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\(A=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)
a) \(A^2=\left(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\right)^2\)
\(=\left(\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\right)^2\)
\(=\left(2+\sqrt{3}+2-\sqrt{3}\right)^2\)
\(=4^2\)
\(=16\)
b) \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2+\sqrt{3}+2-\sqrt{3}\)
\(=4\)
a, \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\Rightarrow7+4\sqrt{3}+7-4\sqrt{3}=14\)
Bài làm:
a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)
\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)
\(A=4+2\sqrt{3}+5\sqrt{3}-1\)
\(A=3+7\sqrt{3}\)
b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)
\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)
\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)
\(A=2\)
Phần b mình viết nhầm tên thành A, bn sửa thành B nhé
c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(C=\sqrt{3}-1-2-\sqrt{3}\)
\(C=-3\)
a) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}\)
\(=-2\sqrt{3}\)
a ) Ta có : \(A=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)
\(\Rightarrow A^2=\left(\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\right)^2\)
\(A^2=\left(\sqrt{7+4\sqrt{3}}\right)^2+2\left(\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\right)+\left(\sqrt{7-4\sqrt{3}}\right)^2\)
\(A^2=7+4\sqrt{3}+2\sqrt{49-16.3}+7-4\sqrt{3}\)
b ) Ta có thể thấy : \(7+4\sqrt{3}=4+4\sqrt{3}+3=\left(2+\sqrt{3}\right)^2\)
Tương tự : \(7-4\sqrt{3}=4-4\sqrt{3}+3=\left(2-\sqrt{3}\right)^2\)
Vậy \(A=2+\sqrt{3}+2-\sqrt{3}=4\)