Tính giá trị của các biểu thức sau :
a) \(\dfrac{6^2.6^3}{3^5}\) b) \(\dfrac{25^2.4^2}{5^5.\left(-2\right)^5}\)
c) \(\dfrac{\left(0,125\right)^5.\left(2,4\right)^5}{\left(-0,3\right)^5.\left(0,01\right)^3}\)
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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) \(\left(-2\dfrac{3}{4}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{-11}{4}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{-11}{4}+\dfrac{2}{4}\right)^2\)
\(=\left(\dfrac{-9}{4}\right)^2\)
\(=\dfrac{81}{16}\)
\(1,\\ a,=\left(\dfrac{1}{4}\right)^3\cdot32=\dfrac{1}{64}\cdot32=\dfrac{1}{2}\\ b,=\left(\dfrac{1}{8}\right)^3\cdot512=\dfrac{1}{512}\cdot512=1\\ c,=\dfrac{2^6\cdot2^{10}}{2^{20}}=\dfrac{1}{2^4}=\dfrac{1}{16}\\ d,=\dfrac{3^{44}\cdot3^{17}}{3^{30}\cdot3^{30}}=3\\ 2,\\ a,A=\left|x-\dfrac{3}{4}\right|\ge0\\ A_{min}=0\Leftrightarrow x=\dfrac{3}{4}\\ b,B=1,5+\left|2-x\right|\ge1,5\\ A_{min}=1,5\Leftrightarrow x=2\\ c,A=\left|2x-\dfrac{1}{3}\right|+107\ge107\\ A_{min}=107\Leftrightarrow2x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{1}{6}\)
\(d,M=5\left|1-4x\right|-1\ge-1\\ M_{min}=-1\Leftrightarrow4x=1\Leftrightarrow x=\dfrac{1}{4}\\ 3,\\ a,C=-\left|x-2\right|\le0\\ C_{max}=0\Leftrightarrow x=2\\ b,D=1-\left|2x-3\right|\le1\\ D_{max}=1\Leftrightarrow x=\dfrac{3}{2}\\ c,D=-\left|x+\dfrac{5}{2}\right|\le0\\ D_{max}=0\Leftrightarrow x=-\dfrac{5}{2}\)
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