Ôn tập chương 1: Căn bậc hai. Căn bậc ba

NT
4 tháng 7 2021 lúc 11:13

e) Ta có: \(E=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}-\sqrt{6}\)

\(=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-2\sqrt{3}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1-2\sqrt{3}}{\sqrt{2}}=0\)

f) Ta có: \(F=\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)\)

\(=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2\)

=0

g) Ta có: \(G=4\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)\)
\(=2\sqrt{2}\cdot\sqrt{4+2\sqrt{3}}+2\sqrt{2}\cdot\sqrt{6-2\sqrt{5}}-2\sqrt{10}\)

\(=2\sqrt{2}\left(\sqrt{3}+1\right)+2\sqrt{2}\left(\sqrt{5}-1\right)-2\sqrt{10}\)

\(=2\sqrt{6}+2\sqrt{2}+2\sqrt{10}-2\sqrt{2}-2\sqrt{10}\)

\(=2\sqrt{6}\)

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AT
4 tháng 7 2021 lúc 11:18

\(E=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}-\sqrt{6}\)

\(=\sqrt{\dfrac{4+2\sqrt{3}}{2}}+\sqrt{\dfrac{4-2\sqrt{3}}{2}}-\sqrt{6}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}{2}}-\sqrt{6}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}-\sqrt{6}\)

\(=\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}+\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{2}}-\sqrt{6}\)

\(=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}-\sqrt{6}=\dfrac{2\sqrt{3}}{\sqrt{2}}-\sqrt{6}=\sqrt{6}-\sqrt{6}=0\)

\(F=\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\right)\)

\(=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)

\(=\sqrt{\left(\sqrt{7}\right)^2+2.\sqrt{7}.1+1^2}-\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}.1+1^2}-2\)

\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}-2=\left|\sqrt{7}+1\right|-\left|\sqrt{7}-1\right|-2\)

\(=\sqrt{7}+1-\sqrt{7}+1-2=0\)

\(G=4\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\dfrac{5}{2}}\right)\)

\(=2\sqrt{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}-\sqrt{2.\dfrac{5}{2}}\right)\)

\(=2\sqrt{2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{5}\right)\)

\(=2\sqrt{2}\left(\sqrt{3}+1+\sqrt{5}-1-\sqrt{5}\right)=2\sqrt{6}\)

\(H=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}\)

\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\)

\(=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\sqrt{2}}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{2}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

 

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