1/6 uyhvnb

Ý ban là:

\(\frac{1}{6}\)

cách giải nè.

Ta có: \(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

=\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.1.2+3.2.2.4+3.2.3.6+3.2.4.8+3.2.5.10}\)

=\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.\left(1.2+2.4+3.6+4.8+5.10\right)}\)

=\(\frac{1}{6}\)

\(\frac{1}{6}\)

\(\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(=\frac{1.2+2.2.2+3.3.2+4.4.2+5.5.2}{3.4+2.3.2.4+3.3.3.4+4.3.4.4+5.3.5.4}\)

\(=\frac{2.\left(1+2.2+3.3+4.4+5.5\right)}{3.4.\left(1+2.2+3.3+4.4+5.5\right)}\)

\(=\frac{1}{6}\)

\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\)

\(=\dfrac{1.2.\left(4+3^2+2.8+5^2\right)}{3.4.\left(4+3^2+2.8+5^2\right)}\)

\(=\dfrac{1}{6}\)

\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

\(=\dfrac{1.2+1.2.2^2+1.2.3^2+1.2.4^2+1.2.5^2}{3.4+3.4.2^2+3.4.3^2+3.4.4^2+3.4.5^2}\)

\(=\dfrac{1.2\left(1+2^2+3^2+4^2+5^2\right)}{3.4\left(1+2^2+3^2+4^2+5^2\right)}\\ =\dfrac{2}{12}=\dfrac{1}{6}\)

\(\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)

= \(\dfrac{2\left(1+4+9+16+25\right)}{4\left(3+12+27+48+65\right)}\)

= \(\dfrac{2.55}{4.155}\)

= \(\dfrac{1.55}{2.155}\)

= \(\dfrac{55}{310}=\dfrac{11}{62}\)

\(=\frac{1}{6}\)