\(A=\dfrac{2\sqrt{3}-5\sqrt{27}+4\sqrt{12}}{\sqrt{3}}=2-15+8=-5\)
\(B=\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}=\dfrac{\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}{\sqrt{2+\sqrt{3}}}=1\)
Ta có: \(B-3\sqrt{2x-7}=A\Rightarrow1-3\sqrt{2x-7}=-5\)
\(\Rightarrow3\sqrt{2x-7}=6\Rightarrow\sqrt{2x-7}=2\Rightarrow2x-7=4\Rightarrow x=\dfrac{11}{2}\)
a)
A=\(\dfrac{2\sqrt{3}-5\sqrt{27}+4\sqrt{12}}{\sqrt{3}}\)
=\(\dfrac{2\sqrt{3}-15\sqrt{3}+8\sqrt{3}}{\sqrt{3}}\)
= \(\dfrac{-5\sqrt{3}}{\sqrt{3}}\)= -5
B= \(\dfrac{\left(2+\sqrt{3}\right)\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}}\)
= \(\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)
= \(\sqrt{\left(2+\sqrt{3}\right).\left(2-\sqrt{3}\right)}\)
= \(\sqrt{4-3}\) =1
b)
B - \(3\sqrt{2x-7}\) = A (x ≥ \(\dfrac{7}{2}\) )
⇔ 1- \(3\sqrt{2x-7}\) = -5
⇔ \(3\sqrt{2x-7}\) = 6
⇔ \(\sqrt{2x-7}\) = 2
⇔ 2x - 7 = 4
⇔ 2x = 11
⇔ x = \(\dfrac{11}{2}\) (t/m)
