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ta có: \(\left(\sqrt{2017^2-1}-\sqrt{2016^2-1}\right)\left(\sqrt{2017^2-1}+\sqrt{2016^2-1}\right)\)
= 20172-1 - (20162-1)
= 20172-20162
= 2017+2016 > 2.2016
=> \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)\(>\) \(\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
So sánh \(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)và \(\frac{2\cdot2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Ta có :
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}=\frac{2017^2-1-2016^2+1}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\frac{2017+2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
\(>\frac{2016+2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Vậy \(\sqrt{2017^2-1}-\sqrt{2016^2-1}>\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
so sánh
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}\)và \(\frac{2\cdot2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
Ta có:
\(\frac{1-\sqrt{n}+\sqrt{n+1}}{1+\sqrt{n}+\sqrt{n+1}}=\frac{\left(1-\sqrt{n}+\sqrt{n+1}\right)^2}{\left(1+\sqrt{n}+\sqrt{n+1}\right)\left(1-\sqrt{n}+\sqrt{n+1}\right)}=\frac{2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}}{2\left(1+\sqrt{n+1}\right)}\)
\(=\frac{\left[2n+2-2\sqrt{n}+2\sqrt{n+1}-2\sqrt{n\left(n+1\right)}\right]\left(1-\sqrt{n+1}\right)}{2\left(1+\sqrt{n+1}\right)\left(1-\sqrt{n+1}\right)}=\frac{-2n\sqrt{n+1}+2n\sqrt{n}}{-2n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra:
\(Q=\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2017}-\sqrt{2016}=\sqrt{2017}-\sqrt{2}< \sqrt{2017}-1=R\)
Vậy Q < R.
\(A=\sqrt{\left(2017-1\right)\left(2017+1\right)}-\sqrt{\left(2016-1\right)\left(2016+1\right)}\)
\(=\sqrt{2016.2018}-\sqrt{2015.2017}< \sqrt{2018.2018}-\sqrt{2015.2015}\)
\(=2018-2015=3\)
\(\Rightarrow\frac{1}{A}>\frac{1}{3}\)
\(B=\frac{2.2016}{A}>\frac{2.2016}{3}=1344>3>A\)
Vậy ta được B lớn hơn A rất nhiều :))
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:
$S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}$
$=1-\frac{1}{\sqrt{2017}}$
1/ Tính: \(A=\dfrac{\sqrt{15-10\sqrt{2}}+\sqrt{13+4\sqrt{10}}-\sqrt{11+2\sqrt{10}}}{2\sqrt{3+2\sqrt{2}}+\sqrt{9-4\sqrt{2}}+\sqrt{12+8\sqrt{2}}}=\dfrac{\sqrt{\left(\sqrt{10}-\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{10}+1\right)^2}}{2\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}+\sqrt{\left(2\sqrt{2}+2\right)^2}}=\dfrac{\sqrt{10}-\sqrt{5}+2\sqrt{2}+\sqrt{5}-\sqrt{10}-1}{2\sqrt{2}+2+2\sqrt{2}-1+2\sqrt{2}+2}=\dfrac{2\sqrt{2}-1}{6\sqrt{2}-3}=\dfrac{2\sqrt{2}-1}{3\left(2\sqrt{2}-1\right)}=\dfrac{1}{3}\)
\(B=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2}+\sqrt{3}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2}-\sqrt{3}}=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2}-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}+\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2}-\sqrt{3}\right)}=\dfrac{2\sqrt{2}-2\sqrt{2}-2\sqrt{3}+\sqrt{6}-\sqrt{6}-3+2\sqrt{2}+2\sqrt{2}+2\sqrt{3}-\sqrt{6}-\sqrt{6}-3}{2-\left(\sqrt{2}+\sqrt{3}\right)^2}=\dfrac{4\sqrt{2}-2\sqrt{6}-6}{2-2-3-2\sqrt{6}}=\dfrac{2\left(2\sqrt{2}-\sqrt{6}-3\right)}{-3-2\sqrt{6}}\)
Với mọi \(n\in N.\)ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}.\)Do đó
\(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}.=1-\frac{1}{\sqrt{2017}}=\frac{\sqrt{2017}-1}{\sqrt{2017}}.\)
\(\sqrt{2017^2-1}-\sqrt{2016^2-1}=\dfrac{2017^2-1-2016^2+1}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\dfrac{\left(2017-2016\right)\left(2017+2016\right)}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}=\dfrac{1+2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}>\dfrac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
vcl hai quay lai cay hoc 24 a