tính bằng cách hợp lý D=(1/9+-5/17)-(12/17-(-1)/2)
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a)=\(\frac{-2.4}{5.7}+\frac{-3.2}{5.7}+\frac{-3}{5}\)
=\(\frac{-2.4}{5.7}+\frac{-2.3}{5.7}+\frac{-3}{5}\)
=\(\frac{-2}{5}\left(\frac{4+3}{7}\right)+\frac{-3}{5}\)
=\(\frac{-2}{5}.1+\frac{-3}{5}\)
=-1
b)
\(a)\)
\(\left(-31\right)+\left(50-19\right)-\left(150-31\right)\)
\(=\left(-31\right)+50-19-150+31\)
\(=\left(-150\right)-19\)
\(=-169\)
\(b)\)
\(25.\left(45-17\right)-45.\left(25-17\right)\)
\(=25.45-25.17-45.25+45.17\)
\(=0\)
\(c)\)
\(\frac{-1}{12}+\frac{4}{3}=\frac{5}{4}\)
\(d)\)
\(3+\frac{-5}{20}+\frac{30}{75}+\frac{-7}{4}\)
\(=\left(\frac{3}{5}+\frac{30}{75}\right)-\left(\frac{5}{20}+\frac{7}{4}\right)\)
\(=1-2\)
\(=-1\)
a) (-31)+(50-19)-(150-31)
= (-31)+50+(-19)-150+(-31)
= (-31)+50-150+(-19)-(-31)
= (-31)+(-100)+12
= -119
b) 25(45-17)-45(25-17)
= 25.45-25.17-45.25-45.17
= 25(45-45)-25(17-17)
= 0
c) -1/12 + 4/3
= -1/12 + 16/12
= 15/12
= 5/4
d) 3/5+(-5)/20+30/75+(-7)/4
= 45/75+30/75+(-5)/20+(-35)/20
= 1+(-2)
= -1
a: \(=\dfrac{-5}{7}-\dfrac{2}{7}+\dfrac{3}{4}+\dfrac{1}{4}-\dfrac{1}{5}=-\dfrac{1}{5}\)
b: \(=\dfrac{-3}{31}-\dfrac{28}{31}+\dfrac{-6}{17}-\dfrac{11}{17}+\dfrac{1}{29}-\dfrac{1}{5}=\dfrac{-24}{145}\)
Bài 2:
a. $=62-81-12+59-9=(62-12)+(59-9)-81$
$=50+50-81=100-81=19$
b. $=39+13-26-62-39=(39-39)+13-(26+62)$
$=0+13-88=-(88-13)=-75$
c. $=(32-42)+(36-34)+(40-38)=10+2+2=14$
d. $=92-55+8-45=(92+8)-(55+45)=100-100=0$
Bài 1:
a. $=(387-87)-224=300-224=76$
b. $=-(75+35)+379=-110+379=379-110=269$
c. $=(11+15)-(13+17)=25-30=-5$
d. $=(31-21)-(27-24)=10-3=7$
\(\dfrac{5}{3}\times\dfrac{3}{17}+\dfrac{3}{17}\times\dfrac{1}{3}-\dfrac{3}{17}\times2\)
\(=\dfrac{3}{17}\times\left(\dfrac{5}{3}+\dfrac{1}{3}+2\right)=\dfrac{3}{17}\times4=\dfrac{12}{17}\)