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a,32
b,\(-\frac{1}{10}\)
c,-1000000
d,\(\frac{9}{16}\)
Có \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)=\(\left(\dfrac{-5}{4}+\dfrac{2}{4}\right)^2\)=\(\left(\dfrac{-3}{4}\right)^2\)=\(\dfrac{\left(-3\right)^2}{4^2}=\dfrac{9}{16}\)
Có \(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,125.2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}=\dfrac{\left(0,3\right)^5}{\left(-0.3\right)^5.\left(0,01\right)^3}=\dfrac{1}{-1.\left(0,01\right)^3}=\dfrac{1}{-\left(0,01\right)^3}\)
b) \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{-11}{4}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{-11}{4}+\dfrac{2}{4}\right)^2\)
\(=\left(\dfrac{-9}{4}\right)^2\)
\(=\dfrac{81}{16}\)
\(=\dfrac{3^5}{-\left(0.003\right)^3\cdot0.09}=-10^{11}\)
\(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0.01\right)^3}=\dfrac{\left(0,125\right)^5.\left(-0,3\right)^5.\left(-8\right)^5}{\left(-0,3\right)^5.\left(0.01\right)^3}=\dfrac{\left(0,125\right)^5.\left(-8\right)^5}{\left(0.01\right)^3}=\dfrac{\left(-8.0,125\right)^5}{\left(0.01\right)^3}=\dfrac{-1}{0.000001}=-1000000\)
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