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\(C=\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x...x\frac{9999}{10000}\)
\(C=\frac{3}{4}x\frac{4x2}{3x3}x\frac{3x5}{2x8}x...x\frac{99x101}{100x100}\)
\(C=...\) ( 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 )
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 !!! ^_^
Dựa vào câu hỏi trên ta có dãy số 1+3+7+...........................+97+99
Xin lỗi, mình chỉ làm được câu 1 thôi
\(A=\frac{1}{7}\left(\frac{555}{222}+\frac{4444}{12221}+\frac{33333}{244442}+\frac{11}{330}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5.111}{2.111}+\frac{4.1111}{11.1111}+\frac{3.11111}{22.11111}+\frac{11}{11.30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left(\frac{5}{2}+\frac{4}{11}+\frac{3}{22}+\frac{1}{30}+\frac{13}{60}\right)\)
\(A=\frac{1}{7}\left[\left(\frac{5}{2 }+\frac{1}{30}+\frac{13}{60}\right)+\left(\frac{4}{11}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left[\left(\frac{150}{60}+\frac{2}{60}+\frac{13}{60}\right)+\left(\frac{8}{22}+\frac{3}{22}\right)\right]\)
\(A=\frac{1}{7}\left(\frac{11}{4}+\frac{1}{2}\right)\)
\(A=\frac{1}{7}.\frac{13}{4}\)
\(A=\frac{13}{21}\)
\(B=\frac{23^{41}+1}{23^{42}+1}\)
Vì B < 1
\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)
P/s: Hoq chắc
ta có
\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)
\(\Rightarrow B< A\)
1.
\(3^{500}=\left(3^5\right)^{100}\)
\(7^{300}=\left(7^3\right)^{100}\)
\(3^5< 7^3\Leftrightarrow3^{500}< 7^{300}\)
\(3^{500}=\left(3^5\right)^{100}\)
\(7^{300}=\left(7^3\right)^{100}\)
35 < 73 => 3500 <7300
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 \(E< W\)