8^5.8^2+8^3
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a) \(110 – 7^2 + 22:2\) = \(110 – 49 + 11 = 61 + 11 = 72\)
b) \(9.(8^2 – 15)= 9.(64 – 15 ) = 9.49 = 441\)
c) \(5.8 – (17 + 8):5 = 40 – 25:5 = 40 – 5 = 35\)
d) \( 75:3 + 6.9^2= 25 + 6.81 = 25 + 486 = 511\)
\(1.\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
\(2.\dfrac{8^5.3^{15}}{2^{14}.81^4}=\dfrac{2^{15}.3^{15}}{2^{14}.3^{16}}=\dfrac{2}{3}\)
Ý 1:
\(\dfrac{27^4.4^3}{9^5.8^2}=\dfrac{\left(3^3\right)^4.\left(2^2\right)^3}{\left(3^2\right)^5.\left(2^3\right)^2}=\dfrac{3^{12}.2^6}{3^{10}.2^6}=3^2=9\)
Ý 2:
\(\dfrac{8^5.3^{15}}{2^{14}.81^4}=\dfrac{\left(2^3\right)^5.3^{15}}{2^{14}.\left(3^4\right)^4}=\dfrac{2^{15}.3^{15}}{2^{14}.3^{16}}=\dfrac{2^{14}.2.3^{15}}{2^{14}.3^{15}.3}=\dfrac{2}{3}\)
a)2/13 . 5 - 9/11 . 2/13 - 7/11 . 2/13
=2/13.(5-9/11-7/11)
=2/13.39/11
=6/11.
b)(-1/2)2 + 4,25
=1/4+4,25
=1/4+17/4
=9/2.
c)(-2/3)2 : ( --2 2/3) -- ( 5/8 -- 5/6 ) + 5.8--45/5.8
= -4/3 : -8/3 - -5/24 + 40 - 72
= -751/24.
a: \(=\left(1.25\right)^{16}\cdot8^{16}\cdot8=8\cdot10^{16}\)
b: \(=\left(\dfrac{5}{2}\right)^{13}\cdot4^{13}\cdot4^2=10^{13}\cdot4^2\)
c: \(=\left(0.25\right)^4\cdot8^4\cdot8^2=2^4\cdot8^2=64\cdot16=1024\)
d: \(=\left(\dfrac{1}{2}\right)^{15}\cdot2^{18}=2^3=8\)
e: \(=\left(\dfrac{1}{3}\cdot6\right)^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=2^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=\dfrac{1}{2}\)
a: ta có: \(5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5\)
\(=5\cdot\left(2^3\right)^{12}\cdot\left(3^2\right)^5\cdot3^2\cdot\left(-1\right)\cdot2^{20}\)
\(=-5\cdot3^{12}\cdot2^{56}\)
Ta có: \(7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5\)
\(=-7\cdot\left(7^2\right)^3\cdot\left(3^4\right)^5\cdot7^{10}\cdot\left(2^3\right)^5\)
\(=-7\cdot7^6\cdot3^{20}\cdot7^{10}\cdot2^{15}=-7^{17}\cdot2^{15}\cdot3^{20}\)
Ta có: \(B=\frac{5\cdot8^{12}\cdot9^5\cdot\left(-3\right)^2\cdot\left(-16\right)^5}{7\cdot49^3\cdot81^5\cdot\left(-7\right)^{10}\cdot\left(-8\right)^5}\)
\(=\frac{-5\cdot3^{12}\cdot2^{56}}{-7^{17}\cdot2^{15}\cdot3^{20}}=\frac{5\cdot2^{41}}{7^{17}\cdot3^8}\)
b: \(13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4\)
\(=13\cdot2^{14}\cdot3^{20}-3^{14}\cdot2^{16}\)
\(=3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)\)
\(15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}\)
\(=3\cdot5\cdot3^{12}\cdot2^8+3\cdot\left(3^3\right)^5\cdot\left(2^3\right)^{10}=3^{13}\cdot5\cdot2^8+3^{16}\cdot2^{30}\)
\(=3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)\)
Ta có: \(C=\frac{13\cdot4^7\cdot9^{15}-9^7\cdot\left(-16\right)^4}{15\cdot3^{12}\cdot2^8+3\cdot27^5\cdot\left(-8\right)^{10}}\)
\(=\frac{3^{14}\cdot2^{14}\left(13\cdot3^6-2^2\right)}{3^{13}\cdot2^8\left(5+3^3\cdot2^{22}\right)}=\frac{3\cdot2^6\cdot\left(13\cdot3^6-2^2\right)}{5+3^3\cdot2^{22}}\)
\(8^5.8^2+8^3=8^7+8^3\)