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9 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{21^2\cdot14\cdot125}{35^2\cdot125}\)
`=`\(\dfrac{3^2\cdot7^2\cdot2\cdot7\cdot5^2}{5^2\cdot7^2\cdot5^2}\)
`=`\(\dfrac{3^2\cdot2\cdot7\cdot5^2\cdot7^2}{5^2\cdot5^2\cdot7^2}\)
`=`\(\dfrac{3^2\cdot2\cdot7}{5^2}=\dfrac{126}{25}\)
30 tháng 10 2018
\(A=\frac{4^6.3^4.9^5}{6^{12}}=\frac{\left(2^2\right)^6.3^3.\left(3^2\right)^5}{6^{12}}\)
\(=\frac{2^{12}.3^3.3^{10}}{6^{12}}=3^{13}.3^{12}=3^{25}\)
30 tháng 10 2018
\(A=\frac{4^6.3^4.9^5}{6^{12}}\)
\(A=\frac{2^6.2^6.3^4.3^5.3^5}{2^{12}.3^{12}}\)
\(A=\frac{3^3.3^5}{1}\)
\(A=3^8\)
\(B=\frac{21^2.14.125}{35^3.6}\)
\(B=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}\)
\(B=\frac{3.1.1.1.1}{1.1.1.1}\)
\(B=3\)
2 tháng 7 2017
\(\dfrac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)
\(=\dfrac{1.3.5+2^3.1.3.5+2^6.1.3.5+7^3.1.3.5}{1.5.7+2^3.1.5.7+2^6.1.5.7+7^3.1.5.7}\)
\(=\dfrac{1.3.5\left(1+2^3+2^6+7^3\right)}{1.5.7\left(1+2^3+2^6+7^3\right)}\)
\(=\dfrac{1.3.5}{1.5.7}\)
QD
19 tháng 2 2017
\(\frac{2.6.10+6.10.14+10.14.18+...+194.198.202}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3.1.3.5+2^3.3.5.7+2^3.97.99.101}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3\left(1.3.5+3.5.7+...+97.99.101\right)}{1.3.5+3.5.7+...+97.99.101}\)
\(=\frac{2^3}{1}=8\)
Vậy A = 8
HJ
7 tháng 8 2017
\(a,\left(10\frac{2}{9}.2\frac{3}{5}\right)-6\frac{2}{9}=\frac{1196}{45}-\frac{56}{9}=\frac{1196}{45}-\frac{280}{45}=\frac{916}{45}\)
\(b,\frac{6}{7}+\frac{1}{7}.\frac{2}{7}+\frac{1}{7}.\frac{5}{7}=\frac{1}{7}\left(6+\frac{2}{7}+\frac{5}{7}\right)=\frac{1}{7}.7=1\)
\(c,3.136.8+4.14.6-14.150=3264+336-2100=1500\)
\(d,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)\(=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)
\(e,\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
S
12 tháng 3 2017
\(A=\frac{8056}{2012.16-1982}\)= \(\frac{2014.4}{2012.15+2012-1982}\)=\(\frac{2014.4}{2012.15+30}\)=\(\frac{2014.4}{2012.15+2.15}\)=\(\frac{2014.4}{15.\left(2012+2\right)}=\frac{2014.4}{15.2014}=\frac{4}{15}\)
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
Ta có :
\(\frac{21^2.14.125}{35^3.6}=\frac{3^2.7^2.2.7.5^3}{5^3.7^3.2.3}=\frac{2.3^2.5^3.7^3}{2.3.5^3.7^3}=\frac{3}{1}=3\)
Vậy \(\frac{21^2.14.125}{35^3.6}=3\)
\(\frac{21^2.14.125}{35^3.6}\)= \(\frac{21^2.2.7.125}{42875.2.3}\)= \(\frac{21^2.7.125}{125.343.3}\)= \(\frac{21^2.7.125}{125.7.49.3}\)= \(\frac{21^2}{49.3}\)= \(\frac{441}{147}\)
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