Tính bằng cách hợp lý
(450×34×10):15
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a. \(\dfrac{10}{21}>1\)
\(\dfrac{9}{23}< 1\)
\(\Rightarrow\dfrac{10}{21}>\dfrac{9}{23}\)
b. \(\dfrac{32}{33}>\dfrac{31}{33}>\dfrac{31}{34}\Rightarrow\dfrac{32}{33}>\dfrac{31}{34}\)
c. \(\dfrac{44}{47}< \dfrac{45}{47}< \dfrac{45}{46}\Rightarrow\dfrac{44}{47}< \dfrac{45}{46}\)
\(d.\dfrac{70}{117}< \dfrac{70}{115}=\dfrac{14}{23}\Rightarrow\dfrac{70}{117}< \dfrac{14}{23}< \dfrac{15}{23}\)
\(\Rightarrow\dfrac{70}{117}< \dfrac{15}{23}\)
a, 34.(15-10)-15.(34-10)
=34.15-34.10-15.34-15.10
=(34.15-15.34)-(34.10-15.10)
=0-(34-15).10
=0-190
=-190
a) 34.(15-10)-15.(34-10)
= (34.15-34.10)-(15.34-15.10)
= (510-340)-(510-150)
= 170-360
= -190
A=(1+2+3+...+200) : (6+8+10+...+34)
A=\(\frac{\left[\left(200-1+1\right).\left(200+1\right)\right]}{2}:\frac{\left[\left(34-6\right):2+1\right].\left(34+6\right)}{2}\)
A=20100 : 300
A=67.
B=( 41x66+34x66):(3+7+11+...+79)
B=[(41+34).66]:\(\frac{\left[\left(79-3\right):4+1\right].\left(79+3\right)}{2}\)
B=(75.66):820
B=4950:820
B=\(\frac{495}{82}\)
\(\dfrac{\left[\left(\dfrac{3}{10}-\dfrac{4}{15}-\dfrac{7}{20}\right)\cdot\dfrac{5}{19}\right]}{\left[\dfrac{1}{14}+\dfrac{1}{7}-\dfrac{\left(-3\right)}{35}\right]\cdot\left(-\dfrac{4}{3}\right)}\)
\(=\dfrac{\left(\dfrac{18}{60}-\dfrac{16}{60}-\dfrac{21}{60}\right)\cdot\dfrac{5}{19}}{\left(\dfrac{5}{70}+\dfrac{10}{70}+\dfrac{6}{70}\right)\cdot\dfrac{-4}{3}}=\dfrac{\dfrac{-19}{60}\cdot\dfrac{5}{19}}{\dfrac{21}{70}\cdot\dfrac{-4}{3}}\)
\(=-\dfrac{1}{12}:\dfrac{-2}{5}=\dfrac{1}{12}\cdot\dfrac{5}{2}=\dfrac{5}{24}\)
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. $=(31-21)-(27-24)=10-3=7$
=30*10*34
=300*34
=10200
=30×10×34
=300×34
=10200