tính nhanh
a)\(\frac{12\cdot4+12\cdot6}{24}\) b)\(\frac{16\cdot8-16\cdot2}{12\cdot4}\)
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tử số : 2.4 + 4.8 + 8.12 + 12.16 + 16.20
= 2.(1.2+2.4+4.6+6.8+8.10)
ta được 2. A=( 1.2+2.4+4.6+6.8+8.10) / ( 1.2+2.4+4.6+6.8+8.10)
=> A=2
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
\(B=\frac{18^6\cdot2^{12}\cdot4^3\cdot9^3}{16^3\cdot6^9\cdot27^3}\)
\(=>B=\frac{\left(3^2\cdot2\right)^6\cdot2^{12}\cdot\left(2^2\right)^3\cdot\left(3^2\right)^3}{\left(2^4\right)^3\cdot\left(2\cdot3\right)^9\cdot\left(3^3\right)^3}\)
\(=>B=\frac{3^{12}\cdot2^6\cdot2^{12}\cdot2^6\cdot3^6}{2^{12}\cdot2^9\cdot3^9\cdot3^9}\)
\(=>B=\frac{\left(3^{12}\cdot3^6\right)\cdot\left(2^6\cdot2^{12}\cdot2^6\right)}{\left(2^{12}\cdot2^9\right)\cdot\left(3^9\cdot3^9\right)}\)
\(=>B=\frac{3^{18}\cdot2^{24}}{2^{21}\cdot3^{18}}\)
\(=>B=\frac{2^{24}}{2^{21}}\)
\(=>B=2^{24-21}\)
\(=>B=2^3\)
\(=>B=8\)
\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1.2}{3.4}\)
\(A=\frac{1}{6}\)
Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy B > A
Theo đề bài, ta có:
\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)
\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1\times2}{3\times4}\)
\(A=\frac{1}{6}\)
Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy \(B>A\)
a.\(\frac{3\cdot4\cdot7}{12\cdot8\cdot9}\)= \(\frac{3\cdot4\cdot7}{3\cdot4\cdot8\cdot9}\)= \(\frac{7}{72}\)
b. \(\frac{4\cdot5\cdot6}{12\cdot10\cdot8}\)= \(\frac{4\cdot5\cdot2\cdot3}{3\cdot4\cdot5\cdot2\cdot8}\)= \(\frac{1}{8}\)
c.\(\frac{5\cdot6\cdot7}{12\cdot14\cdot15}\)= \(\frac{5\cdot6\cdot7}{2\cdot6\cdot2\cdot7\cdot3\cdot5}\)= \(\frac{1}{12}\)
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