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Dựa vào câu hỏi trên ta có dãy số 1+3+7+...........................+97+99
\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(C=\frac{3}{4}x\frac{4x2}{3x3}x\frac{3x5}{2x8}x...x\frac{99x101}{100x100}\)
\(C=...\) ( Tự làm tiếp )
\(E=1\frac{1}{3}x1\frac{1}{8}x1\frac{1}{15}x1\frac{1}{24}x...x1\frac{1}{99}\)
\(E=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x...x\frac{100}{99}\)
\(E=....\)( tương tự câu C )
a) \(49^{12}\)và \(5^{40}\)
\(49^{12}=\left(49^3\right)^4=\left(\left(7^2\right)^3\right)^4=\left(7^6\right)^4\)
\(5^{40}=\left(5^{10}\right)^4\)
\(7^6=\left(7^3\right)^2>\left(5^5\right)^2\)vì \(7^2\cdot7>5^3\cdot5^2\)
\(\Rightarrow49^{12}< 5^{40}\)
\(\left(-\frac{1}{16}\right)^{100}=\left(-\left(\frac{-1}{2}\right)^4\right)^{100}\)
\(=\left(-\frac{1}{2}\right)^{400}< \left(-\frac{1}{2}\right)^{500}\)
1-1/2+1/3-1/4+......-1/1000
=(1+1/3+1/5+......+1/999)-(1/2+1/4+.......+1/1000)
=(1+1/2+1/3+1/4+.....+1/1000)-2(1/2+1/4+.......+1/1000)
=(1+1/2+1/3+.........+1/1000)-(1+1/2+.....+1/500)
=1/501 +1/502+1/503+.....+1/1000 ;
mat khác:
500-500/501-501/502-.....-999/1000
=(1-500/501)+(1-501/502)+.....+(1-999/1000)=1/501+1/502+....+1/1000
=>D=1
\(\dfrac{1}{2^{500}}=\dfrac{1}{\left(2^5\right)^{100}}=\dfrac{1}{32^{100}}\\ \dfrac{1}{5^{200}}=\dfrac{1}{\left(5^2\right)^{100}}=\dfrac{1}{25^{100}}\)
mà `32^(100)>25^(100)`
nên \(\dfrac{1}{2^{500}}>\dfrac{1}{5^{200}}\)
Mk chỉ làm được phần f) thui
f) Ta có :
\(\left(-\frac{1}{16}\right)^{100}=\left(-\frac{1}{2^4}\right)^{100}=\left(-\frac{1}{2}\right)^{400}=\left(\frac{1}{-2}\right)^{400}\)
\(\left(-\frac{1}{2}\right)^{500}=\left(\frac{1}{-2}\right)^{500}\)
Vì \(\left(\frac{1}{-2}\right)^{400}>\left(\frac{1}{-2}\right)^{500}\)nên \(\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)
Ủng hộ mk nha !!! ^_^
Xin lỗi, mình chỉ làm được câu 1 thôi
\(A=\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5.111}{2.111}+\frac{4.1111}{11.1111}+\frac{3.11111}{22.11111}+\frac{11}{11.30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left[\left(\frac{5}{2 }+\frac{1}{30}+\frac{13}{60}\right)+\left(\frac{4}{11}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left[\left(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}\right)+\left(\frac{8}{22}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left(\frac{11}{4}+\frac{1}{2}\right)\)
\(A=\frac{1}{7}.\frac{13}{4}\)
\(A=\frac{13}{21}\)
Ta có:
\(E=\frac{500^{40}+1}{500^{41}+1}\Leftrightarrow10E=\frac{500^{41}+10}{500^{41}+1}=1+\frac{9}{500^{41}+1}\)
\(W=\frac{500^{39}+1}{500^{40}+1}\Leftrightarrow10W=\frac{500^{40}+10}{500^{40}+1}=1+\frac{9}{500^{40}+1}\)
Hay ta đang so sánh: \(E=\frac{9}{500^{41}};W=\frac{9}{500^{40}}\)
Vì \(500^{41}>500^{40}\)nên \(\frac{9}{500^{41}}< \frac{9}{500^{40}}\)hay \(\frac{500^{40}+1}{500^{41}+1}< \frac{500^{39}+1}{500^{40}+1}\).
Vậy \(E< W\)