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a )\(\sqrt{6+\sqrt{8}+\sqrt{12}+\sqrt{24}}\)
=\(\sqrt{2+3+1+2\sqrt{2.1+2\sqrt{3}.1+2\sqrt{2}.\sqrt{3}}}\)
=\(\sqrt{\left(\sqrt{2}+\sqrt{3}+1\right)^2}\)
=\(\sqrt{2}+\sqrt{3}+1\)
Xét với x > 0 : \(\sqrt{1+\left(x-1\right)^2+\frac{\left(x-1\right)^2}{x^2}}+\frac{x-1}{x}=\sqrt{\frac{\left(x^2-x+1\right)^2}{x^2}}+\frac{x-1}{x}\)
\(=\frac{x^2-x+1}{x}+\frac{x-1}{x}=\frac{x^2}{x}=x\)
Áp dụng với x = 2017 suy ra biểu thức cần tính có giá trị bằng 2017
Đặt 2017 = a thì ta có
A = \(\sqrt{1+\left(a-1\right)^2+\frac{\left(a-1\right)^2}{a^2}}+\frac{a-1}{a}\)
= \(\sqrt{\frac{\left(a^2-a+1\right)^2}{1a^2}}+\frac{a-1}{a}\)
= a
Vậy cái đó bằng 2017
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
\(P=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}.=1-\frac{1}{\sqrt{2017}}=\frac{\sqrt{2017}-1}{\sqrt{2017}}.\)
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.
\(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2016}}-\frac{1}{\sqrt{2017}}=1-\frac{1}{\sqrt{2007}}=\frac{\sqrt{2007}-1}{\sqrt{2007}}\)
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 \(\sqrt{2017^2-1}-\sqrt{2016^2-1}>\frac{2.2016}{\sqrt{2017^2-1}+\sqrt{2016^2-1}}\)
\(A=\sqrt{2016^2+\frac{2017}{2017}+\frac{2016^2-1}{2017^2}-\frac{1}{2017^2}}+\frac{2016}{2017}\)
\(A=\sqrt{2016^2+\frac{1}{2017^2}+\frac{2015.2017}{2017^2}+\frac{2017}{2017}}+\frac{2016}{2017}\)
\(A=\sqrt{2016^2+2.2016.\frac{1}{2017}+\frac{1^2}{2017^2}}+\frac{2016}{2017}\)
\(A=\sqrt{\left(2016+\frac{1}{2017}\right)^2}+\frac{2016}{2017}\)
\(A=\left(2016+\frac{1}{2017}\right)+\frac{2016}{2017}\)
A = 2017
Chúc bạn làm bài tốt