tính A:
A = 1/2016.2015 + 1/2015.2014 + 1/2013.2014 + 1/2013.2016 + ... + 1/1.2
LƯU Ý : TRÌNH BÀY LỜI GIẢI RÕ RÀNG
DẤU / BIỂU THỊ PHÂN SỐ; DẤU . BIỂU THỊ DẤU NHÂN
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\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}\)
\(=\frac{2015}{2016}\)
Phép tính trên có thể ghi ngược lại
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
=\(1-\frac{1}{2016}\)
=\(\frac{2015}{2016}\)
A= 1/1.2 + 1/2.3 +...........+ 1/2016.2015
= 1 - 1/2 +1/2 - 1/3 + ............+1/2015 - 1/2016
= 1 - 1/2016
= 2015/2016
5/6+3/4 < 19/11 4/9 x 6/5 = 8/15
17/12-9/8 < 1/3 15/4:3/8 > 8
Ta có: \(\frac{5}{6}+\frac{3}{4}=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
Vì \(12>11\)
\(\Rightarrow\frac{19}{12}< \frac{19}{11}\)
\(\Rightarrow\frac{5}{6}+\frac{3}{4}< \frac{19}{11}\)
Ta có: \(\frac{17}{12}-\frac{9}{8}=\frac{34}{24}-\frac{27}{24}=\frac{7}{24}\)
Ta lại có: \(\frac{1}{3}=\frac{8}{24}\)
Vì \(8>7\)
\(\Rightarrow\frac{7}{24}< \frac{8}{24}\)
\(\Rightarrow\frac{17}{12}-\frac{9}{8}< \frac{1}{3}\)
Ta có: \(\frac{4}{9}\times\frac{6}{5}=\frac{24}{45}=\frac{8}{15}\)
Vì \(\frac{8}{15}=\frac{8}{15}\)
\(\Rightarrow\frac{4}{9}\times\frac{6}{5}=\frac{8}{15}\)
Ta có: \(\frac{15}{4}:\frac{3}{8}=\frac{15}{4}\times\frac{8}{3}=10\)
Vì \(10>8\)
\(\Rightarrow\frac{15}{4}:\frac{3}{8}>8\)
HOK TOT
bai nay ban viet nguoc day so lai roi giai nhu binh thuong la duoc
\(\frac{1}{2}-\frac{1}{2016.2015}-\frac{1}{2015.2014}-...-\frac{1}{3.2}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2016}\)
\(=\frac{1}{2016}\)
\(\frac{1}{2}-\frac{1}{2016.2015}-\frac{1}{2015.2014}-...-\frac{1}{3.2}\)
\(=\frac{1}{2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\left(\frac{1}{2}-\frac{1}{2016}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2016}\)
\(=0+\frac{1}{2016}=\frac{1}{2016}\)
=\(-\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2015}+\frac{1}{2014}-...-\frac{1}{2}+1\)
=\(-\frac{1}{2016}+1=\frac{2015}{2016}\)
Ta có :\(\frac{-1}{2016.2015}-\frac{1}{2015.2014}-\frac{1}{2014.2013}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(-\left(\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
= \(-\left(\frac{1}{2016}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2013}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-\frac{1}{1}\right)\)
= \(-\left(\frac{1}{2016}-1\right)\)
= \(-\left(-\frac{2015}{2016}\right)\)
= \(-\frac{2015}{2016}\)
Mk làm kĩ lắm rồi. ko tích nữa mk cũng chịu bạn luôn @@
Bài 3 : Tính :
A = \(\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+....+\frac{1}{1.2}\)
\(A=\frac{1}{2016.2015}+\frac{1}{2015.2014}+\frac{1}{2014.2013}+...+\frac{1}{1.2}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(=1-\frac{1}{2016}=\frac{2015}{2016}\)
Vậy \(A=\frac{2015}{2016}\).
Mình viết ngược lại cho dễ làm xD
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}+\frac{1}{2015\cdot2016}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\)
\(A=\frac{1}{1}-\frac{1}{2016}\)
\(A=\frac{2015}{2016}\)
Sai thì bỏ quá :3
\(\frac{1}{2016.2015}+\frac{1}{2015.2014}+...+\frac{1}{1.2}\)
\(=\frac{1}{1.2}+...+\frac{1}{2014.2015}+\frac{1}{2015.2016}\)
\(=\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2014}-\frac{1}{2015}+\frac{1}{2015}-\frac{1}{2016}\)
\(=\frac{1}{1}+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{2015}-\frac{1}{2015}\right)-\frac{1}{2016}\)
\(=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
~ Hok tốt ~
=1/2016-1/2015+1/2015-1/2014+...+1-1/2
=1/2016-1/2
=-1007/2016
\(\left(3^2\right)^{2010}=9^{2010}=81^{1005}\)
Vì 1 khi lũy thừa lên bao nhiêu thì số tận cùng vẫn là 1 vì 1 x 1 x 1 x 1... = ......1
Nên \(81^{1005}\)có số tận cùng là 1
Vậy \(\left(3^2\right)^{2010}\)có số tận cùng là 1
A=1/2015-1/2016+1/2014-1/2015+1/2013-1/2014+.............+1-1/2
A=1/2016+1
A=2017/2016
chúc học tốt
Đáp án :
A = 1/1.2 + ... + 1/2013.2014 + 1/2014.2015 + 1/2015.2016
= 1 + 1/2 - 1/2 + .... + 1/2013 - 1/2014 + 1/2014 - 1/2015 + 1/2015 - 1/2016
= 1 + 0 + .... + 0 + 0 + 0 - 1/2016
= 1 - 1/2016
= 2015/2016
Vậy A = 2015/2016