1²+2²+3²+4²+....+12²=650

=>(1+2+3+4+...+12)²=650

=>((1+2+3+4...+12)²)x2=650x2

=>(2+4+6+...12+24)²=1300

=>2²+4²+6²+....+24²=1300

Có : \(2^2+4^2+6^2+...+12^2+24^2\)

          =\(1^2.2^2+2^2.2^2+3^2.2^2+4^2.2^2+...+12^2.2^2\)

          =\(2^2.\left(1^2+2^2+3^2+4^2+...+12^2\right)\)

Theo đề bài ta có : \(2^2+4^2+6^2+...+12^2+24^2=2^2.650\)

                                                                                                     =2600
 

2mũ2+4mũ2+...+12mũ2+24mũ2

2.2+4.4+...+12.3.4+24.4.6

4+4.4+...+4.36+4.144

4.(1+4+...+36+144)

4.(1mũ2+2mũ2+...+12mũ2)

4.650=2600

mình không chắc là nó đúng vì mới có lớp 6 nhưng hi vọng nó không sai..!

2² + 4² + 6² +... + 24²

=(2.1)² + (2.2)² + (2.3)² +...+ (2.12)²

= 2². 1² + 2² . 2² + 2² . 3² +...+ 2² . 12²

⁼ 2² (1² + 2² + 3² +...+12²)

=4 . 650 = 2600

2² + 4² + 6² + ....... + 24² = 2( 1² + 2² + 3² + ..... + 12²) = 2 . 650 = 1300

\(2^2+4^2+6^2+...+24^2=2^2.\left(1^2+2^2+3^2+...+12^2\right)\)

a)
\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(\frac{2.3\left(1.2\right)+2.3\left(2.4\right)+2.3\left(3.6\right)+2.3\left(4.8\right)+2.3\left(5.10\right)}{3.4\left(3.4+6.8+9.12+12.16+15.20\right)}\)
\(=\frac{\left(3.4+6.8+9.12+12.16+15.20\right)}{2.3\left(3.4+6.8+9.12+12.16+15.20\right)}=\frac{1}{2.3}=\frac{1}{6}\)

 

Ta có:

\(1^2\)+\(2^2\)+\(3^2\)+...+\(12^2\)=650

=> \(2^2\).(\(1^2\)+\(2^2\)+\(3^2\)+...+\(12^2\))=\(2^2\).650

\(2^2\)+\(4^2\)+\(6^2\)+....+\(24^2\)=2600

Vậy \(2^2\)+\(4^2\)+\(6^2\)+....+\(24^2\)=2600.

2² + 4² + 6² + ... + 24²

= 2² . ( 1² + 2² + 3² + ... + 12² ) 

= 2² . 650

= 2600

Ta có : 2² + 4² + 6² + ... + 24²

          = 1²  . 2² + 2² . 2² + 3² . 2² + .... + 12² . 2²

          = (1² + 2² + 3² + ... + 12²) . 2²

          = 650 . 4 = 2600

   =2²+4²+6²+...+24²

   =2(1²⁺2²+3²+...+12²)

   =2.650

   =1300