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x= ...... - ....... = a -b
P=(a-b)^3 + 3(a-b) +2018 = a^3-3a^2b+3ab^2-b^3 +3a-3b+2018
=a^3-b^3 -3a(ab-1) -3b(ab -1) +2018 = a^3-b^3 - 3(ab-1)(a+b) +2018
a.b = 1 => ab-1 =0 => P =a^3 -b^3 +2018=\(\sqrt{2}\)-1 -\(\frac{1}{\sqrt{2}-1}\)+2018
=\(\frac{2+1-2\sqrt{2}-1+2018\sqrt{2}-2018}{\sqrt{2}-1}\)=\(\frac{2016\sqrt{2}-2016}{\sqrt{2}-1}\)=2016
Vậy P=2016
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
Ta có: Đặt \(A=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\)=> \(A^2=\frac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{\sqrt{5}+1}\)
=> \(A^2=\frac{2\sqrt{5}+2\sqrt{5-4}}{\sqrt{5}+1}=\frac{2\left(\sqrt{5}+1\right)}{\sqrt{5}+1}=2\)=> \(A=\sqrt{2}\)
\(\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)
==> \(x=\sqrt{2}-\left(\sqrt{2}+1\right)=-1\)
Do đó: N = (-1)2019 + 3.(-1)2020 - 2.(-1)2021 = -1 + 3 + 2 = 4
a) Ta có:
\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)
\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)
Ta có : \(x=\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}\sqrt{\dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}\sqrt{\dfrac{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}.\dfrac{\sqrt{\sqrt{3}+\sqrt{2}}}{\sqrt{\sqrt{3}-\sqrt{2}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{1}}=\sqrt{2}\)
Thay \(x=\sqrt{2}\) vào biểu thức A ta được :
\(A=\left(\sqrt{2}^3-2\sqrt{2}+1\right)^{2012}=1^{2012}=1\)
Vậy \(A=1\)
Ta có:
\(x=\sqrt{3+\sqrt{5+2\sqrt{3}}}+\sqrt{3-\sqrt{5+2\sqrt{3}}}\) ( x> 0 )
\(\Rightarrow x^2=6+2\sqrt{\left(3+\sqrt{5+2\sqrt{3}}\right)\left(3-\sqrt{5+2\sqrt{3}}\right)}\)
\(=6+2\sqrt{9-5-2\sqrt{3}}\)
\(=6+2\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=6+2\sqrt{3}-2=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow x=\sqrt{3}+1\)
Vậy :
\(A=x^2-2x-2=4+2\sqrt{3}-2\sqrt{3}-2-2\)
\(=0\)
a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
b) Ta có: \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
Thay x=2 vào A, ta được:
\(A=\dfrac{-3}{3+\sqrt{2}}=\dfrac{-9+3\sqrt{2}}{7}\)