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đặt
\(A=\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}\)
=>\(\sqrt{2}A=\sqrt{2}\sqrt{7+\sqrt{13}}+\sqrt{2}\sqrt{7-\sqrt{13}}\)
\(=\sqrt{14+2\sqrt{13}}+\sqrt{14-2\sqrt{13}}\)
\(=\sqrt{13+2\sqrt{13}+1}+\sqrt{13-2\sqrt{13}+1}\)
\(=\sqrt{\left(\sqrt{13}+1\right)^2}+\sqrt{\left(\sqrt{13}-1\right)^2}\)
\(=\sqrt{13}+1+\sqrt{13}-1=2\sqrt{13}\)
=>\(A=\frac{2\sqrt{13}}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}\sqrt{13}}{\sqrt{2}}=\sqrt{2}\sqrt{13}=\sqrt{26}\)
suy ra : ĐPCM
Để phân thức có nghĩa thì \(13x+26>0\Rightarrow x>-2\)
Vậy phân thức có nghĩa khi x>-2.
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
