a)=(-14).(-5)
=70
b)=(32-2):(-5)
=30:(-5)
=-6
c)=-72-12
=-84
d)=-35+(-4)-39
=-39-39
=-78
e)=78:(-30+36)
=78:6
=13
f)=-3.(-3)-12
=9-12
=-3
g)=45-5-2+12
=40-2+12
=38+12
=50

mình giải luôn chứ mik hông chép đề đâu nha^^

a) (-6 - 8) . (-9 + 4)

= (-14) . (-5)

= 70.

b) (8. 4 - 2) : (-5)

= (32 - 2) : (-5)

= 30 : (-5)

= -6.

c) -8. 9 - (-9 + 21)

= -8. 9 - 12

= -72 - 12

= -84.

d) -5. 7 + (-32) : 8 - (-15 + 54)

= -5. 7 + (-32) : 8 - 39

= -35 + (-32) : 8 - 39

= -35 + (-4) - 39

= -39 - 39

= -78.

e) 78 : (-6. 5 + 36)

= 78 : (-30 + 36)

= 78 : 6

= 13.

f) 15 : (-5). (-3) - 12

= -3. (-3) - 12

= 9 - 12

= -3.

g) 45 - 15 : +3 - 18 : 6 + 12

= 45 - 5 - 18 : 6 + 12

= 45 - 5 - 3 + 12

= 40 - 3 + 12

= 37 + 12

= 49.

bài 1

a)(-6-8).(-9+4)=14.(-5)=-70

b)(8.4-2):(-5)=30:(-5)=-6

c)-8.9-(-9+21)

= -8.9-12

= -72-12

= -84

d)-5.7+(-32):8-(-15+54)

= -5.7+(-32):8-39

=-35+(-4)-39

=39+(-39)

=0

e)78:(-6.5+36)

=78:6

=13

f)15:(-5).(-3)-12

=-3.(-3)-12

=9-12

=-3

g)45-15:3-18:+6+12

=45-5-3+12

=40-3+12

=37+12

=49

a) \(\dfrac{2727-101}{3.303+404}=\dfrac{2626}{909+404}=\dfrac{2626}{1313}=2\)

b) \(\dfrac{8.9-4.15}{12.7-180}=\dfrac{72-60}{84-180}=\dfrac{12}{-96}=\dfrac{-1}{8}\)

c) \(\dfrac{-19}{3^2.7.11}=\dfrac{-19}{9.7.11}=\dfrac{-19}{63.11}=\dfrac{-19}{693}\)

d) \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+120.6^9}{2^{12}.3^{12}-6^{11}}=\dfrac{2^2.6^{10}+20.6.6^9}{6^{12}-6^{11}}=\dfrac{4.6^{10}+20.6^{10}}{6^{11}\left(6-1\right)}=\dfrac{\left(4+20\right).6^{10}}{5.6^{11}}=\dfrac{24}{30}=\dfrac{4}{5}\)

A) \(\dfrac{4^5.4^2}{16^4}=\dfrac{4^7}{\left(2^4\right)^4}=\dfrac{2^{14}}{2^{16}}=\dfrac{1}{4}\)

b)\(\dfrac{2^8.9^4}{6^6.8^3}=\dfrac{2^8.\left(3^2\right)^4}{2^6.3^6.\left(2^3\right)^3}=\dfrac{2^8.3^8}{2^{15}.3^6}=\dfrac{9}{128}\)

c) \(\dfrac{6^3+3.6^2+3^3}{-13}=\dfrac{2^3.3^3+3.2^2.3^2+3^3}{-13}=\dfrac{2^3.3^3+3^3.2^2+3^3}{-13}=\dfrac{3^3.\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3.13}{-13}=-9\)

Tự làm tiếp đi nhé , 2 câu sau dạng cơ bản , hồi lớp 6 học rồi

`#` `\text{dkhanhqlv}`

`4)`

`a)3.(-5/11)`

`=-15/11`

`b)3/5+4/7 . 14/6`

`=3/5 + 4/3`

`=9/15+20/15`

`=29/15`

`c) 10/21-3/8 . 4/15`

`=10/21-1/10`

`=100/210-21/210`

`=79/100`

`d)(2/3+3/4)(5/7+5/14)`

`=(8/12+9/12)(10/14+5/14)`

`=17/12 . 15/14`

`=85/56`

`5)`

`a)x-1/2=3/10 . 5/6`

`=>x-1/2=1/4`

`=>x=1/4+1/2`

`=>x=1/4+2/4`

`=>x=3/4`

`b)x/5 = -3/14`

`=>x : 5 = -3/14`

`=>x=-3/14 . 5`

`=>x=-15/14`

`c)x+2/3=9/15 . 5/27`

`=>x+2/3=1/9`

`=>x=1/9-2/3`

`=>x=1/9-6/9`

`=>x=-5/9`

b) Ta có: \(B=\dfrac{7}{8}\cdot\dfrac{4}{9}+\dfrac{7}{8}\cdot\dfrac{5}{9}\)

\(=\dfrac{7}{8}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)\)

\(=\dfrac{7}{8}\cdot1=\dfrac{7}{8}\)

1. Tính:

a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)

b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)

c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)

d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)

2. Tính :

a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)

b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)

c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)

d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)

3. Tính :

a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)

b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)

c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)

d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{1}+\dfrac{1}{10}\)

\(=\dfrac{10}{10}-\dfrac{1}{10}\)

= \(\dfrac{9}{10}\)

Chế Kazuto Kirikaya thử tham khảo thử đi !!!

Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya

d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

đây ko phải toán 7

`@` `\text {Ans}`

`\downarrow`

`a)`

\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)

`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)

`=`\(\dfrac{2}{9}\)

Vậy, \(A=\dfrac{2}{9}\)

`b)`

\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)

`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)

Vậy, \(B=\dfrac{4}{25}\)

`c)`

\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)

`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)

Vậy, \(C=\dfrac{99}{100}\)

`x/8=4/(-32)`

`(4x)/32=(-4)/32`

`4x=-4`

`x=-1`

Vậy `x=-1`.

`b)x-7/18=3 5/12`

`x-7/18=41/12`

`x=41/12+7/18`

`x=137/36`

Vậy `x=137/36`

`c)-3/8. 2/5+(-3)/8. 9/5-(-3)/8. 6/5`

`=-3/8. 2/5-3/8. 9/5+3/8. 6/5`

`=3/8(-2/5-9/5+6/5)`

`=3/8.(-1)`

`=-3/8`