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a,\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\\ =\sqrt{2+3+1+2\sqrt{2}.1+2\sqrt{3}.1+2\sqrt{2}.\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}=\sqrt{2}+\sqrt{3}+1\)
Lời giải:
Xét số hạng tổng quát:
$\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n+1})}$
$=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}$
$=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}$
Do đó:
$S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}$
$=1-\frac{1}{\sqrt{2017}}$
Lời giải:
Trong TH này ta thêm điều kiện $x$ là số nguyên dương.
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x(x+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{(x+1)-x}{x(x+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\)
\(=1-\frac{1}{x+1}=\frac{x}{x+1}\)
Vậy \(\frac{x}{x+1}=\frac{\sqrt{2017-x}+2016}{\sqrt{2016-x}+2017}\)
\(\Rightarrow x\sqrt{2016-x}+2017x=(x+1)\sqrt{2017-x}+2016(x+1)\)
\(\Leftrightarrow x\sqrt{2016-x}=(x+1)\sqrt{2017-x}+2016-x\)
\(\Leftrightarrow x(\sqrt{2017-x}-\sqrt{2016-x})+\sqrt{2017-x}+2016-x=0\)
\(\Leftrightarrow \frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}+\sqrt{2017-x}+(2016-x)=0\)
Hiển nhiên ta thấy:
\(\frac{x}{\sqrt{2017-x}+\sqrt{2016-x}}>0\)
\(\sqrt{2017-x}\geq 0\)
\(2016-x\geq 0\)
Do đó pt trên vô nghiệm
Tức là không tìm đc $x$ thỏa mãn.
So Sánh : \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\) và \(\dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}=\dfrac{2017^2-1-2016^2+1}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\dfrac{\left(2017-2016\right)\left(2017+2016\right)}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\dfrac{1+2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}>\dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
\(\sqrt{1+2016^2+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}\)(1)
Đặt x=2016
(1)\(\Leftrightarrow\)\(\sqrt{1+a^2+\dfrac{a^2}{\left(a+1\right)^2}}\)+\(\dfrac{2016}{2017}\)
\(\Leftrightarrow\)\(\sqrt{\dfrac{\left(a+1\right)^2+\left(a+a^2\right)^2+a^2}{\left(a+1\right)^2}}\)+\(\dfrac{2016}{2017}\)(2)
Xét:\(\left(a+1\right)^2+\left(a^2+a\right)^2+a^2\)\(=\)\(\left(a^2+a\right)^2+a^2+2a+1+a^2=\left(a^2+a\right)^2+1+2\left(a^2+a\right)=\left(a^2+a+1\right)^2\)
(2)\(\Leftrightarrow\)\(\sqrt{\dfrac{\left(a^2+a+1\right)^2}{\left(a+1\right)^2}}\)+\(\dfrac{2016}{2017}\)=\(\dfrac{a^2+a+1}{a+1}+\dfrac{2016}{2017}=\dfrac{a^2+a+1}{a+1}+\dfrac{a}{a+1}=\dfrac{a^2+2a+1}{a+1}=\dfrac{\left(a+1\right)^2}{a+1}=a+1=2017\)
Ta chứng minh được công thức \(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\dfrac{1}{a}+\dfrac{1}{b}-\dfrac{1}{a+b}\)
\(\sqrt{\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{\left(a+b\right)^2}}=\sqrt{\dfrac{a^4+2a^3b+a^2b^2+2ab^3+b^4}{a^2b^2\left(a+b\right)^2}}\)
\(=\sqrt{\left(\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\right)^2}=\dfrac{a^2+ab+b^2}{ab\left(a+b\right)}\)
\(=\dfrac{1}{b}+\dfrac{1}{a}-\dfrac{1}{a+b}\)
\(A=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2016^2}+\dfrac{1}{2017^2}}+\sqrt{\dfrac{1}{1^2}+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}\)
\(=\dfrac{1}{1}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{1}+\dfrac{1}{3}-\dfrac{1}{4}+1+\dfrac{1}{2016}-\dfrac{1}{2017}+1+\dfrac{1}{2017}-\dfrac{1}{2018}\)
=>A là số hữu tỉ (ĐPCM)
Khá phổ biến!
\(\sqrt{1+2016^2+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}=\sqrt{\left(2016+1\right)^2-2.2016+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}\) \(=\sqrt{2017^2-2.2016+\dfrac{2016^2}{2017^2}}+\dfrac{2016}{2017}=\sqrt{\left(2017-\dfrac{2016}{2017}\right)^2}+\dfrac{2016}{2017}\)
\(=2017-\dfrac{2016}{2017}+\dfrac{2016}{2017}=2017\)