\(\dfrac{6^3.7-6^3.12}{6^3.\left(5^3-5^2\right)}\)
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ND
Nguyễn Đăng Nhân
CTVHS
22 tháng 2 2023
a) \(\dfrac{\left(-3\right)^7\cdot2^8}{6^7}\)
\(=\dfrac{-1\cdot3^7\cdot2^8}{\left(2\cdot3\right)^7}=\dfrac{-1\cdot3^7\cdot2^7\cdot2}{2^7\cdot3^7}=-1\cdot2=-2\)
b) \(\dfrac{-3\cdot7^4+7^3}{7^5\cdot6-7^3\cdot2}\)
\(=\dfrac{-3\cdot7\cdot7^3+7^3}{7^3\cdot7^2\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\left(-3\cdot7+1\right)}{7^3\left(7^2\cdot6-2\right)}=\dfrac{-3\cdot7+1}{7^2\cdot6-2}\)
\(=\dfrac{-21+1}{294-2}=\dfrac{-20}{290}=\dfrac{-2}{29}\)
b) \(\dfrac{5^3\cdot3^5}{5^3\cdot0,5+125\cdot2\cdot5}\)
\(=\dfrac{5^3\cdot3^5}{5^3\cdot0,5+5^3\cdot2\cdot5}=\dfrac{5^3\cdot3^5}{5^3\left(0,5+2\cdot5\right)}\)
\(=\dfrac{3^5}{0,5+2\cdot5}=\dfrac{243}{10,5}=\dfrac{162}{7}\)
NT
1
17 tháng 7 2018
Ta có :
\(\dfrac{5\left(3.7^{15}-19.7^{14}\right)}{7^6+3.7^{15}}\)
\(=\dfrac{5.7^6\left(3.7^9-19.7^8\right)}{7^6\left(1+3.7^9\right)}\)
\(=\dfrac{5.7^8\left(3.7-19\right)}{1+3.7^9}\)
\(=\dfrac{5.7^8.2}{1+3.7^9}\)
\(=\dfrac{10.7^8}{1+3.7^8.7}\)
\(=\dfrac{10.7^8}{1+7^8.21}\)
27 tháng 8 2017
a)56+48=104
b)343-216-125=2
c)1296-32*27=1296-864=432
d)=0(vì các số *với 0 đều =0(\(2^4\)-4\(^2\)=0)
HL
27 tháng 8 2017
a) \(2^3.7+3^2.6=8.7+9.6\)
\(=56+54\)
\(=110\)
b) \(7^3-6^3-5^3=343-216-125\)
\(=2\)
c) \(6^4-2^5.3^3=1296-32.27\)
\(=1296-864\)
\(=432\)
d) \(\left(7^9-9^7\right)\left(6^8-8^6\right)\left(3^5-5^3\right)\left(2^4-4^2\right)\)
\(=\left(7^9-9^7\right)\left(6^8-8^6\right)\left(3^5-5^3\right).0\)
\(=0\)
NHỚ K CHO MÌNH NHÉ !
13 tháng 6 2023
\(A=\dfrac{\left(17+\dfrac{1}{4}-4-\dfrac{3}{16}-13-\dfrac{5}{6}\right)\cdot\left(-\dfrac{4}{7}\right)+\dfrac{27}{4}}{\left(5+\dfrac{2}{7}-5-\dfrac{1}{3}\right):\left(6+\dfrac{2}{3}-4-\dfrac{1}{2}\right)}\)
\(=\dfrac{\dfrac{37}{84}+\dfrac{27}{4}}{-\dfrac{1}{21}:\dfrac{13}{6}}=\dfrac{-1963}{6}\)
\(\dfrac{6^3.7-6^{13}.12}{6^3\left(5^3-5^2\right)}=\dfrac{6^3\left(7-12\right)}{5^3\left(125-25\right)}=\dfrac{-5}{100}=-\dfrac{1}{20}\)
\(=\dfrac{6^3\left(7-12\right)}{6^3\left(125-25\right)}=\dfrac{-5}{100}=-\dfrac{1}{20}\)