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a) Ta có :
290 = 29.10
536 = 59.4
So sánh tiếp 29.10 và 59.4, ta có :
29.10 = (210)9
59.4 = (54)9
So sánh tiếp 210 và 54
210 = 22.5
54 = 52.2
So sánh tiếp 25 và 52
25 = 32
52 = 25
Vì 32 > 25 nên 290 > 536
b) Ta có :
227 = 23.9 = (23)9 = 89
318 = 32.9 = (32)9 = 99
\(\begin{array}{l}0,49 = {\left( {0,7} \right)^2};\\\,\frac{1}{{32}} =\frac{1^5}{2^5}={\left( {\frac{1}{2}} \right)^5};\\\,\frac{{ - 8}}{{125}} =\frac{(-2)^3}{5^3}= {\left( {\frac{{ - 2}}{5}} \right)^3};\end{array}\)
\(\frac{{16}}{{81}} =\frac{4^2}{9^2}= {\left( {\frac{4}{9}} \right)^2} (hoặc \,\frac{{16}}{{81}} =\frac{2^4}{3^4}= {\left( {\frac{2}{3}} \right)^4});\\\,\frac{{121}}{{169}} =\frac{11^2}{13^2}= {\left( {\frac{{11}}{{13}}} \right)^2}\)
a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)
b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)
c) \(27^{40}=3^{3^{40}}=3^{120}\)
\(64^{60}=8^{2^{60}}=8^{120}\)
Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)
con ai kooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)
b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)
c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)
\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)
\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)
\(27\cdot5^3\cdot3^3\cdot32^{-1}:125=3^3\cdot5^3\cdot3^3\cdot\dfrac{1}{32}:5^3=\dfrac{3^6}{32}=\dfrac{3^6}{2^5}\)
a) Có: 290=25.18=(25)18=3218
536=52.18=2518
mà 32>25=> 3518>2518<=> 290> 536
b)Có 227=23.9=89
318=32.9=99
c) Có (x-1)5=-32
<=> (x-1)5=-25
<=> x-1=-2
<=> x=-1
bài 1 :
Ta có : 290 = 25.18 = ( 25 )18 = 3218
\(5^{36}=5^{2.18}=\left(5^2\right)^{18}=25^{18}\)
Vì 32 > 25 => \(32^{18}>25^{18}\) hay \(2^{90}>5^{36}\)
Bài 2 :
\(2^{27}=2^{3.9}=\left(2^3\right)^9=8^9\)
\(3^{18}=3^{2.9}=\left(3^2\right)^9=9^9\)
Bài 3 :
\(\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=-2^5\)
\(x-1=-2\)
\(x=-2+1\)
\(x=-1\)