Biểu diễn các số thực $\genfrac{}{}{0ex}{}{3}{4};\genfrac{}{}{0ex}{}{-2}{3}$ trên trục số.

vvv

$3-1+\genfrac{}{}{0ex}{}{5-3}{3.5}+\genfrac{}{}{0ex}{}{7-5}{5.7}+\dots +\genfrac{}{}{0ex}{}{101-99}{99.101}$

$=\genfrac{}{}{0ex}{}{3}{1.3}-\genfrac{}{}{0ex}{}{1}{1.3}+\genfrac{}{}{0ex}{}{5}{3.5}-\genfrac{}{}{0ex}{}{3}{3.5}+\genfrac{}{}{0ex}{}{7}{5.7}-\genfrac{}{}{0ex}{}{5}{5.7}+\dots +\genfrac{}{}{0ex}{}{101}{99.101}-\genfrac{}{}{0ex}{}{99}{99.101}$ 

$=1-\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{7}+\dots +\genfrac{}{}{0ex}{}{1}{99}-\genfrac{}{}{0ex}{}{1}{101}$

$=1-\genfrac{}{}{0ex}{}{1}{101}=\genfrac{}{}{0ex}{}{100}{101}$

Vậy $\genfrac{}{}{0ex}{}{2}{1.3}+\genfrac{}{}{0ex}{}{2}{3.5}+\genfrac{}{}{0ex}{}{2}{5.7}+\dots +\genfrac{}{}{0ex}{}{2}{99.101}=\genfrac{}{}{0ex}{}{100}{101}$.

a) $\genfrac{}{}{0ex}{}{-2}{7}+\genfrac{}{}{0ex}{}{2}{7}:\genfrac{}{}{0ex}{}{3}{5}$

$=\genfrac{}{}{0ex}{}{-2}{7}+\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{5}{3}$

$=\genfrac{}{}{0ex}{}{-2}{7}+\genfrac{}{}{0ex}{}{10}{21}$

$=\genfrac{}{}{0ex}{}{-6}{21}+\genfrac{}{}{0ex}{}{10}{21}$

$=21$

b)$\genfrac{}{}{0ex}{}{-8}{19}+\genfrac{}{}{0ex}{}{-4}{21}-\genfrac{}{}{0ex}{}{17}{21}+\genfrac{}{}{0ex}{}{27}{19}$

$=\genfrac{}{}{0ex}{}{-8}{19}+\genfrac{}{}{0ex}{}{-4}{21}+\genfrac{}{}{0ex}{}{-17}{21}+\genfrac{}{}{0ex}{}{27}{19}$

$=\left(\genfrac{}{}{0ex}{}{-8}{19}+\genfrac{}{}{0ex}{}{27}{19}\right)+\left(\genfrac{}{}{0ex}{}{-4}{21}+\genfrac{}{}{0ex}{}{-17}{21}\right)$

$=\genfrac{}{}{0ex}{}{-8+27}{19}+\genfrac{}{}{0ex}{}{(-4)+(-17)}{21}$

$=\genfrac{}{}{0ex}{}{19}{19}+\genfrac{}{}{0ex}{}{-21}{21}$

$=1-1=0$

c) $\genfrac{}{}{0ex}{}{6}{5}.\genfrac{}{}{0ex}{}{3}{13}-\genfrac{}{}{0ex}{}{6}{5}.\genfrac{}{}{0ex}{}{16}{13}$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(\genfrac{}{}{0ex}{}{3}{13}-\genfrac{}{}{0ex}{}{16}{13}\right)$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(\genfrac{}{}{0ex}{}{3-16}{13}\right)$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(\genfrac{}{}{0ex}{}{-13}{13}\right)$

$=\genfrac{}{}{0ex}{}{6}{5}.\left(-1\right)$

$=\genfrac{}{}{0ex}{}{-6}{5}.$

$4$

a) Theo hình vẽ, ta có: $AI+IB=AB$

Hay $4+IB=9$

$IB=9-4=5$ cm

b) Vì $E$ là trung điểm của $IB$ nên

$EI=EB=\genfrac{}{}{0ex}{}{IB}{2}=\genfrac{}{}{0ex}{}{5}{2}=2,5$ (cm)

Theo hình vẽ, ta có: $AE=AI+IE=4+2,5=6,5$ (cm)

   a) Theo hình vẽ, ta có: $AI+IB=AB$

Hay $4+IB=9$

$IB=9-4=5$ cm

b) Vì $E$ là trung điểm của $IB$ nên

$EI=EB=\genfrac{}{}{0ex}{}{IB}{2}=\genfrac{}{}{0ex}{}{5}{2}=2,5$ (cm)

Theo hình vẽ, ta có: $AE=AI+IE=4+2,5=6,5$ (cm)