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\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
\(B=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(B=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(B=\sqrt{3}+1\)
\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\) =\(\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)+\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\) =\(\frac{6}{6}=1\)
\(\Rightarrow A=\sqrt{2}\)
\(A=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{4}+\sqrt{6}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\frac{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
\(A=\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(\sqrt{2}A=\sqrt{5}+\sqrt{5}+1-1\)
\(\sqrt{2}A=2\sqrt{5}\)
\(A=\sqrt{10}\)
P/s tham khảo nha
\(\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{6-2\sqrt{5}}\)
\(=3-\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=3-\sqrt{5}+\sqrt{5}-1=2\)
\(\sqrt{9+4\sqrt{5}}=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{5}\)
\(=\sqrt{5}+2-\sqrt{5}=2\)
Chúc học tốt!!!!!!!!!!!!!
1 bài thôi nhé, tui còn phải xem World Cup :vv
\(\sqrt{x^4-4x+4}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\Leftrightarrow x^4-4x+4=\sqrt{5}-\sqrt{5}+1\)
\(\Leftrightarrow x^4-4x+3=0\)
\(\Leftrightarrow x^4+2x^3+3x^2-2x^3-4x^2-6x+x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(x^2+2x+3\right)-2x\left(x^2+2x+3\right)+\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(x^2+2x+3\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+2x+3\right)=0\)
Vì: \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2>0\)
=> \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) (thỏa mãn)
Vậy pt có nghiệm x = 1
p/s: đkxđ là x thuộc R nên tui k ghi vào :v
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