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\(\left(\dfrac{1}{2}\right)^{50}=\left[\left(\dfrac{1}{2}\right)^5\right]^{10}=\left(\dfrac{1}{32}\right)^{10}\)
1/12>1/32
=>(1/12)^10>(1/32)^10
=>(1/12)^10>(1/2)^50
Có: \(\left(\dfrac{1}{12}\right)^{10}=\dfrac{1}{12^{10}}\)
\(\left(\dfrac{1}{2}\right)^{50}=\dfrac{1}{2^{50}}=\dfrac{1}{\left(2^5\right)^{10}}=\dfrac{1}{32^{10}}\)
Do \(12< 32\Rightarrow12^{10}< 32^{10}\)
\(\Rightarrow\dfrac{1}{12^{10}}>\dfrac{1}{32^{10}}\) hay \(\left(\dfrac{1}{12}\right)^{10}>\left(\dfrac{1}{2}\right)^{50}\)
a) Ta có: \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
Mà \(\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\Rightarrow m=5\)
b)Ta có: \(\frac{343}{125}=\left(\frac{7}{5}\right)^3\)
Mà \(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\Rightarrow n=3\)
\(a)\) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Leftrightarrow\)\(\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(\Leftrightarrow\)\(\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(\Leftrightarrow\)\(m=5\)
Vậy \(m=5\)
\(b)\) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\)\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\)\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\)\(n=3\)
Vậy \(n=3\)
Chúc bạn học tốt ~
Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề của bạn hơn.
\(\left(\dfrac{3}{5}\right)^{10}\cdot\left(\dfrac{5}{3}\right)^{10}-\dfrac{13^4}{39^4}+2014^0\)
=1-(1/3)^4+1
=4/3-1/81
=107/81
(1/2)^m = 1/32
mà 1/32 = (1/2)^5 nên m = 5
343/125= (7/5)^n
mà 343/125 = (7/5)^3 nên n=3
(\(\frac{1}{5}\))2 .n = (\(\frac{1}{125}\))3 - n
<=> \(\frac{1}{25}\)n +n = \(\frac{1}{5^9}\)
<=> \(\frac{26}{25}\)n = \(\frac{1}{5^9}\)
<=> n = \(\frac{1}{5^9}\): \(\frac{26}{25}\)= \(\frac{1}{2031250}\)
ta có:\(\left(\dfrac{-1}{16}\right)^{10}=\left(\dfrac{1}{16}\right)^{10}=\left(\dfrac{1^4}{2^4}\right)^{10}=\left[\left(\dfrac{1}{2}\right)^4\right]^{10}=\left(\dfrac{1}{2}\right)^{40}=\dfrac{1^{40}}{12^{40}}=\dfrac{1}{2^{40}}\)
ta có:
\(\left(\dfrac{-1}{2}\right)^{500}=\left(\dfrac{1}{2}\right)^{500}=\dfrac{1^{500}}{2^{500}}=\dfrac{1}{2^{500}}\)
Vì 40<500
⇒2\(^{40}< 2^{500}\)
⇒\(\dfrac{1}{2^{40}}>\dfrac{1}{2^{500}}\)
⇒\(\left(\dfrac{-1}{16}\right)^{10}>\left(\dfrac{-1}{2}\right)^{500}\)
Vậy \(\left(\dfrac{-1}{16}\right)^{10}>\left(\dfrac{-1}{2}\right)^{500}\)
\(+,\left(\dfrac{-1}{16}\right)^{10}=\left(\dfrac{\left(-1\right)^4}{2^4}\right)^{10}=\left[\left(\dfrac{-1}{2}\right)^4\right]^{10}=\left(\dfrac{-1}{2}\right)^{40}\)
Vì 40<500→\(\left(\dfrac{-1}{2}\right)^{40}< \left(\dfrac{-1}{2}\right)^{500}hay\left(\dfrac{-1}{16}\right)^{10}< \left(\dfrac{-1}{2}\right)^{500}\)
\(\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{10\cdot4}=\left(\frac{1}{16}\right)^{10}\)
Mà ta có
\(\left(\frac{1}{32}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{32}\right)^{10}\)