\(-1\) : căn bậc hai của \(1\)

\(-2\) : căn bậc hai của \(4\)

\(-3\) : căn bậc hai của \(9\)

\(4\) : căn bậc hai của \(16\)

\(5\) : căn bậc hai của \(25\)

\(6\) : căn bậc hai của \(36\)

\(7\) : căn bậc hai của \(49\)

\(0\) : căn bậc hai của \(0\)

\(11\) căn bậc hai của \(121\)

Đặt x²⁵ = t

=> x¹⁰⁰ + x⁵⁰ + 1 = t⁴ + t² + 1 

= t⁴ + 2t² + 1 - t² 

= (t² + 1)² - t² 

= (t² - t + 1)(t² + t + 1) 

= (x⁵⁰ - x²⁵ + 1)(x⁵⁰ + x²⁵ + 1)

\(0< a_i< 10,\forall i=1,2,3,...,n\)

\(\overline{a_1a_2}+\overline{a_2a_3}+\overline{a_3a_4}+...+\overline{a_{n-1}a_n}+\overline{a_na_1}\)

\(=10a_1+a_2+10a_2+a_3+10a_3+a_4+...+10a_{n-1}+a_n+10a_n+a_1\)

\(=11a_1+11a_2+11a_3+...+11a_{n-1}+11a_n\)

\(=11\left(a_1+a_2+...+a_{n-1}+a_n\right)\left(đpcm\right)\)

em đăng lại câu hỏi vào box Toán học nhé.

\(a,A=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{5+2\sqrt{6}}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{3+2\sqrt{3}.\sqrt{2}+2}\)

\(=\left(\sqrt{3}-\sqrt{2}\right)\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

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

\(=3-2=1\)

\(b,B=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)

\(=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{\dfrac{9}{4}.8}-2\sqrt{\dfrac{9}{4}.12}+\sqrt{\dfrac{9}{4}.20}}\)

\(=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3.\dfrac{3}{4}\sqrt{8}-2.\dfrac{3}{4}\sqrt{12}+\dfrac{3}{4}\sqrt{20}}\)

\(=\dfrac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{\dfrac{3}{4}\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right)}\)

\(=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)

\(A=\dfrac{x}{x-5}-\dfrac{10x}{x^2-25}-\dfrac{5}{x+5}\left(ĐKXĐ:x\ne\pm5\right)\)

\(=\dfrac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\dfrac{10x}{\left(x-5\right)\left(x+5\right)}-\dfrac{5\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+5x-10x-5x+25}{\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{x+5}\)