(1/2-1)(1/3-1)(1/4-1)......(1/2015-1)
Những câu hỏi liên quan
tính a: [1-1/2*2]*[1-1/3*3]*[1-1/4*4]*[1-1/5*5]*......*[1-1/2015*2015]*[1-1/2016*2016]
Tính C= 1+1/2(1+2)+1/3(1+2+3)+........+1/2015(1+2+3+4+...+2015)
C = 1+1/2(1+2)+1/3(1+2+3)+........+1/2015(1+2+3+4+...+2015)
C = 1 + \(\frac{1}{2}\cdot\frac{2.3}{2}\)+ \(\frac{1}{3}\cdot\frac{3.4}{2}\)+ ... + \(\frac{1}{2015}\cdot\frac{2015.2016}{2}\)
C = \(\frac{2}{2}\) + \(\frac{3}{2}+\frac{4}{2}+...+\frac{2016}{2}\)
C = \(\frac{2+3+4+...+2016}{2}\)
Đặt D = 2 + 3 + 4 + ... + 2016
Số số hạng của D là : (2016 - 2) : 1 + 1 = 2015
Tổng D là : (2 + 2016) . 2015 : 2 = 2033135
Thay D vào biểu thức C ta được : \(\frac{2033135}{2}\)
Vậy C = ... .
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Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
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Tính: (1*2015+2*2014+3*2013+...+2015*1)/(1*2+2*3+3*4+4*5+...+2015*2016)
Tìm x biết:
1. 9/4:(1/3x-1/2)=4 và 1/2
2. 1/2015.x=(1-1/2) (1-1/3) (1-1/4)...(1-1/2014) (1-1/2015)
\(\frac{1}{2015}x=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2014}\right)\left(1-\frac{1}{2015}\right)\)
\(\frac{1}{2015}x=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times.....\times\frac{2013}{2014}\times\frac{2014}{2015}\)
\(\frac{1}{2015}x=\frac{1}{2015}\)
\(x=1\)
Chúc bạn học tốt
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Tinh tong S=1/1×2+1/2×3+1/3×4+...+1/2014×2015+1/2015×2016
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)
\(S=1-\frac{1}{2016}=\frac{2015}{2016}\)
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\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-........+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+......+\left(-\frac{1}{2015}+\frac{1}{2015}\right)-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
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A= ( 1/2017+ 2/2016+ 3/2015+...+ 2015/3+ 2016/2+ 2017) : ( 1/2+1/3+1/4+...+1/2017+1/2018)
a. [(-2)^5*2014-4^2*2015]-(-2015^0 + 3^2 - 2^3)
b. [9-(1/2+1/3+1/4+...+1/10)] : (1/2+2/3+3/4+...+9/10)
a. \(\left[\left(-2\right)^5.2014-4^2.2015\right]-\left(-2015^0+3^2-2^3\right)\)
\(=-64448-32240+1-9+8=-96688\)
Tớ lm lại nhé:
SBC = 9-1/2-1/3-1/4-...-1/10
=1+1+...+1(9 số 1) -1/2-1/3-1/4-1/5-...-1/10.
=(1-1/2)+(1-1/3)+...+(1-1/10)
=1/2+2/3+...+9/10= SC
=> phép chia có thương là 1(vì SBC=SC)
SBC = (1-1/2) + (1-1/3) + ... + (1-1/10)
=1/2 + 2/3 +3/4 +... +9/10 = SC
Vậy thương là 1.