Bài 1:
\(a,=\frac{2}{3}-\frac{16}{3}=\frac{-14}{3}\)
\(b,=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{6}{19}+\frac{-13}{19}\right)=1-1=0\)
\(c,=\frac{3}{5}.\left(\frac{8}{9}-\frac{7}{9}+\frac{26}{9}\right)=\frac{3}{5}.3=\frac{9}{5}\)
a,\(\dfrac{1}{2}\).\(\dfrac{4}{3}\)-\(\dfrac{20}{3}\).\(\dfrac{4}{5}\)=\(\dfrac{2}{3}\)-\(\dfrac{16}{3}\)=-\(\dfrac{14}{3}\)
2/3 - 16/3= -14/3
(3/4+4/7)+[(-6/19)+(-13/19)= 1+ (-1)=0
a) 2/3 - 16/3 = -14/3
a,2/3-16/3=-14/3
b,(3/7+4/7)+(-6/19+-13/19)=1+(-1)=0
c,3/5.(8/9-7/9+26/9)=3/5.3=9/5
a)\(\dfrac{2}{3}-\dfrac{16}{3}=\dfrac{-14}{3}\).
b)\(\left(\dfrac{3}{7}+\dfrac{4}{7}\right)+\left(\dfrac{-6}{19}+\dfrac{-13}{19}\right)=1+\left(-1\right)=0\)
c)\(\dfrac{3}{5}.\left(\dfrac{8}{9}-\dfrac{7}{9}+\dfrac{26}{9}\right)=\dfrac{3}{5}.3=\dfrac{9}{5}\)
A,=2/3-16/3
=-14/3
B,
Bài 1. (3 điểm) Thực hiện phép tính.
a) \dfrac{1}{2} \cdot \dfrac{4}{3}-\dfrac{20}{3} \cdot \dfrac{4}{5}21⋅34−320⋅54 ;
b) \dfrac{3}{7}+\dfrac{-6}{19}+\dfrac{4}{7}+\dfrac{-13}{19}73+19−6+74+19−13 ;
c) \dfrac{3}{5} \cdot \dfrac{8}{9}-\dfrac{7}{9} \cdot \dfrac{3}{5}+\dfrac{3}{5} \cdot \dfrac{26}{9}53⋅98−97⋅53+53⋅926.
Hướng dẫn giải:
a) \dfrac{1}{2} \cdot \dfrac{4}{3}-\dfrac{20}{3} \cdot \dfrac{4}{5}21⋅34−320⋅54
=\dfrac{2}{3}-\dfrac{16}{3}=32−316
=-\dfrac{14}{3}=−314 ;
b) \dfrac{3}{7}+\dfrac{-6}{19}+\dfrac{4}{7}+\dfrac{-13}{19}73+19−6+74+19−13
=\left( \dfrac{3}{7}+\dfrac{4}{7} \right)+\left( \dfrac{-6}{19}+\dfrac{-13}{19} \right)=(73+74)+(19−6+19−13)
=\dfrac{7}{7}+\dfrac{-19}{19}=77+19−19
=1+\left( -1 \right)=0=1+(−1)=0 ;
c) \dfrac{3}{5} \cdot \dfrac{8}{9}-\dfrac{7}{9} \cdot \dfrac{3}{5}+\dfrac{3}{5} \cdot \dfrac{26}{9}53⋅98−97⋅53+53⋅926
=\dfrac{3}{5} \cdot \left( \dfrac{8}{9}-\dfrac{7}{9}+\dfrac{26}{9} \right)=53⋅(98−97+926)
=\dfrac{3}{5} \cdot \dfrac{27}{9}=53⋅927
=\dfrac{3}{5} \cdot 3=53⋅3
=\dfrac{9}{5}=59.
a) \dfrac{1}{2} \cdot \dfrac{4}{3}-\dfrac{20}{3} \cdot \dfrac{4}{5}
=\dfrac{2}{3}-\dfrac{16}{3}
=-\dfrac{14}{3} ;
b) \dfrac{3}{7}+\dfrac{-6}{19}+\dfrac{4}{7}+\dfrac{-13}{19}
=\left( \dfrac{3}{7}+\dfrac{4}{7} \right)+\left( \dfrac{-6}{19}+\dfrac{-13}{19} \right)
=\dfrac{7}{7}+\dfrac{-19}{19}
=1+\left( -1 \right)=0 ;
c) \dfrac{3}{5} \cdot \dfrac{8}{9}-\dfrac{7}{9} \cdot \dfrac{3}{5}+\dfrac{3}{5} \cdot \dfrac{26}{9}
=\dfrac{3}{5} \cdot \left( \dfrac{8}{9}-\dfrac{7}{9}+\dfrac{26}{9} \right)
=\dfrac{3}{5} \cdot \dfrac{27}{9}
=\dfrac{3}{5} \cdot 3
=\dfrac{9}{5}.
A) -14/3 b) 0 c) 9/5