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\(A=\sqrt[]{1+2015^2+\dfrac{2015^2}{2016^2}}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=\sqrt[]{\left(1+2015\right)^2-2.2015+\dfrac{2015^2}{\left(2015+1\right)^2}}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=\sqrt[]{\left(1+2015-\dfrac{2015}{2015+1}\right)^2}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=\left|1+2015-\dfrac{2015}{2016}\right|+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=1+2015-\dfrac{2015}{2016}+\dfrac{2015}{2016}\)
\(\Leftrightarrow A=1+2015=2016\)
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
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)^2k-k^2\left(k+1\right)}\)
=\(\frac{\left(k+1\right)\sqrt{k}-k\left(\sqrt{k+1}\right)}{\left(k+1\right)k\left(k+1-k\right)}\)
=\(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
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
=>|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)
Với mọi n>0 ta có:\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Áp dụng đẳng thức trên vào D ta được:
\(D=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}=1-\frac{\sqrt{2016}}{2016}=\frac{2016-\sqrt{2016}}{2016}\)
\(\dfrac{\sqrt{x-2015}-1}{x-2015}+\dfrac{\sqrt{y-2016}-1}{y-2016}=\dfrac{1}{2}\)
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:
\(a-a^2+b-b^2=\dfrac{1}{2}\)
\(\Leftrightarrow\left(2a^2-2a+\dfrac{1}{2}\right)+\left(2b^2-2b+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left(\sqrt{2}a-\dfrac{1}{\sqrt{2}}\right)^2+\left(\sqrt{2}b-\dfrac{1}{\sqrt{2}}\right)^2=0\)
\(\Leftrightarrow a=b=\dfrac{1}{4}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-2015}}=\dfrac{1}{4}\\\dfrac{1}{\sqrt{y-2016}}=\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2019\\y=2020\end{matrix}\right.\)
\(\dfrac{8}{\sqrt{5}-1}-\dfrac{22}{4+\sqrt{5}}+\dfrac{\sqrt{15}+2\sqrt{5}}{2+\sqrt{3}}\)
\(=\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}-\dfrac{22\left(4-\sqrt{5}\right)}{\left(\sqrt{5}+4\right)\left(4-\sqrt{5}\right)}+\dfrac{\sqrt{5}\left(\sqrt{3}+2\right)}{2+\sqrt{3}}\)
\(=\dfrac{8\sqrt{5}+8}{5-1}-\dfrac{88-22\sqrt{5}}{16-5}+\sqrt{5}\)
\(=\dfrac{8\sqrt{5}+8}{4}-\dfrac{88-22\sqrt{5}}{11}+\sqrt{5}\)
\(=2\sqrt{5}+2-8+2\sqrt{5}+\sqrt{5}=5\sqrt{5}-6\)
Ta thấy: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+.....+\dfrac{1}{\sqrt{2015}}-\dfrac{1}{\sqrt{2016}}\)
\(=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2016}}=\dfrac{\sqrt{2016}-1}{\sqrt{2016}}\)
mình chưa hiểu dòng đầu tiên, bạn giải thích cho mình được không?