27 : x =  62 - 59

27 : x =  3

x = 29 : 3

x = 9

X x 3 = 46 - 16

X x 3 = 30

X = 30 : 3

X = 10

\(a,\Leftrightarrow27:x=3\)

\(\Leftrightarrow x=27:3\)

\(\Leftrightarrow x=9\)

\(b,\Leftrightarrow x\times3=30\)

\(\Leftrightarrow x=10\)

27:x= 62-59

=>27:x=3

=>x=29:3

x=9

x*3+16=46

=>x*3 = 46 - 16

=>X*3 = 30

=>x = 30 : 3

=> x =10

a, -(-12) + (+19) - (+12) + 8 - 19

= 12 + 19 - 12 + 8 - 19

= ( 12 - 12) + ( 19- 19) + 8

= 0 + 0 + 8

=  8

b, (59 - 78) - (42 - 78 + 59)

= 59 - 78 - 42 + 78 - 59

= (59 - 59) - 42 - ( 78 - 78)

= 0  - 42 - 0

 = -42

c, ( - 68 + 103) - (-50 - 68 + 103)

= -68 + 103 + 50 + 68 - 103

= (-68 + 68) + ( 103 - 103) + 50

= 0 + 0 + 50

= 50 

a) \(...=12+19-12+8-19=8\)

b) \(...=-19-23=-42\)

c) \(...=35-\left(-15\right)=35+15=50\)

=46+[72+59+11]

=46+262

=272

tính hơn nhanh nên ko có chắc đấp án đúng nha

46 + 72 + 46 + 59 + 46 + 11 = 280

Tk nha!!

46 + 72 + 46 + 59 + 46 + 11

= ( 46 + 46 + 46 ) + ( 72 + 59 + 11 )

= ( 46 × 3 ) + 142 

= 138 + 142 

= 280

a:1,25*59*1,25*41

=1,25*(59*41)

=1,25*2419

=3023,75

b:77,28*20,2+22,72*20,2

=20,2*(77,28+22,72)

=20,2*100

=2020

c

:1,25*59*1,25*41

=1,25*(59*41)

=1,25*2419

=3023,75

d:2,5*46+54*2,5

=2,5*(46+54)

=2,5*100

=250

Do 16^x +24 đồng dư 0 ( mod10) suy ra  824y đồng dư 0 ( mod 10) nên y chia hết 5, y nguyên tố  cho nên y = 5 và x =3.

a) \(A=1-\dfrac{2002}{2003}\)

\(A=\dfrac{1}{2003}\)

b) \(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\)

\(B=\left(\dfrac{179}{30}-\dfrac{59}{30}\right)-\dfrac{3}{5}\)

\(B=\dfrac{120}{30}-\dfrac{3}{5}\)

\(B=4-\dfrac{3}{5}\)

\(B=\dfrac{17}{5}\)

c) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right).11\)

\(C=\dfrac{501}{55}.11\)

\(C=\dfrac{501}{5}\)

a) \(A=1-\dfrac{2002}{2003}=\dfrac{2003}{2003}-\dfrac{2002}{2003}=\dfrac{1}{2003}\)

b) \(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)=\dfrac{179}{30}-\dfrac{59}{30}+\dfrac{3}{5}=4+\dfrac{3}{5}=\dfrac{23}{5}\)

c) \(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11=\dfrac{46}{5}\cdot11-\dfrac{1}{11}\cdot11=\dfrac{506}{5}-1=\dfrac{501}{5}\)

a)

\(A=1-\dfrac{2002}{2003}\\ \Rightarrow A=\dfrac{2003}{2003}-\dfrac{2002}{2003}\\ \Rightarrow A=\dfrac{1}{2003}\)

b)

\(B=\dfrac{179}{30}-\left(\dfrac{59}{30}-\dfrac{3}{5}\right)\\ \Rightarrow B=\dfrac{179}{30}-\dfrac{59}{30}+\dfrac{3}{5}\\ \Rightarrow B=\left(\dfrac{179}{30}-\dfrac{59}{30}\right)+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{120}{30}+\dfrac{3}{5}\\ \Rightarrow B=4+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{20}{5}+\dfrac{3}{5}\\ \Rightarrow B=\dfrac{23}{5}\)

c)

\(C=\left(\dfrac{46}{5}-\dfrac{1}{11}\right)\cdot11\\ C=\left(\dfrac{506}{55}-\dfrac{5}{55}\right)\cdot11\\ \Rightarrow C=\dfrac{501}{55}\cdot11\\ \Rightarrow C=\dfrac{501}{5}\)

A

A

D

1. B
2. A
3. D
4; 5. bị lỗi hình
6. lỗi nốt:))