Đề bài người ta cho nhầm em, do số người phải là số nguyên mà câu b ko tính ra số nguyên => đề bài sai

Bài 1:

a: \(\dfrac{-7}{27}+\dfrac{-8}{27}=\dfrac{-7-8}{27}=-\dfrac{15}{27}=-\dfrac{5}{9}\)

b: \(\dfrac{1}{2}+\dfrac{13}{19}-\dfrac{4}{9}+\dfrac{6}{19}+\dfrac{5}{18}\)

\(=\left(\dfrac{1}{2}-\dfrac{4}{9}+\dfrac{5}{18}\right)+\left(\dfrac{13}{19}+\dfrac{6}{19}\right)\)

\(=\dfrac{9-8+5}{18}+1=\dfrac{1}{3}+1=\dfrac{4}{3}\)

c: \(\left(\dfrac{23}{41}-\dfrac{15}{82}\right):\dfrac{15}{41}\)

\(=\left(\dfrac{46}{82}-\dfrac{15}{82}\right):\dfrac{15}{41}\)

\(=\dfrac{31}{82}\cdot\dfrac{41}{15}=\dfrac{31}{30}\)

d: \(\dfrac{4}{3}+\dfrac{-11}{31}+\dfrac{3}{10}-\dfrac{20}{31}-\dfrac{2}{5}\)

\(=\left(\dfrac{4}{3}+\dfrac{3}{10}-\dfrac{2}{5}\right)+\dfrac{-11-20}{31}\)

\(=\dfrac{40+9-12}{30}-1=\dfrac{49-12-30}{30}=\dfrac{7}{30}\)

e: \(\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)

f: \(\dfrac{2}{11}\cdot\dfrac{-5}{4}+\dfrac{-9}{11}\cdot\dfrac{5}{4}+1\dfrac{3}{4}\)

=\(\dfrac{-5}{4}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{7}{4}\)

\(=\dfrac{-5}{4}+\dfrac{7}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)

Bài 2:

a: \(60,7+25,5-38,7\)

\(=\left(60,7-38,7\right)+25,5\)

=22+25,5

=47.5

b: \(\left(-9,207\right)+3.8+\left(-1,5030\right)-2,8\)

\(=\left(-9,207-1,503\right)+\left(3,8-2,8\right)\)

=-10,71+1

=-9,71

c: \(\left(-12,5\right)+17,55+\left(-3,5\right)-\left(-2,45\right)\)

\(=\left(-12,5-3,5\right)+\left(17,55+2,45\right)\)

=20-16

=4

d: \(721,9\cdot75+25\cdot721,9\)

\(=721,9\cdot\left(75+25\right)\)

\(=721,9\cdot100=72190\)

e: \(\left(2,07+3,005\right)-\left(12,005-4,23\right)\)

\(=2,07+3,005-12,005+4,23\)

=6,3-9

=-2,7

f: \(4,35-\left(2,67-1,65\right)+\left(3,54-6,33\right)\)

\(=4,35-2,67+1,65+3,54-6,33\)

\(=\left(4,35+1,65\right)+\left(-2,67-6,33\right)+3,54\)

=6-9+3,54

=3,54-3

=0,54

g: \(\left(-0.4\right)\cdot\left(-0,5\right)\cdot\left(-0,8\right)\)

\(=-0,4\cdot0,5\cdot0,8\)
\(=-0,4\cdot0,4=-0,16\)

h: \(\left(-1,6\right)\cdot\left(-0,125\right)\cdot\left(-0,5\right)\)

\(=-1,6\cdot0,125\cdot0,5\)

\(=-0,8\cdot0,125=-0,1\)

\(\dfrac{x+1}{3}=\dfrac{25}{-5}\)

\(\Rightarrow\dfrac{x+1}{3}=-5\)

\(\Rightarrow x+1=-5\cdot3\)

\(\Rightarrow x+1=-15\)

\(\Rightarrow x=-15-1\)

\(\Rightarrow x=-16\)

Vậy \(x=-16\).

=>x+1=-5x3

x+1=-15

x=-15-1

x=-16

\(\dfrac{x-2}{3}=\dfrac{x+1}{4}\)

\(4\left(x-2\right)=3\left(x+1\right)\)

\(4x-8=3x+3\)

\(4x-3x=3+8\)

\(x=11\)

\(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

=>\(x\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=\dfrac{5}{12}\)

=>\(x\cdot\dfrac{1}{6}=\dfrac{5}{12}\)

=>\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

h: \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)

\(=6+\dfrac{3}{10}-3-\dfrac{4}{7}-2-\dfrac{3}{10}\)

\(=1-\dfrac{4}{7}=\dfrac{3}{7}\)

i: \(\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}\)

\(=7+\dfrac{4}{9}+4+\dfrac{7}{11}-3-\dfrac{4}{9}\)

\(=8+\dfrac{7}{11}=\dfrac{95}{11}\)

\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)

\(=10+\dfrac{2}{9}+2+\dfrac{2}{5}-7-\dfrac{2}{9}\)

\(=5+\dfrac{2}{5}=\dfrac{27}{5}\)

a: C nằm giữa A và B

=>CA+CB=AB

=>CB+4=8

=>CB=4(cm)

b: C nằm giữa A và B

mà CA=CB(=4cm)

nên C là trung điểm của AB

Câu 1:

\(a,\dfrac{x}{15}+\dfrac{7}{20}=\dfrac{73}{60}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{73}{60}-\dfrac{7}{20}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{13}{15}\)

\(\Rightarrow x=13\)

b, \(x+\dfrac{1}{3}=\dfrac{12}{20}\)

\(\Rightarrow x=\dfrac{3}{5}-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{4}{15}\)

c, \(\dfrac{11}{8}+\dfrac{13}{6}=\dfrac{85}{x}\)

\(\Rightarrow\dfrac{85}{24}=\dfrac{85}{x}\)

\(\Rightarrow x=24\)

Câu 2:

a, \(60\%\cdot x+\dfrac{2}{3}\cdot x=-76\)

\(\Rightarrow x\cdot\left(60\%+\dfrac{2}{3}\right)=-76\)

\(\Rightarrow x\cdot\dfrac{19}{15}=-76\)

\(\Rightarrow x=-76:\dfrac{19}{15}\)

\(\Rightarrow x=-60\)

b, \(-\dfrac{5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)

\(\Rightarrow-\dfrac{5}{6}-x=\dfrac{1}{4}\)

\(\Rightarrow x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(\Rightarrow x=-\dfrac{13}{12}\)

c, \(x-4=\dfrac{-14}{35}:\dfrac{7}{5}\)

\(\Rightarrow x-4=\dfrac{-2}{5}\cdot\dfrac{5}{7}\)

\(\Rightarrow x-4=-\dfrac{2}{7}\)

\(\Rightarrow x=-\dfrac{2}{7}+4\)

\(\Rightarrow x=\dfrac{26}{7}\)

d, \(\left(\dfrac{2}{7}\cdot x+\dfrac{3}{7}\right):\dfrac{11}{5}-\dfrac{3}{7}=1\)

\(\Rightarrow\left(\dfrac{2}{7}\cdot x+\dfrac{3}{7}\right)\cdot\dfrac{5}{11}=1+\dfrac{3}{7}\)

\(\Rightarrow\dfrac{2}{7}\cdot x\cdot\dfrac{5}{11}+\dfrac{3}{7}\cdot\dfrac{5}{11}=\dfrac{10}{7}\)

\(\Rightarrow\dfrac{10}{77}\cdot x+\dfrac{15}{77}=\dfrac{10}{7}\)

\(\Rightarrow\dfrac{10}{77}\cdot x=\dfrac{110}{77}-\dfrac{15}{77}\)

\(\Rightarrow\dfrac{10x}{77}=\dfrac{95}{77}\)

\(\Rightarrow10x=95\)

\(\Rightarrow x=95:10\)

\(\Rightarrow x=\dfrac{19}{2}\)

$\text{#}Toru$

`1a)x/15+7/20=73/60`

`=>x/15=73/60-7/20`

`=>x/15=13/15`

`=>x=13`

Vậy `x=13`

_

`b)x+1/3=12/20`

`=>x=12/20-1/3`

`=>x=4/5`

Vậy `x=4/5`

_

`c)11/8+13/6=85/x`

`=>85/24=85/x`

`=>x=24`

Vậy `x=24`

_

`2a)60%x+2/3x=-7/6`

`=>3/5x+2/3x=-7/6`

`=>19/15x=-7/6`

`=>x=-7/6:19/15`

`=>x=-105/114`

Vậy `x=-105/114`

_

`b)-5/6-x=7/12+(-1/3)`

`=>-5/6-x=1/4`

`=>x=-5/6-1/4`

`=>x=-13/12`

Vậy `x=-13/12`

_

`c)x-4=-14/35:7/5`

`=>x-4=-2/7`

`=>x=-2/7+4`

`=>x=26/7`

Vậy `x=26/7`

_

`d)(2/7x+3/7):11/5-3/7=1`

`=>(2/7x+3/7):11/5=3/7`

`=>2/7x+3/7=3/7 . 11/5`

`=>2/7x=3/7 . 11/5-3/7`

`=>2/7x=3/7(11/5-1)`

`=>2/7x=3/7 . 6/5`

`=>2/7x=18/35`

`=>x=18/35:2/7`

`=>x=9/5`