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\(A=\sqrt{\sqrt{7}-\sqrt{3}}-\sqrt{\sqrt{7}+\sqrt{3}}\)
=>\(A^2=\sqrt{7}-\sqrt{3}+\sqrt{7}+\sqrt{3}-2\sqrt{4}\)
=>A^2=2căn 7-4
=>A=2căn 7-4
=>\(M=\dfrac{2\left(\sqrt{7}-2\right)}{\sqrt{7}-2}=2\)
Xét \(\sqrt{2}.A=\sqrt{\dfrac{4+2\sqrt{3}}{2}}-\sqrt{\dfrac{4-2\sqrt{3}}{2}}\)
= \(\sqrt{\dfrac{\left(1+\sqrt{3}\right)^2}{2}}-\sqrt{\dfrac{\left(1-\sqrt{3}\right)^2}{2}}\)
= \(\dfrac{1+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}\)
<=> A = 1
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)
\(=\frac{1}{4}\)
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) \(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(=\sqrt{4-4\sqrt{3}+3}-\sqrt{4+4\sqrt{3}+3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)
\(=2-\sqrt{3}-2-\sqrt{3}\)
\(=-2\sqrt{3}\)
\(\frac{4}{\sqrt{7}-\sqrt{3}}+\frac{6}{3+\sqrt{3}}+\frac{\sqrt{7}-7}{\sqrt{7}-1}\)
\(=\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}+\frac{6\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\sqrt{7}\left(1-\sqrt{7}\right)}{\sqrt{7}-1}\)
\(=\frac{4\left(\sqrt{7}+\sqrt{3}\right)}{7-3}+\frac{6\left(3-\sqrt{3}\right)}{9-3}-\sqrt{7}\)
\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)
\(=3\)
\(\frac{\sqrt{3}+\sqrt{7}}{\sqrt{3}-\sqrt{7}}+\frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)+\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)^2+\left(\sqrt{3}-\sqrt{7}\right)^2}{3-7}\)
\(=\frac{3+2\sqrt{3}.\sqrt{7}+7+3-2\sqrt{3}.\sqrt{7}+7}{-4}\)
\(=\frac{3+7+3+7}{-4}\)
\(=\frac{20}{-4}=-5\)
Bài này đơn giản chỉ quy đồng về HDT thoi