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a) 52 – 42 + 37 – 28 + 38 + 63 = (52 – 42) + (37 + 63) + (38 – 28) = 10 + 100 + 10 = 120
b) 130 + 35.19 – 9.70 – 30
= (35.19 – 9.70) + (130 – 30) = (35.19 – 9.35.2) + 100 = 35.(19 – 18) + 100 = 35 + 100 = 135
c) (11 + 13 + 15 + … + 19).(6.8 – 48) = (11+13+15+…+19).0 = 0
d) (72 – 8.9):(20+22+24 – 19 +21+23) = 0:(20+22+24 – 19 +21+23) = 0
a) \(\left(-25\right).5.\left(-4\right).2\)
\(=\left[\left(-25\right).\left(-4\right)\right].\left(2.5\right)\)
\(=100.10=1000\)
b) \(\left(-456\right)-\left(67-456\right)\)
\(=-456-67+456\)
\(=\left(456-456\right)-67\)
\(=67\)
Lời giải:
a. $=-(37+63)+[25+(-25)]+(-9)=-100+0+(-9)=-(100+9)=-109$
b. $=[1+(-3)]+[5+(-7)]+....+[21+(-23)]$
$=\underbrace{(-2)+(-2)+....+(-2)}_{6}=(-2).6=-12$
c. $=-(280+20)+[-(79+21)]=-300+(-100)=-(300+100)=-400$
d. $=[-(27+43)]+[-(208+102)]=-70+(-310)=-(70+310)=-380$
e. $=(38+120)-(12+46)=158-58=100$
f. $=9+15+11+24=(9+11)+(15+24)=20+39=59$
Bài 1:
a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)
\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)
hay \(x=-\dfrac{1}{3}\)
Vậy: \(x=-\dfrac{1}{3}\)
b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)
hay \(x=\dfrac{50}{9}\)
Vậy: \(x=\dfrac{50}{9}\)
c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)
\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)
\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)
hay \(x=\dfrac{22}{15}\)
Vậy: \(x=\dfrac{22}{15}\)
d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)
\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)
\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)
hay \(x=\dfrac{15}{19}\)
Vậy:\(x=\dfrac{15}{19}\)
a) \(\left[\left(-6\right)+15\right].\left(-2\right)=9.\left(-2\right)=-18\)
b) \(\left[\left(-7\right)-13\right]\div\left(-5\right)=\left(-20\right):\left(-5\right)=4\)
c) \(\left(-37\right)-\left(-28\right)+150+\left(-63\right)+72\)
\(\left(-37\right)+28+150-63+72=\left[\left(-37\right)-63\right]+150\)