两个数相除,商9余4,被除数、除数、商、余数之和是867,被除数和除数各是多少?

解：设除数为$ x $，则被除数为$ 9x + 4 $。
$ 867 - 9 - 4 = 854 $
$ x + (9x + 4) = 854 $
$ 10x + 4 = 854 $
$ 10x = 854 - 4 $
$ 10x = 850 $
$ x = 85 $
被除数：$ 9×85 + 4 = 769 $
答：被除数是769，除数是85。

1. 直接写出得数。
$578+216=$ $18.25-3.3=$ $3.2-\frac{1}{2}=$ $\frac{1}{9}×8.1=$
$\frac{1}{2}+\frac{1}{3}=$ $\frac{1}{4}÷3=$ $\frac{3}{4}+\frac{1}{2}÷\frac{1}{2}=$ $\frac{1}{4}×\frac{1}{27}=$

$578+216=794$
$18.25-3.3=14.95$
$3.2-\frac{1}{2}=3.2-0.5=2.7$
$\frac{1}{9}×8.1=0.9$
$\frac{1}{2}+\frac{1}{3}=\frac{3}{6}+\frac{2}{6}=\frac{5}{6}$
$\frac{1}{4}÷3=\frac{1}{4}×\frac{1}{3}=\frac{1}{12}$
$\frac{3}{4}+\frac{1}{2}÷\frac{1}{2}=\frac{3}{4}+1=1\frac{3}{4}$
$\frac{1}{4}×\frac{1}{27}=\frac{1}{108}$

2. 递等式计算。
$( \frac{7}{9}+\frac{5}{6}-\frac{1}{4} )×72$ $\frac{18}{23}÷\frac{8}{5}+\frac{18}{23}×\frac{3}{8}$
$50.4×2.7+24.48÷24$ $36.5×99+36.5$

$(\frac{7}{9}+\frac{5}{6}-\frac{1}{4})×72$
$=\frac{7}{9}×72+\frac{5}{6}×72-\frac{1}{4}×72$
$=56+60-18$
$=116-18$
$=98$
$\frac{18}{23}÷\frac{8}{5}+\frac{18}{23}×\frac{3}{8}$
$=\frac{18}{23}×\frac{5}{8}+\frac{18}{23}×\frac{3}{8}$
$=\frac{18}{23}×(\frac{5}{8}+\frac{3}{8})$
$=\frac{18}{23}×1$
$=\frac{18}{23}$
$50.4×2.7+24.48÷24$
$=136.08+1.02$
$=137.1$
$36.5×99+36.5$
$=36.5×99+36.5×1$
$=36.5×(99+1)$
$=36.5×100$
$=3650$