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\(x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3+2\sqrt{2}}\)
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
\(x=\sqrt{\sqrt{2}-1}\Leftrightarrow x^2+1=\sqrt{2}\Leftrightarrow x^4+2x^2-1=0\)
\(Q=x^2\left(x^4+2x^2-1\right)+x\left(x^4+2x^2-1\right)+2019=2019\)
\(x=\sqrt{\sqrt{2}-1}\)\(\Leftrightarrow x^2=\sqrt{2}-1\)\(\Leftrightarrow x^2+1=\sqrt{2}\)\(\Leftrightarrow\left(x^2+1\right)^2=2\)
rồi chuyển vế
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 :
\(A=x^2-2x-2=4+2\sqrt{3}-2\sqrt{3}-2-2\)
\(=0\)
Do tại \(x=2019^4>1\Rightarrow x-1>0\)
\(A=\sqrt{x+2\sqrt{x\left(x-1\right)}+x-1}+\sqrt{x-2\sqrt{x\left(x-1\right)}+x-1}\)
\(A=\sqrt{\left(\sqrt{x}+\sqrt{x-1}\right)^2}+\sqrt{\left(\sqrt{x}-\sqrt{x-1}\right)^2}\)
\(A=\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\)
\(A=2\sqrt{x}=2\sqrt{2019^4}=2.2019^2\)