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GIẢI RA THÌ HƠI DÀI NHG KO SEO, MK WILL CỐ GẮNG ^_^
= ( 163 . 25 : 240 ) . ( 94 . 275 . 81 . 312 . 924 : 2718 ) . ( 1255 . 252 ) ( Chỗ nài nhiều số quá ko bt mk có bỏ sót ko )
= ( 212 . 25 : 240 ) . ( 38 . 315 . 34 . 312 . 348 : 354 ) . ( 515 . 54 )
= 212 + 5 - 40 . 38 + 15 + 4 + 12 + 48 - 54 . 515 + 4
= 2-23 . 333 . 519
Đó, chắc sai r bn đừng chép vào, nhiều số đâm ra mk hay bj lú
(y + \(\dfrac{1}{3}\)) + ( y + \(\dfrac{1}{9}\)) + ( y + \(\dfrac{1}{27}\)) + ( y + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)
( y + y + y + y ) + (\(\dfrac{1}{3}\)+ \(\dfrac{1}{9}\) + \(\dfrac{1}{27}\) + \(\dfrac{1}{81}\)) = \(\dfrac{56}{81}\)
4\(y\) + ( \(\dfrac{27}{81}\) + \(\dfrac{9}{81}\) + \(\dfrac{3}{27}\) + \(\dfrac{1}{81}\) ) = \(\dfrac{56}{81}\)
4y + \(\dfrac{40}{81}\) = \(\dfrac{56}{81}\)
4y = \(\dfrac{56}{81}\) - \(\dfrac{40}{81}\)
4y = \(\dfrac{16}{81}\)
y = \(\dfrac{16}{81}\) : 4
y = \(\dfrac{4}{81}\)
\(\left(y+\dfrac{1}{3}\right)+\left(y+\dfrac{1}{9}\right)+\left(y+\dfrac{1}{27}\right)+\left(y+\dfrac{1}{81}\right)=\dfrac{56}{81}\)
\(\Rightarrow y+\dfrac{1}{3}+y+\dfrac{1}{9}+y+\dfrac{1}{27}+y+\dfrac{1}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y+\dfrac{40}{81}=\dfrac{56}{81}\)
\(\Rightarrow4\times y=\dfrac{56}{81}-\dfrac{40}{81}\)
\(\Rightarrow4\times y=\dfrac{16}{81}\)
\(\Rightarrow y=\dfrac{16}{81}:4\)
\(\Rightarrow y=\dfrac{4}{81}\)
Ta có \(9.27^2.81^3=3^2.\left(3^3\right)^2.\left(3^4\right)^3=3^2.3^{3.2}.3^{4.3}=3^2.3^6.3^{12}=3^{2+6+12}=3^{20}\)
9 × 272 × 813
= 9 x 729 x 531441
= 6561 x 531441
= 3486784401
\(a,81^3=\left(9^2\right)^3=9^6\)
Vì \(9^{27}>9^6\) nên \(9^{27}>81^3\)
\(b,5^{14}=\left(5^2\right)^7=25^7\)
Vì \(25^7< 27^7\) nên \(5^{14}< 27^7\)
\(c,10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(1000^{10}< 1024^{10}\) nên \(10^{30}< 2^{100}\)