a,

\(\dfrac{\sqrt{7}-5}{2}-\dfrac{6-2\sqrt{7}}{4}+\dfrac{6}{\sqrt{7}-2}-\dfrac{5}{4+\sqrt{7}}\)

\(=\dfrac{\sqrt{7}-5}{2}-\dfrac{3-\sqrt{7}}{2}+\dfrac{6\sqrt{7}+12}{3}-\dfrac{20-5\sqrt{7}}{9}\)

\(=\dfrac{2\sqrt{7}-8}{2}+\dfrac{18\sqrt{7}+36}{9}-\dfrac{20-5\sqrt{7}}{9}\)

\(=\sqrt{7}-4+\dfrac{23\sqrt{7}+16}{9}\)

\(=\dfrac{9\sqrt{7}-36}{9}+\dfrac{23\sqrt{7}+16}{9}=\dfrac{32\sqrt{7}-20}{9}\)

\(\dfrac{4}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}+3}\)

\(=\dfrac{4\left(\sqrt{3}+1\right)}{3-1}+\dfrac{\sqrt{3}+2}{3-4}+\dfrac{6\left(\sqrt{3}-3\right)}{3-9}\)

\(=2\sqrt{3}+2-\sqrt{3}+2-\sqrt{3}+3\)

\(=7\)

\(\sqrt{5+\sqrt{21}}+\sqrt{5-\sqrt{21}}\) đề thế này phải k bn

ta có :

\(A=\cos^228+\cos^241+\cos^262+\cos^249\)

\(A=\cos^228+\cos^241+\sin^228+\sin^241\)

\(A=1+1=2\)

chúc bn hc tốt

Nam Ka !!!!!

ực nhìn thèm quá

ta có :

\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\ge2\sqrt{\dfrac{1}{R_1}.\dfrac{1}{R_2}}\)

\(=\dfrac{2}{\sqrt{R_1R_2}}\)

\(\Rightarrow\dfrac{1}{R_{tđ}}\ge\dfrac{2}{\sqrt{R_1R_2}}\)

\(\Rightarrow R_{tđ}\le\dfrac{\sqrt{R_1R_2}}{2}\)

Dấu " = " xảy ra khi \(R_1=R_2\)

Vậy để R_tđ lớn nhất thì R₁ = R₂