第62页 - 苏教版数学补充习题五年级上下册答案

$\frac{2}{5}$

$\frac{1}{3}$

1

$\frac{2}{9}$

$\frac{12}{35}$

$\frac{1}{3}$

$\begin{aligned} 解： x&=\frac 12-\frac 14\\x&=\frac 14\end{aligned}$

$\begin{aligned} 解： x&=1-\frac 35\\x&=\frac 25\end{aligned}$

$\begin{aligned} 解： x&=\frac 56+\frac 12\\x&=\frac 43\end{aligned}$

$\begin{aligned} 解： x&=\frac 25+\frac 47\\x&=\frac {34}{35}\end{aligned}$

$=\frac {7}{8}+\frac {1}{8}+\frac {4}{7}$

$=1+\frac {4}{7}$

$=\frac {11}{7}$

$=(\frac {1}{6}+\frac {1}{6})+(\frac {3}{5}+\frac {2}{5})$

$=\frac {1}{3}+1$

$=\frac {4}{3}$

$=\frac {9}{5}-(\frac {2}{7}+\frac {5}{7})$

$=\frac {9}{5}-1$

$=\frac {4}{5}$

$=(\frac {5}{9}+\frac {4}{9})-(\frac {3}{8}+\frac {3}{8})$

$=1-\frac {6}{8}$

$=\frac {1}{4}$

$=\frac {5}{6}-\frac {1}{4}$

$=\frac {7}{12}$

$=\frac {2}{5}-\frac {2}{5}+\frac {2}{3}$

$=\frac {2}{3}$

$\frac 14+x=\frac 12$

(1) 解： $x=\frac12-\frac14$

$x=\frac14$

$x+\frac 35=1$

(2) 解： $x=1-\frac35$

$x=\frac25$

$x-\frac 12=\frac 56$

(3) 解： $x=\frac56+\frac12$

$x=\frac43$

$x-\frac 47=\frac 25$

(4) 解： $x=\frac25+\frac47$

$x=\frac{34}{35}$

$\dfrac{7}{8}+\dfrac{4}{7}+\dfrac{1}{8}$

$=\dfrac{7}{8}+\dfrac{1}{8}+\dfrac{4}{7}$

$=1+\dfrac{4}{7}$

$=1\dfrac{4}{7}$

$\dfrac{1}{6}+\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{1}{6}$

$=\left(\dfrac{1}{6}+\dfrac{1}{6}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)$

$=\dfrac{1}{3}+1$

$=1\dfrac{1}{3}$

$\dfrac{9}{5}-\dfrac{2}{7}-\dfrac{5}{7}$

$=\dfrac{9}{5}-\left(\dfrac{2}{7}+\dfrac{5}{7}\right)$

$=\dfrac{9}{5}-1$

$=\dfrac{4}{5}$

$\dfrac{5}{9}-\dfrac{3}{8}+\dfrac{4}{9}-\dfrac{3}{8}$

$=\left(\dfrac{5}{9}+\dfrac{4}{9}\right)-\left(\dfrac{3}{8}+\dfrac{3}{8}\right)$

$=1-\dfrac{3}{4}$

$=\dfrac{1}{4}$

$\dfrac{5}{6}-\left(\dfrac{3}{4}-\dfrac{1}{2}\right)$

$=\dfrac{5}{6}-\dfrac{1}{4}$

$=\dfrac{7}{12}$

$\dfrac{2}{5}+\dfrac{2}{3}-\dfrac{2}{5}$

$=\dfrac{2}{5}-\dfrac{2}{5}+\dfrac{2}{3}$

$=0+\dfrac{2}{3}$

$=\dfrac{2}{3}$