Ta có:

\(\left(\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\right)+-\frac{1}{2}=\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\)\(-\frac{1}{2}\)

=\(\frac{6}{30}+\frac{10}{30}+\frac{9}{30}-\frac{15}{30}=\frac{6+10+9-15}{30}=\frac{10}{30}=\frac{1}{3}\)

a, Áp dụng t/c dtsbn:

\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{2}{-2}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-7\\y=-5\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{y}=\dfrac{7}{2}\Rightarrow\dfrac{x}{7}=\dfrac{y}{2}=\dfrac{x+y}{7+2}=\dfrac{-27}{9}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-21\\y=-6\end{matrix}\right.\)

c, \(\dfrac{x}{32}=\dfrac{2}{x}\Rightarrow x^2=2\cdot32=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

d, \(\left|x+\dfrac{1}{3}\right|-2=\dfrac{1}{2}\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{2}\\x+\dfrac{1}{3}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{17}{6}\end{matrix}\right.\)

\(A.12:\left(x+4\right)=2\)

\(\Rightarrow x+4=12:2\)

\(\Rightarrow x+4=6\)

\(\Rightarrow x=2\)

\(B,12:\left(x+4\right).3+18=24\)

\(12:\left(x+4\right).3=6\)

\(\Rightarrow12:\left(x+4\right)=2\)

\(\Rightarrow x+4=6\)

\(\Rightarrow x=2\)

a) 12 : ( x + 4 ) = 2

x + 4 = 12 : 2

x + 4 = 6

x = 6 - 4 

x = 2

Vậy x = 2

b) 12 : ( x + 4 ) . 3 + 18 = 24

12 : ( x + 4) . 3 = 24 - 18

12 : ( x + 4 ) . 3 = 6

12 : ( x + 4 ) = 6 : 3

12 : ( x + 4 ) = 2

x + 4 = 12 : 2 

x + 4 = 6

x = 6 - 4

x = 2

Vậy x = 2

Học tốt !!!

=))

\(12:\left(x+4\right)=2\)

\(x+4=12:2=6\)

\(\Rightarrow x=6-4=2\)

Vậy \(x=2\)

1/2-(4/12+9/12)<x<1/24-(3/24-8/24)

1/2-13/12<x<1/24-(-5/24)

-7/12<x<1/4

=>x\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) E{0}

ta có:\(\frac{1}{2}-\left(\frac{1}{3}+\frac{3}{4}\right)=\frac{-1}{12}=-0,08333333\)

mà \(\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)=\frac{1}{4}=0.25\)

nên suy ra không có số nguyên x nào thỏa mãn đề bài.

Sao ai cũng giải khó hiểu thế vậy trời:))

     Bài 1:

\(x\) \(\times\) \(\dfrac{15}{16}\) - \(x\) \(\times\) \(\dfrac{4}{16}\) = 2

\(x\) \(\times\) (\(\dfrac{15}{16}\) - \(\dfrac{4}{16}\)) = 2

\(x\) \(\times\) \(\dfrac{11}{16}\) = 2

\(x\) = 2 : \(\dfrac{11}{16}\)

\(x\) = 2 x \(\dfrac{16}{11}\)

\(x\) = \(\dfrac{32}{11}\)

Bài 2: 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\) + \(\dfrac{1}{10}\) + ... + \(\dfrac{1}{x\left(x+1\right):2}\) = 1 : \(\dfrac{2011}{2012}\)

1 + 2\(\times\) ( \(\dfrac{1}{2\times3}\) + \(\dfrac{1}{2\times6}\) + \(\dfrac{1}{2\times10}\) + ... + \(\dfrac{2}{2\times x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\)(\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+...+ \(\dfrac{1}{x\times\left(x+1\right)}\)) = \(\dfrac{2012}{2011}\)

1 + 2 \(\times\) (\(\dfrac{1}{2\times3}\) + \(\dfrac{1}{3\times4}\) + ... + \(\dfrac{1}{x\times\left(x+1\right)}\)) = 1 + \(\dfrac{1}{2011}\)

1 + 2\(\times\)(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{x}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

 1 + 2 \(\times\) (\(\dfrac{1}{2}\) - \(\dfrac{1}{x+1}\)) = 1 + \(\dfrac{1}{2011}\)

  1 + 1 - \(\dfrac{2}{x+1}\) = 1 + \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) = 1 + 1 - 1  - \(\dfrac{1}{2011}\)

    \(\dfrac{2}{x+1}\) =  2 - 1 - \(\dfrac{1}{2011}\)

  \(\dfrac{2}{x+1}\)  = 1 - \(\dfrac{1}{2011}\)

     \(\dfrac{2}{x+1}\) = \(\dfrac{2010}{2011}\)

       \(x\) + 1 = 2 : \(\dfrac{2010}{2011}\)

        \(x\) + 1 = \(\dfrac{2011}{1005}\)

         \(x\) = \(\dfrac{2011}{1005}\) - 1  = \(\dfrac{1006}{1005}\)(loại vì \(\dfrac{1006}{1005}\) không phải là số tự nhiên)

Vậy không có giá trị nào của \(x\) là số tự nhiên thỏa mãn đề bài. 

Bài 3:

\(\dfrac{x}{16}\) \(\times\) (2017 - 1) = 2

\(\dfrac{x}{16}\) \(\times\) 2016 = 2

 \(\dfrac{x}{16}\)             = 2 : 2016

 \(\dfrac{x}{16}\)            = \(\dfrac{1}{1008}\)

   \(x\)             = \(\dfrac{1}{1008}\) x 16

  \(x\) = \(\dfrac{1}{63}\)

a) Ta có: \(\frac{3+x}{5+y}=\frac{3}{5}\)

=> (3 + x).5 = 3(5 + y)

=> 15 + 5x = 15 + 3y

=> 5x = 3y

=> x = 3/5y

Mà x + y = 16

hay 3/5y + y = 16

=> (3/5 + 1).y = 16

=> 8/5.y = 16

=> y = 16 : 8/5

=> y = 10

=> x = 16 - 10 = 6

Vậy x = 6; y = 10

b) Ta có: \(\frac{x-7}{y-6}=\frac{7}{6}\)

=> (x - 7).6 = 7.(y - 6)

=> 6x - 42 = 7y - 42

=> 6x = 7y

=> x = 7/6y

Mà x - y = -4

hay 7/6y - y = -4

=> 1/6y = -4

=> y = -4 : 1/6

=> y = -24

=> x = -4 - 24 = -28

Vậy x =  -28; y = -24

2 : \(buổi \) \(sáng\) \(bán\) \(dc :\)

\((360 - 142 : 2 =109 l\)

\(buổi\) \(chiều\) \(bán\) \(dc :\)

\(360 - 109 = 251 l\)

\(1,\)

\(a,x\times\dfrac{3}{9}=\dfrac{9}{15}\)            \(b,x:\dfrac{1}{2}=\dfrac{5}{6}\)

\(x=\dfrac{9}{15}:\dfrac{3}{9}\)                    \(x=\dfrac{6}{5}\times\dfrac{1}{2}\)

\(x=\dfrac{81}{45}=\dfrac{9}{5}\)                  \(x=\dfrac{6}{10}=\dfrac{3}{5}\)

\(2,\) có bạn làm rồi nhé ;>

\(3,\)

\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\left(\dfrac{3}{4}\times\dfrac{2}{9}\right)=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{7+2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\left(\dfrac{4}{15}:\dfrac{2}{5}\right)=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{8-6}{9}=\dfrac{2}{9}\)

a.x=9/15:3/9

x=9/5

b.x=5/6x1/2

x=5/12

2.ngày đầu bán được số dầu là:

360x2/5=144(l)

cửa hàng còn lại số dầu là:

360-144=216(l)

3.a.7/12+3/4x2/9

=7/12+1/6

=7/12+2/12

=9/12=3/4