Rút gọn: A=\(\sqrt{1+2015^{2^{ }}+\dfrac{2015^2}{2016^2}}+\dfrac{2015}{2016}\)
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23 tháng 6 2017
Ta thấy: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{2015}}-\dfrac{1}{\sqrt{2016}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2016}}=\dfrac{\sqrt{2016}-1}{\sqrt{2016}}\)
4 tháng 9 2022
=>|x-1|+|x-2|=2016
TH1: x<1
Pt sẽ là 1-x+2-x=2016
=>-2x+3=2016
=>-2x=2013
=>x=-2013/2(nhận)
TH2: 1<=x<2
Pt sẽ là x-1+2-x=2016
=>1=2016(loại)
TH3: x>=2
Pt sẽ là 2x-3=2016
=>2x=2019
=>x=2019/2(nhận)
DT
17 tháng 4 2017
đặt phân số trên là A
tử là
(1+2015/2)+...+(1+2/2015)+(1+1/2016)+1
=2017/2+....+2017/2015+2017/2016+2017/2017
=2017.(1/2+...+1/2015+1/2016+1/2017)
=>A=\(\dfrac{2017.\left(\dfrac{1}{2}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}\right)}{\dfrac{1}{2}+...+\dfrac{1}{2015}+\dfrac{1}{2016}+\dfrac{1}{2017}}\)
Vậy A=2017
BC
0
23 tháng 6 2017
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
=\(\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}\)
=\(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
áp dụng vào biểu thức ta có\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
=\(1-\frac{1}{\sqrt{2016}}\)
đến đây cậu tự giải nốt nhé
TB
1
HN
10 tháng 6 2017
\(\dfrac{\sqrt{x-2015}-1}{x-2015}+\dfrac{\sqrt{y-2016}-1}{y-2016}=\dfrac{1}{2}\)
Điều kiện \(\left\{{}\begin{matrix}x>2015\\y>2016\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x-2015}}-\dfrac{1}{x-2015}+\dfrac{1}{\sqrt{y-2016}}-\dfrac{1}{y-2016}=\dfrac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=a>0\\\dfrac{1}{\sqrt{y-2016}}=b>0\end{matrix}\right.\) thì ta có:
\(a-a^2+b-b^2=\dfrac{1}{2}\)
\(\Leftrightarrow\left(2a^2-2a+\dfrac{1}{2}\right)+\left(2b^2-2b+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}a-\dfrac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2}b-\dfrac{1}{\sqrt{2}}\right)^2=0\)
\(\Leftrightarrow a=b=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=\dfrac{1}{4}\\\dfrac{1}{\sqrt{y-2016}}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2019\\y=2020\end{matrix}\right.\)
16 tháng 8 2021
Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)
Khi đó phương trình trở thành:
\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)
Tick plz
\(A=\sqrt[]{1+2015^2+\dfrac{2015^2}{2016^2}}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=\sqrt[]{\left(1+2015\right)^2-2.2015+\dfrac{2015^2}{\left(2015+1\right)^2}}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=\sqrt[]{\left(1+2015-\dfrac{2015}{2015+1}\right)^2}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=\left|1+2015-\dfrac{2015}{2016}\right|+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=1+2015-\dfrac{2015}{2016}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=1+2015=2016\)