{2.4}...\dfrac{2020.2020}{2019.2021}=\left(\dfrac{2}{1}.\dfrac{3}{2}...\dfrac{2020}{2019}\right).\left(\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{2020}{2021}\right)=2020.\dfrac{2}{2021}=\dfrac{4040}{2021}\)
a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=2^5\)
b) \(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{5^4.2^4}{5^5.\left(-2\right)^5}=\dfrac{1.1}{5.\left(-2\right)}=\dfrac{1}{-10}=\dfrac{-1}{10}\)
( Mình giải được có 2 bài thôi. ) :)
a) \(\dfrac{6^2.6^3}{3^5}=\dfrac{6^5}{3^5}=3^5\)
b)\(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}=\dfrac{\left(25.4\right)^2}{\left(5.\left(-2\right)\right)^5}=\dfrac{\left(100\right)^2}{\left(-10\right)^5}=\dfrac{1}{-10}\)
Còn phần c mình cx đang tắc
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\)\(\frac{\frac{1}{6}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{5^3}{2^6.3^3}.\left(-1\right)}=\frac{\frac{1}{2.3}}{\frac{5}{2^4.3^3}}=\frac{2^3.3^2}{5}=\frac{72}{5}\)
a/ \(\frac{25^2\times4^2}{5^5\times\left(-2\right)^5}\)= \(\frac{625\times16}{3125\times\left(-32\right)}\)=\(\frac{10000}{-10000}\)= -1