a: x là số dương

=>(3-x)/(x+10)>0

=>(x-3)/(x+10)<0

=>-10<x<3

b: x<0

=>(3-x)/(x+10)>0

=>x>3 hoặc x<-10

c: x=0

=>3-x=0

=>x=3

\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)

TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)

TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)

\(\Rightarrow x\in\left\{2;3;-4\right\}\)

\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)

\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)

\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)

a) (1/7x - 2/7)(-1/5x + 3/5)(1/3x + 4/3) = 0

3 trường hợp:

TH1: 1/7x - 2/7 = 0 <=> 1/7x = 0 + 2/7 <=> 1/7x = 2/7 <=> x = 2.7/7 = 2

=> x = 2

TH2: -1/5x + 3/5 = 0 <=> -1/5x = 0 - 3/5 <=> -1/5x = -3/5 <=> x = (-3/5).(-5) = 3

=> x = 3

TH3: 1/3x + 4/3 = 0 <=> 1/3x = 0 - 4/3 <=> 1/3x = -4/3 <=> x = x = 3.(-4/3) = -4

=> x = -4

Vậy: x = 2, 3, -4

b) 1/6x + 1/10x - 4/15x + 1 = 0

<=> 1/6x + 1/10x - 4/15x = 0 - 1

<=> 1/6x + 1/10x - 4/15x = -1

<=> 1/6x.30 + 1/10x.30 - 4/15x.30 = -1.30

<=> 5x + 3x - 8x = -30

<=> 0 = -30

=> không có x thỏa mãn

Bài 1 : Ta có:

\(\frac{7+\frac{7}{11}+\frac{7}{23}+\frac{7}{31}}{9+\frac{9}{11}+\frac{9}{23}+\frac{9}{31}}\)

= \(\frac{7.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}{9.\left(1+\frac{1}{11}+\frac{1}{23}+\frac{1}{31}\right)}\)

= \(\frac{7}{9}\)

Bài 2 :

 \(\frac{x}{2}+\frac{3x}{4}+\frac{5x}{6}=\frac{10}{24}\)

=> \(\frac{12x+18x+20x}{24}=\frac{10}{24}\)

=> 50x = 10

=> x = 10 : 50

=> x = 1/5

Bài 3 : Để A nhận giá trị nguyên thì 3 \(⋮\)x + 3

                                         <=> x + 3 \(\in\)Ư(3) = {1; -1; 3; -3}

Lập bảng :

Vậy 

b, \(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

        \(0+1=0\)

=> x thuoc rong