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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
\(\frac{A}{\sqrt{2}}=\frac{1+\sqrt{7}}{2+\sqrt{8+2\sqrt{7}}}+\frac{1-\sqrt{7}}{2-\sqrt{8-2\sqrt{7}}}\)
\(=\frac{1+\sqrt{7}}{2+1+\sqrt{7}}+\frac{1-\sqrt{7}}{2-\sqrt{7}+1}\)
\(=\frac{1+\sqrt{7}}{3+\sqrt{7}}+\frac{1-\sqrt{7}}{3-\sqrt{7}}\)
=\(\frac{\left(1+\sqrt{7}\right)\left(3-\sqrt{7}\right)+\left(1-\sqrt{7}\right)\left(3+\sqrt{7}\right)}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}\)
\(=\frac{-8}{2}=-4\)
\(\Rightarrow A=-4\sqrt{2}\)
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 :
\(A=x^2-2x-2=4+2\sqrt{3}-2\sqrt{3}-2-2\)
\(=0\)
Có \(x=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}=\frac{2}{\sqrt[3]{16}+\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}+\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)
\(y=\frac{2}{2\sqrt[3]{2}-2+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{16}-\sqrt[3]{8}+\sqrt[3]{4}}=\frac{\sqrt[3]{8}}{\sqrt[3]{4}\left(\sqrt[3]{4}-\sqrt[3]{2}+1\right)}=\frac{\sqrt[3]{2}}{\sqrt[3]{4}-\sqrt[3]{2}+1}\)
Đặt \(\sqrt[3]{2}=a\)
=> \(x=\frac{a}{a^2+a+1}\) ,\(y=\frac{a}{a^2-a+1}\)
Có: \(x+y=\frac{a}{a^2+a+1}+\frac{a}{a^2-a+1}=\frac{a^3-a^2+a+a^3+a^2+a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{2a^3+2a}{a^4+a^2+1}\)
\(x-y=\frac{a}{a^2+a+1}-\frac{a}{a^2-a+1}=\frac{a^3-a^2+a-a^3-a^2-a}{\left(a^2+a+1\right)\left(a^2-a+1\right)}=\frac{-2a^2}{a^4+a^2+1}\)
Có x2-y2= (x-y)(x+y)=\(\frac{2a^3+2a}{a^4+a^2+1}.\frac{-2a^2}{a^4+a^2+1}=\frac{-2a^2.2a\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4a^3\left(a^2+1\right)}{\left(a^4+a^2+1\right)^2}=\frac{-4.2\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
=\(\frac{-8\left(\sqrt[3]{4}+1\right)}{\left(\sqrt[3]{16}+\sqrt[3]{4}+1\right)^2}\)
Bạn xem lại đề bài 1 và 2.b nhé !
2/ \(A=\sqrt{\left(3-5\sqrt{2}\right)^2}-\sqrt{51+10\sqrt{2}}\)
\(A=5\sqrt{2}-3-\sqrt{\left(5\sqrt{2}+1\right)^2}\)
\(A=5\sqrt{2}-3-5\sqrt{2}-1\)
\(A=-4\)
ĐẶT: \(a=\sqrt[3]{\sqrt{2}-1}\)
=> \(a^3=\sqrt{2}-1\)
=> \(x=a-\frac{1}{a}\)
=> \(x^3=a^3-\frac{1}{a^3}-3a+\frac{3}{a}\)
<=> \(x^3=\sqrt{2}-1-\frac{1}{\sqrt{2}-1}-3\left(a-\frac{1}{a}\right)\)
<=> \(x^3=\frac{\left(\sqrt{2}-1\right)^2-1}{\sqrt{2}-1}-3x\)
<=> \(x^3+3x=\frac{3-2\sqrt{2}-1}{\sqrt{2}-1}\)
<=> \(x^3+3x=\frac{2-2\sqrt{2}}{\sqrt{2}-1}\)
<=> \(x^3+3x=\frac{2\left(1-\sqrt{2}\right)}{\sqrt{2}-1}\)
<=> \(x^3+3x=-2\)
<=> \(x^3+3x+2=0\Rightarrow P=0\)
VẬY \(P=0\)
Đặt \(a=\sqrt[3]{\sqrt{2}-1};b=\frac{1}{\sqrt[3]{\sqrt{2}-1}}\Rightarrow\hept{\begin{cases}x=a-b\\ab=1\end{cases}}\)
Xét \(x^3=\left(a-b\right)^3=a^3-b^3-3ab\left(a-b\right)\)
\(x^3=\left(\sqrt{2}-1\right)-\frac{1}{\sqrt{2}-1}-3x\)
\(\Leftrightarrow x^3=-2-3x\Leftrightarrow x^3+3x+2=0\)
Vậy P=0