\(\frac{2.3+6.9+10.15+14.21}{2.5+6.15+10.25+14.35}\)

Rút gọn:

\(=\frac{3}{5}+\frac{3}{5}+\frac{3}{5}+\frac{3}{5}\)

\(=\frac{12}{5}\)

\(\frac{2.3+6.9+10.15+14.21}{2.5+6.15+10.25+14.35}=\frac{2.3+6.3.3+10.5.3+14.7.3}{2.5+6.3.5+10.5.5+14.7.5}\)

\(=\frac{3\left(2.1+6.3+10.5+14.7\right)}{5\left(2.1+6.3+10.5+14.7\right)}=\frac{3}{5}\)

a)
\(\frac{1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10}{3\cdot4+6\cdot8+9\cdot12+12\cdot16+15\cdot20}\)

\(=\frac{1\cdot2\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}{3\cdot4\cdot\left(1+2\cdot2+3\cdot3+4\cdot4+5\cdot5\right)}\)

\(=\frac{1\cdot2}{3\cdot4}=\frac{1}{6}\)

b) \(\frac{12\cdot13+24\cdot26+36\cdot39}{24\cdot26+48\cdot52+72\cdot78}\)

\(=\frac{12\cdot13\cdot\left(1+2\cdot2+3\cdot3\right)}{24\cdot26\cdot\left(1+2\cdot2+3\cdot3\right)}\)

\(=\frac{12\cdot13}{24\cdot26}=\frac{1}{4}\)

\(\frac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}=\frac{2\cdot3\cdot\left(1+2\cdot2+7\cdot7\right)}{3\cdot5\cdot\left(1+2\cdot2+7\cdot7\right)}=\frac{2\cdot3}{3\cdot5}=\frac{2}{5}\)

\(\frac{1.2+1.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)

=\(\frac{1.2}{2.3}\)+\(\frac{1.4}{4.6}\)+\(\frac{3.6}{6.9}\)+\(\frac{4.8}{8.12}\)

= \(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{3}\)+\(\frac{1}{3}\)

= \(\frac{2}{6}+\frac{1}{6}+\frac{2}{6}+\frac{2}{6}\)

=\(\frac{7}{6}\)

Mình nghĩ đề bài phải là:

          \(\frac{1.2+2.4+3.6+4.8}{2.3+4.6+6.9+8.12}\)   *2.3 + 4.6 + 6.9 + 8.12 = 3.(1.2 + 2.4 + 3.6 + 4.8)*

\(=\)\(\frac{1\left(1.2+2.4+3.6+4.8\right)}{3\left(1.2+2.4+3.6+4.8\right)}\)

\(=\)\(\frac{1}{3}\)

14/45

A =\(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{14}+\frac{1}{14}-\frac{1}{21}+...+\frac{1}{140}-\frac{1}{150}\)

A=\(\left(\frac{1}{3}-\frac{1}{150}\right)\)

A=\(\left(\frac{50}{150}-\frac{1}{150}\right)\)

A=\(\frac{49}{150}\)

\(A=\frac{1}{3x7}+\frac{1}{7x14}+....+\frac{1}{140x150}\)

\(A=\frac{1}{3}-\frac{1}{7}+.....+\frac{1}{140}-\frac{1}{150}\)

\(A=\frac{1}{3}-\frac{1}{150}\)

\(A=\frac{49}{150}\)

\(A=\frac{1}{3x7}+\frac{1}{7x14}+....+\frac{1}{140x150}\)

\(A=\frac{1}{3}-\frac{1}{7}+.....+\frac{1}{140}-\frac{1}{150}\)

\(A=\frac{1}{3}-\frac{1}{150}\)

\(A=\frac{49}{150}\)

= \(\frac{1.2+2.4+3.6+4.8+5.10}{3.2\left(1.2+2.4+3.6+4.8+5.10\right)}=\frac{1}{6}\)
nha :*