1155 nhé bn nhưng mk k bt cách làm

Bn nào bt giúp mk vs

\(A=\left(12^2+14^2+16^2+18^2+20^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)\)

\(A=\left(12^2-1^2\right)+\left(14^2-3^2\right)+\left(16^2-5^2\right)+\left(18^2-7^2\right)+\left(20^2-9^2\right)\)\(A=\left(12+1\right)\left(12-1\right)+\left(14+3\right)\left(14-3\right)+\left(16-5\right)\left(16+5\right)+\left(18-7\right)\left(18+7\right)+\left(20-9\right)\left(20+9\right)\)

\(A=11.13+11.17+11.21+11.25+11.29\)

\(A=11.\left(13+17+21+25+29\right)\)

\(A=11.\left[\left(13+17\right)+\left(21+29\right)+25\right]\)

\(A=11.\left(30+50+25\right)\)

\(A=11.105=1155\)

Ta thấy : 2\(^2\)=4=4.1\(^2\)

  4\(^2\)=16=4.2\(^2\)

   6\(^2\)=36=4.3\(^2\)

     ...

  20\(^2\)=400=4.10\(^2\)

 =>S=2\(^2\)+4\(^2\)+6\(^2\)+...+20\(^2\)=4.(1\(^2\)+2\(^2\)+3\(^2\)+...+10\(^2\))

 = 4.385 

 =1540

  Vậy tổng S = 1540

Ta thấy A gấp 1²+2²+....+10² 4 lần nên Tổng A gấp 4 lần nó

=> A=385.4=1540

\(\left(12^2+14^2+16^2+18^2+20^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)\\\Rightarrow\left[\left(2^2.6^2\right)+\left(2^2.7^2\right)....+\left(2^2.10^2\right)\right]-\left(1^2+3^2+...+9^2\right)\\ \Rightarrow2^2.\left(6^2+7^2....+.10^2\right)-\left(1^2+3^2+5^2+7^2+9^2\right)\\ \Rightarrow4.330-165=1155\)

a) Ta có: \(A=3\sqrt{20}-\sqrt{45}+2\sqrt{18}+\sqrt{72}\)

\(=6\sqrt{5}-3\sqrt{5}+6\sqrt{2}+6\sqrt{2}\)

\(=3\sqrt{5}+12\sqrt{2}\)

b) Ta có: \(B=\dfrac{12}{3-\sqrt{5}}-\dfrac{16}{\sqrt{5}+1}\)

\(=\dfrac{12\left(3+\sqrt{5}\right)}{4}-\dfrac{16\left(\sqrt{5}-1\right)}{4}\)

\(=3\left(3+\sqrt{5}\right)-4\left(\sqrt{5}-1\right)\)

\(=9+3\sqrt{5}-4\sqrt{5}+4\)

\(=13-\sqrt{5}\)

c) Ta có: \(C=10\sqrt{\dfrac{1}{5}}+\dfrac{1}{5}\sqrt{125}-2\sqrt{20}\)

\(=\dfrac{10}{\sqrt{5}}+\dfrac{1}{5}\cdot5\sqrt{5}-2\cdot2\sqrt{5}\)

\(=2\sqrt{5}+\sqrt{5}-4\sqrt{5}\)

\(=-\sqrt{5}\)

e) Ta có: \(E=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}\)

\(=\sqrt{3}+1-2+\sqrt{3}\)

\(=2\sqrt{3}-1\)

f) Ta có: \(F=\sqrt{6+2\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+2\)

=3