Tính bằng hai cách :
a, 3 1 b, 3 1 c, 1 3 d, 1 2
7—- + 4 —- 8 — – 5 —- -7— + 5 — -3— – 5—-
8 4 5 2 3 7 3 11
Đáp án:
$a)\dfrac{93}{8}\\
b)
\dfrac{31}{10}\\
c)
\dfrac{-40}{21}\\
d)
\dfrac{-281}{33}$
Giải thích các bước giải:
$a)C1: 7\dfrac{3}{8}+4\dfrac{1}{4}=\dfrac{7.8+3}{8}+\dfrac{4.4+1}{4}\\
=\dfrac{59}{8}+\dfrac{17}{4}=\dfrac{59}{8}+\dfrac{17.2}{8}=\dfrac{93}{8}\\
C2: 7\dfrac{3}{8}+4\dfrac{1}{4}=(7+4)+\left ( \dfrac{3}{8}+\dfrac{1}{4} \right )\\
=11+\left (\dfrac{3}{8}+\dfrac{2}{8} \right )=11+\dfrac{5}{8}=\dfrac{11.8}{8}+\dfrac{5}{8}=\dfrac{93}{8}\\
b)
C1:8\dfrac{3}{5}-5\dfrac{1}{2}=\dfrac{8.5+3}{5}-\dfrac{5.2+1}{2}\\
=\dfrac{43}{5}-\dfrac{11}{2}=\dfrac{43.2}{10}-\dfrac{11.5}{10}=\dfrac{31}{10}\\
C2:8\dfrac{3}{5}-5\dfrac{1}{2}=(8-5)+\left ( \dfrac{3}{5}-\dfrac{1}{2} \right ) \\
=3+\left ( \dfrac{6}{10}-\dfrac{5}{10} \right )=3+\dfrac{1}{10}=\dfrac{31}{10}\\
c)
C1: -7\dfrac{1}{3}+5\dfrac{3}{7}=\dfrac{-(7.3+1)}{3}+\dfrac{5.7+3}{7}\\
=\dfrac{-22}{3}+\dfrac{38}{7}=\dfrac{(-22).7}{21}+\dfrac{38.3}{21}=\dfrac{-40}{21}\\
C2: -7\dfrac{1}{3}+5\dfrac{3}{7}=(-7+5)+\left ( -\dfrac{1}{3}+\dfrac{3}{7} \right )\\
=-2+\left ( -\dfrac{7}{21}+\dfrac{9}{21} \right )=-2+\dfrac{2}{21}=\dfrac{(-2).21+2}{21}=\dfrac{-40}{21}\\
d)
C1: -3\dfrac{1}{3}-5\dfrac{2}{11}=-\dfrac{3.3+1}{3}-\dfrac{5.11+2}{11}\\
=\dfrac{-10}{3}-\dfrac{57}{11}=\dfrac{(-10).11}{33}-\dfrac{57.3}{33}=\dfrac{-281}{33}\\
C2:-3\dfrac{1}{3}-5\dfrac{2}{11}=(-3-5)+\left ( -\dfrac{1}{3}-\dfrac{2}{11} \right )\\
=-8+\left ( -\dfrac{11}{33}-\dfrac{6}{33} \right )=-8-\dfrac{17}{33}\\
=\dfrac{(-8).33}{33}-\dfrac{17}{33}=\dfrac{-281}{33}$