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\(\)Đặt \(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}...+\frac{1}{205}}{\frac{204}{1}+\frac{203}{2}+\frac{202}{3}+...+\frac{1}{204}}=\frac{B}{C}\)
Biến đổi C:
\(C=\left(\frac{204}{1}+1\right)+\left(\frac{203}{2}+1\right)+\left(\frac{202}{3}+1\right)+...+\left(\frac{1}{204}+1\right)-204\)
\(=205+\frac{205}{2}+\frac{205}{3}+..+\frac{205}{204}+\frac{205}{205}-205\)
\(=205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}}{205.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{205}\right)}=\frac{1}{205}\)
a= 1/2 + 1/4 + 1/8 - 1 x 1 + 8/1 - 4/1 - 2/1=\(1\frac{7}{8}\)=1,875
b=3/1 - 6/3 - 9/6 - 369/1 : 1/3 + 3/6 + 6/9 - 1/963 \(\approx\)186,665628245067
c=1/1 - 1/2 + 3/1 - 1/4 + 5/1 - 1/6 + 7/1 - 1/8 + 9/1 - 1/10=\(\approx\)23,8583333333333
vậy a>b>c
**************************l i k e***********************************8
A = \(\left(-\frac{1}{8}\right)\times\left(-13\right)=\frac{13}{8}\) => 0 < A < 2
B: Tử âm ; mẫu dương => B < 0
C = \(\left(\frac{1}{1}+\frac{3}{1}+\frac{5}{1}+\frac{7}{1}+\frac{9}{1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
= 25 \(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}\right)\)
Dễ có: B < A < C
\(A=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}+\frac{1}{8}.\frac{1}{9}\)
\(A=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\)
\(A=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+\frac{7-6}{6.7}+\frac{8-7}{7.8}+\frac{9-8}{8.9}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(A=\frac{1}{2}-\frac{1}{9}\)
\(A=\frac{7}{18}\)
Vậy \(A=\frac{7}{18}\)
A = 1/2.3 + 1/3.4 + ..... +1/8.9
= 1/2 - 1/3 + 1/3 - 1/4 + ........ + 1/8 - 1/9
= 1/2 - 1/9
= 7/18
Tk mk nha
b; \(3\frac{1}{4}.x-1\frac{1}{6}.x-1\frac{2}{3}=\frac{5}{12}\)
\(\Rightarrow x.\left(3\frac{1}{4}-1\frac{1}{6}-1\frac{2}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{13}{4}-\frac{7}{6}-\frac{5}{3}\right)=\frac{5}{12}\)
\(\Rightarrow x.\left(\frac{39}{12}-\frac{14}{12}-\frac{20}{12}\right)=\frac{5}{12}\)
\(\Rightarrow x.\frac{5}{12}=\frac{5}{12}\)
\(\Rightarrow x=\frac{5}{12}\div\frac{5}{12}=1\)
Vậy x=1
\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)
\(9A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)
\(9A+A=\left(\frac{1}{3}-\frac{1}{3^{^2}}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(10A=\frac{1}{3}-\frac{1}{3^{204}}\)
A = (1/3 - 1/3204) : 10
Vậy A = (1/3 - 1/3204) : 10.
A= \(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\left(1\right)\\ \)
\(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)
\(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\left(2\right)\)
Từ (1) và (2) vế theo vế ta có :\(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)
\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{206}}\right):\frac{8}{9}\)