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23 tháng 7 2017
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\left(\sqrt{20}-3\right)}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(A=\sqrt{\sqrt{5}-\sqrt{5}+1}=\sqrt{1}=1\)
23 tháng 7 2017
\(B=\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3}+1\right)^2}}}=\sqrt{6+2\sqrt{5-\left(2\sqrt{3}+1\right)}}\)
\(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(B=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(B=\sqrt{3}+1\)
11 tháng 7 2018
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3}+9}}=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\) \(B=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}=\sqrt{6+2\sqrt{3-2\sqrt{3}+1}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{3+2\sqrt{3}+1}=\sqrt{3}+1\)
NC
4 tháng 10 2020
a) \(\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{29-6\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-\sqrt{\left(\sqrt{20}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3}-2\sqrt{5}+3}\)
\(=\sqrt{3-\sqrt{3}-\sqrt{5}}\)
24 tháng 6 2018
\(A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-6\sqrt{20}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\sqrt{\sqrt{5}-\sqrt{5}+1}\\ =\sqrt{1}=1\)
\(B=\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12+1+2\sqrt{12}}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\\ =\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\\ =\sqrt{6+2\sqrt{3+1-2\sqrt{3}}}\\ =\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\sqrt{6+2\sqrt{3}-2}\\ =\sqrt{3+1+2\sqrt{3}}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+4\sqrt{3}}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(\sqrt{3}+2\right)^2}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{3}-20}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-10\sqrt{3}}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\\ =\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\\ =\sqrt{4+\sqrt{25}}=\sqrt{4+5}=\sqrt{9}=3\)
\(D=\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\\ \text{Ta có }:\left(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\right)^2\\ =3+\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}+3-\sqrt{5}\\ =6-2\sqrt{9-5}=6-2\sqrt{4}=6-4=2\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}=\sqrt{2}\\ \Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
AV
1
AT
14 tháng 6 2018
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
YK
16 tháng 8 2019
a) Ta có :\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\) =\(\sqrt{\frac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
Tương tự : \(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\) = \(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)
=>\(\sqrt{\frac{5+2\sqrt{6}}{5-2\sqrt{6}}}\)+\(\sqrt{\frac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)=\(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)+\(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)= \(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)=\(\frac{5+2\sqrt{6}+5-2\sqrt{6}}{3-2}\)=10
NB
1
22 tháng 10 2019
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.2\sqrt{5}.3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\)\(\sqrt{\sqrt{5}-\sqrt{3-\left|2\sqrt{5}-3\right|}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left|\sqrt{5}-1\right|}\)
\(=\sqrt{1}=1\)
12 tháng 9 2016
\(4\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
= \(4\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2×3×2\sqrt{5}+9}}}\)
= \(4\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
= \(4\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
= \(4\sqrt{\sqrt{5}-\sqrt{5}+1}\)
= 4
Ta lại có\(\sqrt[3]{10+6\sqrt{3}}×\left(\sqrt{3}-1\right)\)
= \(\sqrt[3]{3\sqrt{3}+3×3+3\sqrt{3}+1}×\left(\sqrt{3}-1\right)\)
= (\(\sqrt{3}+1\))(\(\sqrt{3}-1\))
= 2
Từ đó a = 2
Thế vào P ta được ( 25 - 7×22 - 3)81 + 19
= 20
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.\sqrt{20}.3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-5.\sqrt{5}.1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)