a) `123 - 27 + 27 + 13 - 123`

`= (123 - 123) + (27 - 27) + 13`

`= 0 + 0 + 13`

`= 13`

`---`

b) `175 + 25 + 13 + 15 - 175 - 25`

`= (175-175)+(25-25)+25+13`

`=0+0+25+13`

`=28`

`---`

c) \(53\times39+47\times39+53\times21+47\times21\)

\(=\text{[}\left(53+47\right)\times39]+\left[\left(53+47\right)\times21\right]\)

\(=\left(100\times39\right)+\left(100\times21\right)\)

\(=3900+2100\)

`=6000`

`---`

d) \(49\times35+49\times16-49\times61\)

`=` \(49\times\left(35+16-61\right)\)

`=` \(49\times\left(-10\right)\)

`=-490`

Bạn tự giải

54 x 47 - 47 x 53 - 20 - 27

= 54 x 47 - 47 x 53 - 47

= 47 x (54-53-1)

= 47 x 0

= 0

54 x 47 - 47 x 53 - 20 - 27

= 47 x (54 - 53) - 20 - 27

= 47 x 1 - 20 - 27

= 47 - (20 + 27)

= 47 - 47

= 0

a) 27 x 35 + 73 x 6 - 27 x 29

= 27 x ( 35 - 29 ) + 73 x 6

= 27 x 6 + 73 x 6

= 6 x ( 27 + 73 )

= 6 x 100

= 600

b) 29 x ( 47 + 53 ) - 47 x ( 29 + 53 )

= 29 x 47 + 29 x 53 - 47 x 29 + 47 x 53

= 29 x ( 47 + 53 - 47 ) + 47 x 53

= 29 x 53 + 47 x 53

= 53 x ( 29 + 47 )

= 53 x 76

= 4028

c) 5 x 17 x 4 x 25 x 20

= ( 5 x 20 ) x ( 4 x 25 ) x 17

= 100 x 100 x 17

= 10000 x 17

= 170000

Mới lớp 3!

lớp 3 cút , ng ta nói hơn lớp 6 mak

(145 x 99 + 145 ) - ( 143 x 102 – 143 × 2 ) + 54 x 47 - 47 x 53 - 20 - 27

= 145 × (99 + 1) – 143 × (102 – 2) + 47 × (54 – 53) – (20 + 27)

= 145 × 100 – 143 × 100 + 47 × 1 – 47

= 100 × (145 – 143) + (47 – 47)

= 100 × 2 + 0

= 200

(145 x 99 + 145 ) - ( 143 x 102 – 143 × 2 ) + 54 x 47 - 47 x 53 - 20 - 27

= 145 × (99 + 1) – 143 × (102 – 2) + 47 × (54 – 53) – (20 + 27)

= 145 × 100 – 143 × 100 + 47 × 1 – 47

= 100 × (145 – 143) + (47 – 47)

= 100 × 2 + 0

= 200

a. 54 x 113 + 45 x 113 + 113

= 113 x (54 + 45 + 1)

= 113 x 100 = 11300

Vậy giá trị cần tìmlaf: 11300

b. 54 x 47 - 47 x 53 - 20 - 27

= 54 x 47 - 47 x 53 - (20 + 27)

= 54 x 47 - 47 x 53 - 47

= 47 x (54 - 53 - 1)

= 47 x 0 = 0

Vậy giá trị cần tìm là: 0

\(a,54\times113+45\times113+113\\ =113\times\left(54+45+1\right)\\ =113\times100=11300\)

\(b,54\times47-47\times53-20-27\\ =54\times47-47\times53-47\\ =47\times\left(54-53-1\right)\\ =47\times0=0\)

a 11300

 b  0

(145 x 99 + 145 ) - ( 143 x 102 – 143 × 2 ) + 54 x 47 - 47 x 53 - 20 - 27

= 145 × (99 + 1) – 143 × (102 – 2) + 47 × (54 – 53) – (20 + 27)

= 145 × 100 – 143 × 100 + 47 × 1 – 47

= 100 × (145 – 143) + (47 – 47)

= 100 × 2 + 0

= 200

= 145 x (99+1) - 143 x ( 102-2) + 47 x ( 54 - 53) -20-27

= 145 x 100 - 143 x 100 + 47 x 1 -20-27

= 100 x ( 145 - 143 ) + 47 - 20 - 27

= 200 + 47 - 20 - 27

= 200

một tấm bìa hình thoi có tổng độ dài hai đường chéo bằng 126cm, biết độ dài đường chéo thứ nhất bằng 2/5 độ dài đường chéo thứ hai.Tính diện tích tấm bìa hình thoi đó.

Ta có: \(13\cdot\left(-47\right)-53\cdot13\)

\(=13\cdot\left(-47-53\right)\)

\(=13\cdot\left(-100\right)=-1300\)

Ta có: \(27\cdot\left(-67\right)+33\cdot\left(-27\right)\)

\(=27\cdot\left(-67\right)-33\cdot27\)

\(=27\left(-67-33\right)\)

\(=27\cdot\left(-100\right)=-2700\)

Ta có: \(-49\cdot65+36\cdot\left(-49\right)+49\)

\(=-49\cdot65+\left(-49\right)\cdot36+\left(-49\right)\cdot\left(-1\right)\)

\(=-49\cdot\left(65+36-1\right)\)

\(=-49\cdot100=-4900\)

Ta có: \(136\cdot\left(-47\right)+36\cdot47\)

\(=\left(-136\right)\cdot47+36\cdot47\)

\(=47\cdot\left(-136+36\right)\)

\(=47\cdot\left(-100\right)=-4700\)

Ta có: \(\left(-48\right)\cdot72+36\cdot\left(-304\right)\)

\(=-96\cdot36+36\cdot\left(-304\right)\)

\(=36\cdot\left(-96-304\right)\)

\(=36\cdot\left(-400\right)=-14400\)

\(2b,\)\(54\times47-47\times53-20-27.\)

\(=47\times54-47\times53-47\)

\(=47\times\left(54-53-1\right)\)

\(=47\times0=0\)

\(2c,\)\(10000-47\times72-47\times28\)

\(=100\times100-47\times\left(72+28\right)\)

\(=100\times100-47\times100\)

\(=100\times\left(100-47\right)\)

\(=100\times53=5300\)

2b

54 x 47 - 47 x 53 - 20 - 27 = 47 x ( 54 - 53 ) - 20 - 27 = 47 - 20 - 27 = 47 - ( 20 + 27 ) = 47 - 47 = 0

2c

10000 - 47 x 72 - 47 x 28 = 10000 - 47 x ( 72 - 28 ) = 10000 - 47 x 44 = 10000 - 2068 = 7932