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\(\sqrt{2+\sqrt{3}.}\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right).\left(2-\sqrt{2+\sqrt{3}}\right)}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)
=\(\sqrt{4-3}=1\)
Lời giải:
Đặt \(\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}=a; \sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}=b\)
Có:
\(a^2+b^2=(2+\sqrt{3}+\sqrt{2-\sqrt{3}})+(2+\sqrt{3}-\sqrt{2-\sqrt{3}})=2(2+\sqrt{3})\)
\(=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\)
\(ab=\sqrt{(2+\sqrt{3}+\sqrt{2-\sqrt{3}})(2+\sqrt{3}-\sqrt{2-\sqrt{3}})}\)
\(=\sqrt{(2+\sqrt{3})^2-(2-\sqrt{3})}=\sqrt{5+5\sqrt{3}}\)
Như vậy:
\(\frac{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}+\frac{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{5+5\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt{5}.\sqrt{\sqrt{3}+1}}=\frac{(\sqrt{3}+1)^{1.5}}{\sqrt{5}}\)
a) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=\frac{\left(2+\sqrt{3}\right)^2}{4-3}\)
\(=\left(2+\sqrt{3}\right)^2=7+4\sqrt{3}\)
\(\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=\frac{\left(5+2\sqrt{6}\right)^2}{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}=\frac{\left(5+2\sqrt{6}\right)^2}{25-24}\)
\(=\left(5+2\sqrt{6}\right)^2=49+20\sqrt{6}\)
b) \(\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3-2\sqrt{3}+1}{3-1}\)
\(=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
c) \(\frac{2+\sqrt{3}}{2-\sqrt{3}}+\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2+\left(2-\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{4+4\sqrt{3}+3+4-4\sqrt{3}+3}{4-3}=14\)
d) \(\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}\)
\(=\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}-\left(2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\right)}{2+\sqrt{3}-\left(2-\sqrt{3}\right)}\)
\(=\frac{4\sqrt{4-3}}{2\sqrt{3}}=\frac{4}{2\sqrt{3}}=\frac{2}{\sqrt{3}}\)
\(\dfrac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\dfrac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}=\dfrac{\left(\sqrt{5+3\sqrt{2}}+\sqrt{5-3\sqrt{2}}\right)^2}{2\sqrt{2}}-\dfrac{\left(\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}\right)^2}{2\sqrt{2}}=\dfrac{4+2\sqrt{\left(5+3\sqrt{2}\right)\left(5-3\sqrt{2}\right)}-2\sqrt{\left(3+\sqrt{2}\right)\left(3-\sqrt{2}\right)}}{2\sqrt{2}}\) \(=\dfrac{4+2\sqrt{7}-2\sqrt{7}}{2\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\)
\(=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}=4-3=1\)
Rút gọn biểu thức
√ 2+√3.√2+√2+√3.√2−√2+√2+√3.√2+√2+√2+√3
= √2+√3.√2+√2+√3.√4−(2+√2+√3)=2+3.2+2+3.4−(2+2+3)
=√2+√3.√2+√2+√3.√2−√2+√3=2+3.2+2+3.2−2+3
=√2+√3.√4−(2+√3)=2+3.4−(2+3)
=√2+√3.√2−√3=4−3=1
Sửa đề ~~
\(A=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(A=\sqrt{\left(2+\sqrt{3}\right).\left(2+\sqrt{2+\sqrt{3}}\right).\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right).\left(2-\sqrt{2+\sqrt{3}}\right)}\)
\(A=\sqrt{2+\sqrt{3}.\left(4-\left(2+\sqrt{3}\right)\right).\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(A=\sqrt{2+\sqrt{3}.\left(4-2-\sqrt{3}\right).\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(A=\sqrt{2+\sqrt{3}.\left(2-\sqrt{3}\right).\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(A=\sqrt{\left(4-3\right).\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(A=\sqrt{1.\left(2+\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}\)
\(A=\sqrt{2+\sqrt{2}+\sqrt{2+\sqrt{3}}}\)
\(A=\dfrac{\sqrt{3}-3}{\sqrt{2-\sqrt{3}}+2\sqrt{2}}+\dfrac{\sqrt{3}+3}{\sqrt{2+\sqrt{3}}-2\sqrt{2}}\)
\(A=\dfrac{\sqrt{2}\left(\sqrt{3}-3\right)}{\sqrt{2}.\left(\sqrt{2-\sqrt{3}}+2\sqrt{2}\right)}+\dfrac{\sqrt{2}.\left(\sqrt{3}+3\right)}{\sqrt{2}.\left(\sqrt{2+\sqrt{3}}-2\sqrt{2}\right)}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{4-2\sqrt{3}}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{4+2\sqrt{3}}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{\left(\sqrt{3}-1\right)^2}+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{\left(\sqrt{3}+1\right)^2}-4}\)
\(A=\dfrac{\sqrt{6}-3\sqrt{2}}{\sqrt{3}-1+4}+\dfrac{\sqrt{6}+3\sqrt{2}}{\sqrt{3}+1-4}\)
\(A=\dfrac{\sqrt{3}\left(\sqrt{2}-\sqrt{6}\right)}{\sqrt{3}\left(1+\sqrt{3}\right)}+\dfrac{\sqrt{3}\left(\sqrt{2}+\sqrt{6}\right)}{\sqrt{3}\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}}{1+\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{6}}{1-\sqrt{3}}=\dfrac{\left(\sqrt{2}-\sqrt{6}\right)\left(1-\sqrt{3}\right)+\left(\sqrt{2}+\sqrt{6}\right)\left(1+\sqrt{3}\right)}{\left(1+\sqrt{3}\right)\left(1-\sqrt{3}\right)}\)
\(A=\dfrac{\sqrt{2}-\sqrt{6}-\sqrt{6}+3\sqrt{2}+\sqrt{2}+\sqrt{6}+\sqrt{6}+3\sqrt{2}}{1-3}=\dfrac{8\sqrt{2}}{-2}=-4\sqrt{2}\)
* \(B=\dfrac{\sqrt{11+2\sqrt{30}}-\sqrt{11-2\sqrt{30}}}{\sqrt{5}}\) \(=\dfrac{\sqrt{6+2.\sqrt{6}.\sqrt{5}+5}-\sqrt{6-2.\sqrt{6}.\sqrt{5}+5}}{\sqrt{5}}\)\(=\dfrac{\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}{\sqrt{5}}\)
\(=\dfrac{\sqrt{6}+\sqrt{5}-\sqrt{6}+\sqrt{5}}{\sqrt{5}}=\dfrac{2\sqrt{5}}{\sqrt{5}}=2\)
* \(C=2\sqrt{3+\sqrt{5}}-\left(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\right)\)
Đặt:\(y=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\Rightarrow y^2=4+\sqrt{15}+4-\sqrt{15}+2\sqrt{\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)}=8+2=10\Rightarrow y=\sqrt{10}\)
Suy ra: \(C=\sqrt{12+4\sqrt{5}}-y=\sqrt{\left(\sqrt{10}+\sqrt{2}\right)^2}-\sqrt{10}=\sqrt{10}+\sqrt{2}-\sqrt{10}=\sqrt{2}\)* \(D=\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\dfrac{\left(\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)+\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2-\sqrt{3}}\right)}{\left(\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}\right)}=\dfrac{2+\sqrt{3}+2-\sqrt{3}}{1}=4\)
học ghê v, đến tận đây r à :V
\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)