a) \(A=\left(5\sqrt{2}-2\sqrt{5}\right)\cdot\sqrt{5}-\sqrt{250}\)
\(A=\left(5\sqrt{2}-2\sqrt{5}\right)\cdot\sqrt{5}-\sqrt{5^2\cdot10}\)
\(A=5\sqrt{2}\cdot\sqrt{5}-2\sqrt{5}\cdot\sqrt{5}-5\sqrt{10}\)
\(A=5\sqrt{10}-10+5\sqrt{10}\)
\(A=-10\)
b) \(B=\dfrac{2}{\sqrt{3}-\sqrt{2}}+\dfrac{1}{3+2\sqrt{2}}\)
\(B=\dfrac{2\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{3-2\sqrt{2}}{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\)
\(B=\dfrac{2\cdot\left(\sqrt{3}+\sqrt{2}\right)}{3-2}+\dfrac{3-2\sqrt{2}}{9-8}\)
\(B=2\sqrt{3}+2\sqrt{2}+3-2\sqrt{2}\)
\(B=2\sqrt{3}+3\)
c) \(C=\dfrac{3+\sqrt{5}-\left(3-\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}.\dfrac{\sqrt{5}-1}{\sqrt{5}\left(\sqrt{5}-1\right)}\)
\(=\dfrac{2\sqrt{5}}{3^2-\left(\sqrt{5}\right)^2}.\dfrac{1}{\sqrt{5}}\)
\(=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}\)
\(=\dfrac{1}{2}\)
d)
\(D=\left(\sqrt{5}-1-5\right).\left[\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{1+\sqrt{5}}+6\right]\)
\(=\left(\sqrt{5}-6\right)\left(\sqrt{5}+6\right)=\left(\sqrt{5}\right)^2-6^2=-31\)
a, A =\(\left(5\sqrt{2}-2\sqrt{5}\right).\sqrt{5}-\sqrt{250}\)
= \(\sqrt{5^2.10}-\sqrt{10}-\sqrt{5^2.10}\)
= \(5\sqrt{10}-10-5\sqrt{10}\)
= \(-10\)
giúp e vs ạ ;-;
giúp e vs ạ :<