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Biểu thức B ko bt có sai đề ở căn thứ 2 ko ạ
Nếu nhân B với căn 2 thì cái căn thức nhất tách đc thành hđt (a+b)2 đấy ạ nhưng cái căn thứ 2 thì ko tách đc
a)
\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\\ =\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\\ =\sqrt{5}+1-\sqrt{5}+1\\ =2\)
b) Sửa đề:
\(\sqrt{7+2\sqrt{6}}+\sqrt{7-2\sqrt{6}}-2\sqrt{6}\\ =\sqrt{\left(\sqrt{6}+1\right)^2}+\sqrt{\left(\sqrt{6}-1\right)^2}-2\sqrt{6}\\ =\sqrt{6}+1+\sqrt{6}-1-2\sqrt{6}\\ =0\)
c)
\(\sqrt{9+4\sqrt{5}}+\sqrt{9-4\sqrt{5}}-2\sqrt{5}\\ =\sqrt{9+2\sqrt{20}}+\sqrt{9-2\sqrt{20}}-2\sqrt{5}\\ =\sqrt{\left(\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}-2\sqrt{5}\\ =\sqrt{5}+2+\sqrt{5}-2-2\sqrt{5}\\ =0\)
\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(A=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(A=\sqrt{5}-1-\sqrt{5}-1\)
\(A=-2\)
\(B=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(B=\sqrt{\left(\sqrt{5}+2\right)^2}-\sqrt{\left(\sqrt{5}-2\right)^2}\)
\(B=\sqrt{5}+2-\sqrt{5}+2\)
\(B=4\)
Sửa đề :
\(C=\sqrt{14-6\sqrt{5}}-\sqrt{14+6\sqrt{5}}\)
\(C=\sqrt{\left(3-\sqrt{5}\right)^2}-\sqrt{\left(3+\sqrt{5}\right)^2}\)
\(C=3-\sqrt{5}-3-\sqrt{5}\)
\(C=-2\sqrt{5}\)
a. \(\sqrt{13+6\sqrt{4+\sqrt{9-4\sqrt{2}}}}=\sqrt{13+6\sqrt{4+\sqrt{\left(\sqrt{8}-1\right)^2}}}=\sqrt{13+6\sqrt{4+\sqrt{8}-1}}=\sqrt{13+6\sqrt{3+\sqrt{8}}}=\sqrt{13+6\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{13+6\left(\sqrt{2}+1\right)}=\sqrt{13+6\sqrt{2}+6}=\sqrt{19+6\sqrt{2}}=\sqrt{\left(3\sqrt{2}+1\right)^2}=1+3\sqrt{2}\)
b. \(\left(\sqrt{3}-1\right)\sqrt{2\sqrt{19+8\sqrt{3}}-4}=\left(\sqrt{3}-1\right)\sqrt{2\sqrt{\left(4+\sqrt{3}\right)^2}-4}=\left(\sqrt{3}-1\right)\sqrt{8+2\sqrt{3}-4}=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
c. \(\sqrt{5+2\sqrt{6}}+\sqrt{14-4\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+2\sqrt{3}-\sqrt{2}=3\sqrt{3}\)
d. \(\sqrt{5-2\sqrt{6}}+\sqrt{11-4\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{3}\right)^2}=\sqrt{3}-\sqrt{2}+2\sqrt{2}-\sqrt{3}=\sqrt{2}\)
\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)
\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)