tính :
\(\dfrac{12^{10}\cdot35+2^{10}\cdot65+6^2\cdot12^6\cdot15^2}{2^8\cdot10^4}\)
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NL
Nguyễn Lê Phước Thịnh
CTVHS
15 tháng 3
Ta có: \(5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2\)
\(=5^2\cdot2^{11}\cdot3^{11}\cdot2^8+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot5^2\cdot3^2\)
\(=2^{19}\cdot3^{11}\cdot5^2+2^{14}\cdot3^{10}\cdot5^2\)
\(=2^{14}\cdot3^{10}\cdot5^2\left(2^5\cdot3+1\right)=2^{14}\cdot3^{10}\cdot5^2\cdot97\)
Ta có: \(2\cdot6^{12}\cdot10^4-81^2\cdot960^3\)
\(=2\cdot2^{12}\cdot3^{12}\cdot2^4\cdot5^4-3^8\cdot\left(2^6\cdot3\cdot5\right)^3\)
\(=2^{17}\cdot3^{12}\cdot5^4-2^{18}\cdot3^{11}\cdot5^3\)
\(=2^{17}\cdot3^{11}\cdot5^3\left(3\cdot5-2\right)=2^{17}\cdot3^{11}\cdot5^3\cdot13\)
Ta có: \(\frac{5^2\cdot6^{11}\cdot16^2+6^2\cdot12^6\cdot15^2}{2\cdot6^{12}\cdot10^4-81^2\cdot960^3}\)
\(=\frac{2^{14}\cdot3^{10}\cdot5^2\cdot97}{2^{17}\cdot3^{11}\cdot5^3\cdot13}=\frac{97}{13\cdot2^3\cdot3\cdot5}=\frac{97}{1560}\)
26 tháng 4 2018
\(\dfrac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot45}\)
=\(\dfrac{3+6+12+21\cdot35}{14+28+7\cdot45}\)
=\(\dfrac{450}{119}\)
Vì \(\dfrac{450}{119}>1\) mà \(1>\dfrac{303}{708}\)
\(\Rightarrow\)\(\dfrac{450}{119}>\dfrac{303}{708}\)
HM
27 tháng 4 2017
A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=2.2.2.2.2
A=32
HH
27 tháng 4 2017
\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)
\(2\cdot2\cdot2\cdot2\cdot2=2^5\)
\(=32\)
TA
20 tháng 2 2018
\(\frac{5^2.3^{11}.2^{11}.2^8+2^2.3^2.3^2.5^2}{2.2^{12}.3^{12}.2^4.5^4-3^8.2^{18}.3^3.5^3}\)
\(=\frac{5^2.3^{11}.2^{19}+2^2.3^4.5^2}{2^{17}.3^{12}.5^4-3^{11}.2^{18}.5^3}\)
\(=\frac{5^2.3^4.2^2.\left(3^7.2^{17}+1\right)}{2^{17}.3^{11}.5^3.\left(3.5-2\right)}=\frac{3^7.2^{17}}{2^{15}.3^7.5}+\frac{1}{2^{15}.3^7.5}=\frac{4}{5}+\frac{1}{2^{15}.3^7.5}\)
Ta có: \(12^{10}\cdot35+2^{10}\cdot65+6^2\cdot12^6\cdot15^2\)
\(=2^{20}\cdot3^{10}\cdot5\cdot7+2^{10}\cdot5\cdot13+2^2\cdot3^2\cdot2^{12}\cdot3^6\cdot3^2\cdot5^2\)
\(=2^{20}\cdot3^{10}\cdot5\cdot7+2^{10}\cdot5\cdot13+2^{14}\cdot3^{10}\cdot5^2\)
\(=2^{10}\cdot5\left(3^{10}\cdot7+13+2^4\cdot5\right)\)
Ta có: \(\frac{12^{10}\cdot35+2^{10}\cdot65+6^2\cdot12^6\cdot15^2}{2^8\cdot10^4}\)
\(=\frac{2^{10}\cdot5\left(3^{10}\cdot7+13+2^4\cdot5\right)}{2^8\cdot2^4\cdot5^4}=\frac{2^{10}\cdot5\cdot413436}{2^{12}\cdot5^4}=\frac{413436}{2^2\cdot5^3}=\frac{413436}{500}=\frac{103359}{125}\)