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Theo bà ra ta có :
\(f\left(2\sqrt{3}\right)=\left(m+1\right)x-2=\left(m+1\right)\left(2\sqrt{3}\right)-2\)
\(=\sqrt{12}\left(m+1\right)-2\)
\(f\left(3\sqrt{2}\right)=\left(m+1\right)x-2=\left(m+1\right)3\sqrt{2}-2\)
\(=\sqrt{18}\left(m+1\right)-2\)
vì 12 < 18 => \(\sqrt{12}< \sqrt{18}\)
hay \(f\left(2\sqrt{3}\right)< f\left(3\sqrt{2}\right)\)
a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
\(a,9\sqrt{5}+3\sqrt{20}-7\sqrt{45}=9\sqrt{5}+6\sqrt{5}-21\sqrt{5}=-6\sqrt{5}\\ b,\dfrac{2\sqrt{6}+\sqrt{40}}{\sqrt{3}+\sqrt{5}}=\dfrac{2\sqrt{6}+2\sqrt{10}}{\sqrt{3}+\sqrt{5}}\\ =\dfrac{2\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{2\sqrt{2}\left(5-3\right)}{5-3}=2\sqrt{2}\)
