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\(C=\frac{1.5.3.2+2.10.6.2+4.20.12.2+...+9.45.27.2}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(C=\frac{2\left(1.5.3+2.10.6+4.20.12+...+9.45.27\right)}{1.3.5+2.6.10+4.12.20+...+9.27.45}\) = 2
\(A=\frac{8056}{2012.16-1982}\)= \(\frac{2014.4}{2012.15+2012-1982}\)=\(\frac{2014.4}{2012.15+30}\)=\(\frac{2014.4}{2012.15+2.15}\)=\(\frac{2014.4}{15.\left(2012+2\right)}=\frac{2014.4}{15.2014}=\frac{4}{15}\)
B = \(\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
= \(\frac{1.2.3.2+2.2.2.12+4.4.2.24+7.7.2.42}{1.2.3.9+2.12.2.9+4.24.4.9+7.42.7.9}\)
= \(\frac{2\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}{9\left(1.2.3+2.2.12+4.4.24+7.7.42\right)}\)
= \(\frac{2}{9}\)
Ta có: \(\frac{4}{15}=\frac{4.3}{15.3}=\frac{12}{45};\frac{2}{9}=\frac{2.5}{9.5}=\frac{10}{45}\)
Vì \(\frac{12}{45}>\frac{10}{45}\Rightarrow\frac{4}{15}>\frac{2}{9}\Rightarrow A>B\)
Vậy A > B
\(A=\dfrac{8056}{2012.16-1982}\)
\(A=\dfrac{8056}{32192-1982}\)
\(A=\dfrac{8056}{30210}=\dfrac{12}{45}\)
\(B=\dfrac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)
\(B=\dfrac{12+96+768+4116}{54+432+3456+18522}\)
\(B=\dfrac{4992}{22464}=\dfrac{10}{45}\)
Vậy: \(\dfrac{12}{45}>\dfrac{10}{45}\Rightarrow A>B\)
\(\dfrac{1\cdot2\cdot3+2\cdot4\cdot6+4\cdot8\cdot12}{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20}\\ =\dfrac{1\cdot2\cdot3+2\cdot1\cdot2\cdot2\cdot2\cdot3+4\cdot1\cdot4\cdot2\cdot4\cdot3}{1\cdot3\cdot5+2\cdot1\cdot2\cdot3\cdot2\cdot5+4\cdot1\cdot4\cdot3\cdot4\cdot5}\\ =\dfrac{1\cdot2\cdot3\cdot\left(1+2^3+4^3\right)}{1\cdot3\cdot5\cdot\left(1+2^3+4^3\right)}\\ =\dfrac{1\cdot2\cdot3}{1\cdot3\cdot5}\\ =\dfrac{6}{15}\)
H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
đề là tính nhanh nhé
\(2\cdot31\cdot12+4\cdot6\cdot42+8\cdot27\cdot3\)
\(=\left(2\cdot12\right)\cdot31+\left(4\cdot6\right)\cdot42+\left(8\cdot3\right)\cdot27\)
\(=24\cdot31+24\cdot42+24\cdot27\)
\(=24\cdot\left(31+42+27\right)\)
\(=24\cdot100=2400\)