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a,(-23).(-3).(+4).(-7)
= 69.(+4).(-7)
= 276.(-7)
= -1932
b,2.8.(-14).(-3)
= 16.(-14).(-3)
= -224.(-3)
= 672
a) \(\left(-28\right).\left(-3\right).\left(+4\right).\left(-7\right)=\left(+84\right).\left(+4\right).\left(-7\right)=\left(+336\right).\left(-7\right)=\left(-2352\right)\)
b) \(2.8.\left(-14\right).\left(-3\right)=16.\left(-14\right).\left(-3\right)=\left(-224\right).\left(-3\right)=672\)
1, Thực hiện phép tính
a, (-23).(-3).(+4).(-7)
=69.(-28)
=(-1932)
b, 2.8.(-14).(-3)
=16.42
=672
2, Tìm x
a, 2436 : x = 12 b, 6. x - 5 = 613
x =2436 : 12 6. x = 613 + 5 = 618
x = 203 x =618 : 6 = 103
c, 12. (x - 1) = 0 d, 0 : x = 0
x - 1 = 12 : 0 = 0 x = 0 : 0
x = 0 + 1 = 1 x = 0
1.
a) \(\left(-23\right).\left(-3\right).4.\left(-7\right)=[\left(-23\right).\left(-3\right).\left(-7\right)].4\)
\(=-483.4=-1932\)
b) \(=16.42=672\)
2.
a) \(\Rightarrow x=2436\div12=203\)
b) \(\Rightarrow6x=618\Rightarrow x=103\)
c) \(\Rightarrow x-1=0\Rightarrow x=1\)
d) Ko có gtr nào của x thỏa mãn 0:x=0
~ HỌC TỐT ~
# Q.TRANG #
\(\frac{4}{5}\)+ \(\frac{1}{5}\): \(\frac{-2}{15}\)= \(\frac{4}{5}\)+ \(\frac{1}{5}\)* \(\frac{-15}{2}\)=\(\frac{4}{5}\) + \(\frac{-3}{2}\)= \(\frac{-7}{10}\)
b = \(\frac{-7}{12}\). (\(\frac{3}{11}\)+ \(\frac{-14}{11}\)) + \(\frac{5}{6}\)= \(\frac{-7}{12}\). (-1) + \(\frac{5}{6}\)= \(\frac{7}{12}\)+ \(\frac{10}{12}\)= \(\frac{17}{12}\)
a) \(\dfrac{7}{30}+\dfrac{\left(-12\right)}{37}+\dfrac{23}{30}+\dfrac{\left(-25\right)}{37}=\left(\dfrac{7}{30}+\dfrac{23}{30}\right)+\left(\dfrac{-12}{37}+\dfrac{-25}{37}\right)=1+\left(-1\right)=0\)
b) \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}=\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)=-\dfrac{5}{11}\)
c) \(\dfrac{\left(-5\right)}{7}\cdot\dfrac{3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+1\dfrac{5}{7}=\dfrac{5}{7}\cdot\dfrac{-3}{13}-\dfrac{5}{7}\cdot\dfrac{10}{13}+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(\dfrac{-3}{13}-\dfrac{10}{13}\right)+\dfrac{12}{7}=\dfrac{5}{7}\cdot\left(-1\right)+\dfrac{12}{7}=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=\dfrac{7}{7}=1\)
