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\(-12\left(7-72\right)-25\left(55-43\right)\)
\(=-12\cdot\left(-65\right)-25\cdot12\)
\(=12\cdot65-25\cdot12\)
\(=12\left(65-25\right)\)
\(=12\cdot40=480\)
−12(7−72)−25(55−43)
=−12⋅(−65)−25⋅12
=12⋅65−25⋅12
=12(65−25)
=12⋅40
=480
a) (-12). (7 - 72) - 25. (55 - 43)
= (-12). (- 65) - 25. 12
= 12. 65 – 12. 25
= 12. (65 - 25)
= 12. 40
= 480
b) (39 - 19) : (- 2) + (34 - 22). 5
= 20 : (- 2) + 12. 5
= - 10 + 60
= 60 - 10
= 50.
TL:
\(\text{a) (-12).(7-72)-25.(55-43) = (-12).(-65)-25.12 = 12.(65-25) = 12.40 = 480 b) (39-19):(-2)+(34-22).5 = 20:(-2)+12.5 = -10+60 = 50}\)
HT
a) (-12).(7-72)-25.(55-43)
=(-12).(-65)-25.12
=12.(65-25)
=12.40
=480
b) (39-19):(-2)+(34-22).5
=20:(-2)+12.5
=-10+60
^HT^
(- 47 ) - 53 = (- 47 ) + ( - 53 ) = -100
(- 43 ) - ( -43 ) =0
Ta có : -(4+7) =-11
( -4 ) - 7 = -11
Vì - ( 4+7 ) = -11 , ( -4 ) - 7 =-11 nên -(4 +7) = ( - 4) -7
Ta có : -( 12 -25 ) = 13
( -12 + 25 )=13
Vì -(12 -25 ) = 13; (-12 +25) = 13 nên - (12 - 25 ) = ( -12 +25 )
\(x+\frac{13}{12}+\frac{21}{20}+...+\frac{91}{90}\)
\(=x+1+\frac{1}{12}+1+\frac{1}{20}+...+1+\frac{1}{90}\)
\(=x+\left(1+1+1+1+1+1+1\right)+\left(\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=x+7+\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=x+7+\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=x+7+\left(\frac{1}{3}-\frac{1}{10}\right)\)
\(=x+7+\frac{7}{30}\)
\(=x+7\frac{7}{30}\)
a) 25.(-45)+(-25).55
=25.(-45)+25.(-55)
=25.(-45-55)
=25.(-100)
=-2500
b)(-75).18+18.(-25)+(-72).100
=18.(-75-25)+(-72).100
=18.(-100)+72.(-100)
=(-100).(18+72)
=(-100).100
=-10000
(-12) x (7 - 72) - 25 x (55 - 43)
= -12 x (-65) - 25 x 12
= 12 x [-(-65) - 25]
= 12 x [65 - 25]
= 12 x 40
= 480