a) tập hợp các tia gốc a là: a={Ax,Az,Ay}

b)các điểm thuộc az mà không thuộc ay là: AM,AC

c)am và ma là một đoạn thẳng nên không đối nhau

giá 15 quyển là:

7000.15=105000(đồng)

10% của 105000 là:

105000.10%=10500(đồng)

số tiền đã giảm là:

105000-10500=94500(đồng)

vậy an đủ tiền mua và thùa 5500 đồng

a)\(1-\dfrac{1}{2}+\dfrac{1}{3}\)

=\(\dfrac{1}{2}+\dfrac{1}{3}\)

=\(\dfrac{5}{6}\)

b)\(\dfrac{2}{5}+\dfrac{3}{5}:\dfrac{9}{10}\)

=\(\dfrac{2}{5}+\dfrac{2}{3}\)

=\(\dfrac{16}{15}\)

c)\(\dfrac{7}{11}.\dfrac{3}{4}+\dfrac{7}{11}.\dfrac{1}{4}+\dfrac{4}{11}\)

=\(\dfrac{7}{11}.\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+\dfrac{4}{11}\)

=\(\dfrac{7}{11}.1+\dfrac{4}{11}\)

=\(\dfrac{7}{11}+\dfrac{4}{11}\)

=1

d)\(\left(\dfrac{3}{4}+0,5+25\%\right).2\dfrac{2}{3}\)

=\(\left(0,75+0,5+0,25\right).2\dfrac{2}{3}\)

=1,5.2\(\dfrac{2}{3}\)

=1,5.\(\dfrac{8}{3}\)

=\(\dfrac{3}{2}.\dfrac{8}{3}\)

=4

a) $x-\genfrac{}{}{0ex}{}{2}{3}=\genfrac{}{}{0ex}{}{-5}{12}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{2}{3}$

$x=\genfrac{}{}{0ex}{}{-5}{12}+\genfrac{}{}{0ex}{}{8}{12}$

$x=\genfrac{}{}{0ex}{}{-5+8}{12}$

$x=\genfrac{}{}{0ex}{}{3}{12}$

$x=\genfrac{}{}{0ex}{}{1}{4}$

b) $\genfrac{}{}{0ex}{}{8}{5}:x=\genfrac{}{}{0ex}{}{-2}{3}$

$x=\genfrac{}{}{0ex}{}{8}{5}:\left(\genfrac{}{}{0ex}{}{-2}{3}\right)$

$x=\genfrac{}{}{0ex}{}{8}{5}.\left(\genfrac{}{}{0ex}{}{3}{-2}\right)$

$x=\genfrac{}{}{0ex}{}{-12}{5}$

c) $1-\genfrac{}{}{0ex}{}{3}{7}.x=-\genfrac{}{}{0ex}{}{2}{7}$

$\genfrac{}{}{0ex}{}{3}{7}.x=1-\left(-\genfrac{}{}{0ex}{}{2}{7}\right)$

$\genfrac{}{}{0ex}{}{3}{7}.x=\genfrac{}{}{0ex}{}{9}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}:\genfrac{}{}{0ex}{}{3}{7}$

$x=\genfrac{}{}{0ex}{}{9}{7}.\genfrac{}{}{0ex}{}{7}{3}$

$x=3$

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

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

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

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

$=\genfrac{}{}{0ex}{}{4}{21}$

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

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

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

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

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

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

câu hỏi đâu?

He was friendly when he was young

1.B

2.B