Sửa đề: \(\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\)

Giải:

Đặt:

\(A=\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\)

\(\Leftrightarrow A=\dfrac{13}{29}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{6}\right)\)

\(\Leftrightarrow A=\dfrac{13}{29}.0\)

\(\Leftrightarrow A=0\)

Vậy ...

Giải:

\(A=\dfrac{13}{29}.\dfrac{1}{2}+\dfrac{13}{29}.\dfrac{1}{3}-\dfrac{13}{29}.\dfrac{5}{6}\) (Sửa đề)

\(\Leftrightarrow A=\dfrac{13}{29}\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{5}{6}\right)\)

\(\Leftrightarrow A=\dfrac{13}{29}.0\)

\(\Leftrightarrow A=0\)

Vậy ...

a)29 . (19 – 13) – 19 . (29 – 13)

= 29 . 6 – 19 . 16

= 174 – 304

= –130.

b)31.(-18)+31.(-81)-31       

= 31. [-18 + (-81) - 1 ]

= 31. (-100)

= -3100

c)(7.3-3):(-6)  

(7.3-3):(-6)

= (21-3):(-6)

= 18 : (-6)

= 3

d)72:[(-6).2+4)]       

= 72 : ( -12 + 4 )

= 72 : -8

= -9

Bài 2:

a)(-12).47+(-12).52+(-12)       

= (-12).(47+52+1)

= -1200

b)13.(23+22)-3.(17+28)  

13 . (23 + 22) - 3 . (17 + 28)

= 13 . 45 - 3 . 45

= ( 13 - 3 ) . 45

= 10 . 45

= 450

c)18-10:(-2)-7   

= 18-5+7

= 13+7

= 20

d)99:[(-7).2+5)

1. Tính nhanh

a) 29 . (19 – 13) – 19 . (29 – 13)

= 29 . 6 – 19 . 16

= 174 – 304

= –130.

b) 31 . (-18) + 31 . (-81) - 31       

= 31. [-18 + (-81) - 1 ]

= 31. (-100)

= -3100

c) (7 . 3 - 3) : (-6)

= (21 - 3) : (-6)

= 18 : (-6)

= 3

d) 72 : [(-6) . 2 + 4)]       

= 72 : ( -12 + 4 )

= 72 : -8

= -9

2. Tính nhanh

a) (-12) . 47 + (-12) . 52 + (-12)

= (-12) . (47 + 52 + 1)

= (-12) . 100

= -1200        
b) 13 . (23 + 22) - 3 . (17 + 28)   

= 13 . 45 - 3 . 45

= (13 - 3) . 45

= 10 . 45

= 450
c) 18 - 10 : (-2) - 7   

= 18-5+7

= 13+7

= 20

d) 99 : [(-7) . 2 + 5]

= 99 : -9

= -11

\(E=\frac{\frac{-6}{7}+\frac{6}{13}-\frac{6}{29}}{\frac{9}{7}-\frac{9}{13}+\frac{9}{29}}=\frac{-6.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{29}\right)}{9.\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{29}\right)}=\frac{-6}{9}=\frac{-2}{3}\)

\(F=\frac{\frac{2}{15}-\frac{2}{21}+\frac{2}{39}}{0,25-\frac{5}{28}+\frac{5}{52}}=\frac{\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{13}\right)}{\frac{1}{4}-\frac{5}{28}+\frac{5}{52}}\)

                                          \(=\frac{\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{13}\right)}{\frac{5}{4}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{13}\right)}=\frac{2}{3}:\frac{5}{4}=\frac{2}{3}.\frac{4}{5}=\frac{8}{15}\)

Bạn soyeon ơi! Hình như câu đầu bạn làm sai rồi thì phải. Họ ra đề là + đến - nhưng bạn làm - đến +.

a. bằng77

b. bằng 13

lớp 2 đã học cái này rồi à

a. bằng77

b. bằng 13

a: \(\dfrac{5}{13}\left(\dfrac{6}{29}-\dfrac{26}{39}\right)-\dfrac{6}{29}\cdot\left(\dfrac{5}{13}-\dfrac{29}{6}\right)\)

\(=\dfrac{5}{13}\cdot\dfrac{6}{29}-\dfrac{5}{13}\cdot\dfrac{26}{39}-\dfrac{6}{13}\cdot\dfrac{5}{13}+\dfrac{6}{29}\cdot\dfrac{29}{6}\)

\(=\dfrac{-5}{39}\cdot2+1=1-\dfrac{10}{39}=\dfrac{29}{39}\)

b: \(\dfrac{1\cdot198+2\cdot197+3\cdot196+...+198\cdot1}{1\cdot2+2\cdot3+...+198\cdot199}\)

\(=\dfrac{1\left(199-1\right)+2\left(199-2\right)+...+198\cdot\left(199-198\right)}{1\left(1+1\right)+2\left(1+2\right)+...+198\left(1+198\right)}\)

\(=\dfrac{199\left(1+2+...+198\right)-\left(1^2+2^2+...+198^2\right)}{\left(1+2+...+198\right)+\left(1^2+2^2+...+198^2\right)}\)

\(=\dfrac{199\cdot\dfrac{198\cdot199}{2}-\dfrac{198\cdot\left(198+1\right)\cdot\left(2\cdot198+1\right)}{6}}{198\cdot\dfrac{199}{2}+\dfrac{198\left(198+1\right)\left(2\cdot198+1\right)}{6}}\)

\(=\dfrac{3\cdot198\cdot199^2-198\cdot199\cdot397}{6}:\dfrac{3\cdot198\cdot199+198\cdot199\cdot397}{6}\)

\(=\dfrac{198\cdot199\left(3\cdot199-397\right)}{198\cdot199\left(3+397\right)}\)

\(=\dfrac{200}{400}=\dfrac{1}{2}\)

pls mai mình phải nộp rồi

1, 35 . 18 - 5 . 7 . 28

= 35 . 18 - 35 . 28

= 35 . ( 18 - 28 )

= 35 . ( -10 )

= -350

2, 45 - 5 . ( 12 + 9 )

= 40 . 21

= 840

3, 24 . ( 16 - 5 ) - 16 . ( 24 - 5 )

= 24 . 11 - 16 . 19

= 264 - 304

= -40

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