\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2013}\right).\left(1-\frac{2}{2015}\right)\)
Những câu hỏi liên quan
Giair giúp mik nha,cảm ơn các bạn nhìu.
Tính giá trị của các biểu thức sau:
\(C=\left(1+\frac{2}{3}\right).\left(1+\frac{2}{5}\right).\left(1+\frac{2}{7}\right)........\left(1+\frac{2}{2013}\right).\left(1+\frac{2}{2015}\right)\)
\(N=\frac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^6}{5.2^{28}3^{19}-7.2^{29}.3^{18}}\)
\(D=\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^5+\left(\frac{1}{2}\right)^7+......+\left(\frac{1}{2}\right)^{2015}\)
A = \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right).\)
B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{6}}{\frac{2015}{1}+\frac{2014}{2}+\frac{2013}{3}+...+\frac{1}{2015}}\)
đề là tính các bạn ạ. Mình xin lỗi vì quên ko ghi đề.
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\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2017^2}\right)\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2016.2018}{2017^2}\)
\(=\frac{\left(1.2.3...2016\right)\left(3.4.5...2018\right)}{\left(2.3.4...2017\right)\left(2.3.4...2017\right)}\)
\(=\frac{2018}{2017.2}=\frac{1009}{2017}\)
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\(A=\frac{\left(1-2\right).\left(1+2\right)}{2^2}.\frac{\left(1-3\right).\left(1+3\right)}{3^2}.......\frac{\left(1-2013\right).\left(1+2013\right)}{2013^2}.\frac{\left(1-2014\right).\left(1+2014\right)}{2014^2}\)
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
Chứng tỏ:a) frac{2013}{2014}+frac{2014}{2015}+frac{2015}{2013}3b) left(1+frac{1}{2}right)left(1+frac{1}{2^2}right)left(1+frac{1}{2^3}right)left(1+frac{1}{2^4}right)....left(1+frac{1}{2^{50}}right) 3c) Cfrac{1}{2}.frac{3}{4}.frac{5}{6}.....frac{9999}{10000} frac{1}{100}d) frac{1}{2}-frac{1}{2^2}+.............+frac{1}{2^{99}}-frac{1}{2^{100}} frac{1}{3}
Đọc tiếp
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)....\left(1+\frac{1}{2^{50}}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.............+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
\(a)\) Đặt \(A=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}\) ta có :
\(A=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2013+2}{2013}\)
\(A=\frac{2014}{2014}-\frac{1}{2014}+\frac{2015}{2015}-\frac{1}{2015}+\frac{2013}{2013}+\frac{2}{2013}\)
\(A=1-\frac{1}{2014}+1-\frac{1}{2015}+1+\frac{2}{2013}\)
\(A=\left(1+1+1\right)-\left(\frac{1}{2014}+\frac{1}{2015}-\frac{2}{2013}\right)\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\left(\frac{1}{2013}+\frac{1}{2013}\right)\right]\)
\(A=3-\left[\frac{1}{2014}+\frac{1}{2015}-\frac{1}{2013}-\frac{1}{2013}\right]\)
\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]\)
Mà :
\(\frac{1}{2014}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2014}-\frac{1}{2013}< 0\)
\(\frac{1}{2015}< \frac{1}{2013}\)\(\Rightarrow\)\(\frac{1}{2015}-\frac{1}{2013}< 0\)
Từ (1) và (2) suy ra : \(\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)< 0\) ( cộng theo vế )
\(\Rightarrow\)\(-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>0\)
\(\Rightarrow\)\(A=3-\left[\left(\frac{1}{2014}-\frac{1}{2013}\right)+\left(\frac{1}{2015}-\frac{1}{2013}\right)\right]>3\) ( cộng hai vế cho 3 )
\(\Rightarrow\)\(A>3\) ( điều phải chứng minh )
Vậy \(A>3\)
Chúc đệ học tốt ~
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c,
\(C=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot...\cdot\frac{9999}{10000}\)
vì \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
\(\frac{5}{6}< \frac{6}{7}\)
.............................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
nên \(C^2< \frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{10000}{10001}\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)
\(\Rightarrow C< \frac{1}{100}\)
bt lm mỗi một câu :v
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Chứng tỏ:a) frac{2013}{2014}+frac{2014}{2015}+frac{2015}{2013}3b) left(1+frac{1}{2}right)left(1+frac{1}{2^2}right)left(1+frac{1}{2^3}right)left(1+frac{1}{2^4}right)...left(1+frac{2}{50}right) 3c) Cfrac{1}{2}.frac{3}{4}.frac{5}{6}.....frac{9999}{10000} frac{1}{100}d) frac{1}{2}-frac{1}{2^2}+.........+frac{1}{2^{99}}-frac{1}{2^{100}} frac{1}{3}
Đọc tiếp
Chứng tỏ:
a) \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
b) \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2^2}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{2^4}\right)...\left(1+\frac{2}{50}\right)< 3\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
d) \(\frac{1}{2}-\frac{1}{2^2}+.........+\frac{1}{2^{99}}-\frac{1}{2^{100}}< \frac{1}{3}\)
,mình sửa lại đề:
\(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}< 3\)
xóa các chữ số ở tử và mẫu: 2014 và 2014,2015 và 2015
=\(\frac{2013}{2013}\)
=\(1\)
vì \(1>3\) nên \(\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2013}>3\)
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tinh \(G=\frac{\left(1+\frac{2015}{1}\right)+\left(1+\frac{2015}{2}\right)+...+\left(1+\frac{2015}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)+....+\left(1+\frac{1000}{2015}\right)}\)
a,frac{-1}{24}-left[frac{1}{4}-left(frac{1}{2}-frac{7}{8}right)right]
b,left[frac{5}{7}-frac{7}{5}right]-left[frac{1}{2}-left(frac{-2}{7}-frac{1}{10}right)right]
c,left(frac{-1}{2}right)-left(frac{-3}{5}right)+left(frac{-1}{9}right)+frac{1}{71}-left(frac{-2}{7}right)+frac{4}{35}-frac{7}{8}
d,left(3-frac{1}{4}+frac{2}{3}right)-left(5-frac{1}{3}-frac{6}{5}right)-left(6-frac{7}{4}+frac{3}{2}right)
e,left(frac{1}{2}-frac{13}{14}right):frac{5}{7}-left(frac{-2}{21}+frac{1}{7}right):frac{5}{7}
g,f...
Đọc tiếp
a,\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b,\(\left[\frac{5}{7}-\frac{7}{5}\right]-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
c,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)
d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)