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a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
Ta có \(x=\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}}=\dfrac{1}{2}\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{2-1}}=\dfrac{1}{2}\sqrt{\left(\sqrt{2}-1\right)^2}=\dfrac{1}{2}.\left|\sqrt{2}-1\right|=\dfrac{\sqrt{2}-1}{2}\)
Vậy \(x=\dfrac{\sqrt{2}-1}{2}\Leftrightarrow2x+1=\sqrt{2}\Leftrightarrow\left(2x+1\right)^2=2\Leftrightarrow4x^2+4x+1=2\Leftrightarrow4x^2+4x-1=0\)
Ta lại có \(A=\left(4x^5+4x^4-5x^3+5x-2\right)^{2017}+2019=\left(4x^5+4x^4-x^3-4x^3-4x^2+x+4x^2+4x-1-1\right)^{2017}+2019=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)-1\right]^{2017}+2019=\left(x^3.0-x.0+0-1\right)^{2017}+2019=\left(-1\right)^{2017}+2019=-1+2019=2018\)
Vậy khi x=\(\dfrac{1}{2}\sqrt{\dfrac{\sqrt{2}-1}{\sqrt{2}+1}}\) thì A=2018
\(B=\left(\dfrac{2\sqrt{x}+x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\sqrt{x}+x}{\left(\sqrt{x}-1\right)\cdot\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right)\cdot\dfrac{x+\sqrt{x}+1-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-\left(x+\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x+\sqrt{x}+1-\sqrt{x}-2}{x+\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1}{\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{1}{x+\sqrt{x}+1}\cdot\dfrac{x-1}{x+\sqrt{x}+1}\)
\(=\dfrac{x-1}{\left(x+\sqrt{x}+1\right)^2}\)
Đat 2017,5=t Ta có
\(\sqrt{\dfrac{\left(t+0,5\right)^2+\left(t-0,5\right)^2\cdot\left(t+0,5\right)^2+\left(t-0,5\right)^2}{\left(t+0,5\right)^2}}+\dfrac{t-0,5}{t+0,5}\\ =\sqrt{\dfrac{t^2+t+0,25+t^4-0,5t^2+0,0625+t^2-t+0,25}{\left(t+0,5\right)^2}}+\dfrac{t-0,5}{t+0,5}\\ =\dfrac{\sqrt{t^4+1,5t^2+0,5625}}{t+0,5}+\dfrac{t-0,5}{t+0,5}\\ =\dfrac{t^2+0,75+t-0,5}{t+0,5}\\ =\dfrac{\left(t+0,5\right)^2}{t+0,5}\\ =t+0,5\)thay t=2017,5 vào suy ra A=2017,5+0,5=2018
Giải:
\(\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\dfrac{1}{1^2}+\dfrac{1}{\left(\dfrac{1}{2017}\right)^2}+\dfrac{1}{\left(-\dfrac{2018}{2017}\right)^2}}+\dfrac{2017}{2018}\)
\(=\sqrt{\left(\dfrac{1}{1}+\dfrac{1}{\dfrac{1}{2017}}+\dfrac{1}{-\dfrac{2018}{2017}}\right)^2}+\dfrac{2017}{2018}\) (\(\left\{{}\begin{matrix}1>0\\2017^2>0\\\dfrac{2017^2}{2018^2}>0\end{matrix}\right.\Leftrightarrow1+2017^2+\dfrac{2017^2}{2018^2}>0\ne0\))
\(=1+2017+-\dfrac{2017}{2018}+\dfrac{2017}{2018}\)
\(=2018\)
Vậy ...
.... giúp với nha . Thank trước !
Bài 1:
\(A=\dfrac{2}{\sqrt{2017}+\sqrt{2015}}\)
\(B=\dfrac{2}{\sqrt{2019}+\sqrt{2017}}\)
mà \(\sqrt{2015}< \sqrt{2019}\)
nên A>B
a: Đặt a=2017
\(A=\sqrt{1+\left(\dfrac{1}{a}+\dfrac{1}{a+2}\right)^2}\)
\(=\sqrt{1+\left(\dfrac{2a+2}{a\left(a+2\right)}\right)^2}\)
\(=\sqrt{1+\dfrac{4a^2+8a+4}{a^2\cdot\left(a+2\right)^2}}=\sqrt{\dfrac{\left(a^2+a\right)^2+4a^2+8a+4}{a^2\left(a+2\right)^2}}\)
\(=\sqrt{\dfrac{\left(a^2+a\right)^2+4\left(a+1\right)^2}{a^2\left(a+2\right)^2}}\)
\(=\dfrac{\sqrt{\left(a^2+a\right)^2+4\left(a+1\right)^2}}{a\left(a+2\right)}\)
\(=\dfrac{\sqrt{\left(2017^2+2017\right)^2+4\cdot2018^2}}{2017\cdot2019}\)
b: Đặt 2017=a
\(B=\sqrt{a^2+a^2\cdot\left(a+1\right)^2+\left(a+1\right)^2}\)
\(=\sqrt{2a^2+2a+1+\left(a^2+a\right)^2}\)
\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1\)
\(=2017^2+2017+1=4070307\)