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

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

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

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

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

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

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

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

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

= \(\dfrac{2.55}{4.155}\)

= \(\dfrac{1.55}{2.155}\)

= \(\dfrac{55}{310}=\dfrac{11}{62}\)

Ta có:

\(D=\dfrac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\\ =\dfrac{1.2+1.2.4+1.2.3.3+1.2.2.8+1.2.5.5}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\\ =\dfrac{1.2\left(4+3^2+2.8+5^2\right)}{3.4\left(2^2+3^2+2.8+5^2\right)}\\ =\dfrac{1.2}{3.4}=\dfrac{1}{6}\)

Vậy \(D=\dfrac{1}{6}\)

thiếu 

\(=\frac{1.2\left(1+2.2+3.3+4.4+5.5\right)}{3.4\left(1+2.2+3.3+4.4+5.5\right)}\)

\(=\frac{1.2}{3.4}=\frac{1}{6}\)

\(\frac{1.2+2.4+3.6+4.8+5.10}{3.2.2+2.3.4.2+3.3.6.2+4.3.8.2+5.3.10.2}\)=\(\frac{1}{3.2+3.2+3.2+3.2+3.2}\)=\(\frac{1}{30}\)

vậy bt trên =\(\frac{1}{30}\)

tk nhé

\(B=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\\ =\frac{2+8+18+32+50}{12+48+108+192+300}\\ =\frac{110}{660}=\frac{1}{6}\)

B=\(\dfrac{1.2+1.2.4+1.2.3.3+1.2.2.8+1.2.5.5}{3.4+3.4.2.2+3.4.3.3+3.4.2.8+3.4.5.5}\)

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

B=\(\dfrac{1.2}{3.4}=\dfrac{1}{6}\)

Sau khi rút gọn biểu thức B, ta có:

B= \(\dfrac{\left[1.2\left(1+2+3+4+5\right)\right]}{\left[3.4\left(1+2+3+4+5\right)\right]}\)

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

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

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

1/6 uyhvnb

ảnh có dễ thươg các bạn!

Ý ban là:

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