Ta có : 

\(\dfrac{2}{3}+\dfrac{m}{4}+\dfrac{n}{6}=6\Rightarrow\dfrac{m}{4}+\dfrac{n}{6}=5\dfrac{1}{3}\)

=> m/4 = 2/6 = 2/3

m/4 = 16/3 : ( 3+2 ) x 2 = 32/15

n/6 = 16/3-32/15 = 16/5

=> m x n = 32 x 16 = 512

Ta có : 

23+m4+n6=6⇒m4+n6=513

=> m/4 = 2/6 = 2/3

m/4 = 16/3 : ( 3+2 ) x 2 = 32/15

n/6 = 16/3-32/15 = 16/5

=> m x n = 32 x 16 = 512

why m/4 =2/6=2/3 what what ?? i think you wrong ;(

- If a + b + c \(\ne\)0 then

\(\dfrac{2a}{b+c}=\dfrac{2b}{a+c}=\dfrac{2c}{a+b}=\dfrac{2a+2b+2c}{b+c+a+c+a+b}=\dfrac{2a+2b+2c}{2a+2b+2c}=1\)

=> \(m=1\)

- If a + b + c = 0 then

b + c = -a

a + c = -b

a + b = -c

=> \(\dfrac{2a}{b+c}=\dfrac{2b}{a+b}=\dfrac{2c}{a+b}=\dfrac{2a}{-a}=\dfrac{2b}{-b}=\dfrac{2c}{-c}=-2\)

=> \(m=-2\)

- If a + b + c ≠

0 then

2ab+c=2ba+c=2ca+b=2a+2b+2cb+c+a+c+a+b=2a+2b+2c2a+2b+2c=1

=> m=1

- If a + b + c = 0 then

b + c = -a

a + c = -b

a + b = -c

=> 2ab+c=2ba+b=2ca+b=2a−a=2b−b=2c−c=−2

=> m=−2

\(\dfrac{3-x}{5}+\dfrac{x+y}{10}=\dfrac{x-y}{4}\)

\(\Leftrightarrow\dfrac{12-4x}{20}+\dfrac{2x+2y}{20}=\dfrac{5x-5y}{20}\)

\(\Leftrightarrow12-4x+2x+2y=5x-5y\)

\(\Leftrightarrow2y+5y=5x+4x-2x-12\)

\(\Leftrightarrow7y=7x-12\Leftrightarrow y=x-\dfrac{12}{7}\)

3−x5+x+y10=x−y4

⇔12−4x20+2x+2y20=5x−5y20

⇔12−4x+2x+2y=5x−5y

⇔2y+5y=5x+4x−2x−12

⇔7y=7x−12⇔y=x−127

If x = 3

=> \(\dfrac{a-3}{3}=\dfrac{3b-5}{5}\)

=> \(5\left(a-3\right)=3\left(3b-5\right)\)

=> 5a - 15 = 9b - 15

=> 5a = 9b

=> \(\dfrac{a}{b}=\dfrac{9}{5}\)

=> \(\dfrac{b}{a}=\dfrac{5}{9}\)

=> \(\dfrac{a}{b}-\dfrac{b}{a}=\dfrac{9}{5}-\dfrac{5}{9}=\dfrac{56}{45}\)

If x = 3

=> a−33=3b−55

=> 5(a−3)=3(3b−5)

=> 5a - 15 = 9b - 15

=> 5a = 9b

=> ab=95

=> ba=59

=> ab−ba=95−59=5645

i will you to make this topic :)
replace x=1 into the polution above , we have 
p+4q=161 
=> q= (161-p)/4 
=> q=40-(p-1)/4 => q<40(1)
because the p and q are primes so (p-1) divisible for 4 =>p-1 >=4 => p>=5 (2)
Merged (1) with (2) we have q ={2;3;5;7;11;13;17;19;23;29;31}
then you try to the each of case : it will have 2 cases (correct satisfies) 
this case p=37 <=> q=31 and p=133 <=> q=7

i will you to make this topic :)
replace x=1 into the polution above , we have 
p+4q=161 
=> q= (161-p)/4 
=> q=40-(p-1)/4 => q<40(1)
because the p and q are primes so (p-1) divisible for 4 =>p-1 >=4 => p>=5 (2)
Merged (1) with (2) we have q ={2;3;5;7;11;13;17;19;23;29;31}
then you try to the each of case : it will have 2 cases (correct satisfies) 
this case p=37 <=> q=31 and p=133 <=> q=7 

We have : 7*(x*4) = 7*(3x - 20) = 21 - 5(3x - 20) = 121 - 15x

So 121 - 15x = 1 => 15x = 120 => x = 8

(x + 2) : 2 = (5x - 8) : 4

=> \(\dfrac{x+2}{2}=\dfrac{5x-8}{4}\)

<=> 4(x + 2) = 2(5x - 8)

<=> 4x + 8 = 10x - 16

=> 24 = 6x

=> x = 4

(x + 2) : 2 = (5x - 8) : 4

=> x+22=5x−84

<=> 4(x + 2) = 2(5x - 8)

<=> 4x + 8 = 10x - 16

=> 24 = 6x

=> x = 4

17x - 85 = 85 - 17x

<=> 17x + 17x = 85 + 85

<=> 34x           = 170

<=> x               = 5

17x - 85 = 85 - 17x

<=> 17x + 17x = 85 + 85

<=> 34x           = 170

<=> x               = 5

We have : x*(1*2) = x*(3 - 2) = x*1 = 3x - 1

So 3x - 1 = 2 => 3x = 3 => x = 1

\(\dfrac{x}{1.2}+\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{1999.2000}=1\)

\(x\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1999.2000}\right)=1\)

\(x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\right)=1\)

\(x\left(1-\dfrac{1}{2000}\right)=1\)

\(x.\dfrac{1999}{2000}=1\)

\(x=\dfrac{2000}{1999}\)

x1.2+x2.3+x3.4+...+x1999.2000=1

x(11.2+12.3+13.4+...+11999.2000)=1

x(1−12+12−13+13−14+...+11999−12000)=1

x(1−12000)=1

x.19992000=1

x=20001999

\(\dfrac{a-b}{b}=\dfrac{a}{b}-1\).So\(\dfrac{a}{b}-1=\dfrac{3}{5}\Rightarrow\dfrac{a}{b}=\dfrac{8}{5}\)

Replace x = 2 into the given equations,we have :

\(\left\{{}\begin{matrix}2p+q=91\\p-q=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2p+p-2=91\\q=p-2\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}3p=93\\q=p-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}p=31\\q=29\end{matrix}\right.\)=> pq = 899

Replace x = 2 into the given equations,we have :

{2p+q=91p−q=2⇒{2p+p−2=91q=p−2

⇒{3p=93q=p−2⇒{p=31q=29

=> pq = 899

It is very easy. the done equation is writend as   \(\left(x-3\right)\left(\dfrac{13}{6}-\dfrac{4}{3}-\dfrac{7}{6}\right)=8\)

\(\Leftrightarrow-\dfrac{1}{3}\left(x-3\right)=8\Leftrightarrow x=-21\) .

It is very easy. the done equation is writend as   (x−3)(136−43−76)=8

⇔−13(x−3)=8⇔x=−21

 .

@mathlove: \(3-x\ne x-3\) ???

Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.

With a = 4, the equation to solve become  2[3(2+x) - 2(4-x)] = 16 , then x = 2.

Because (a+1) satisfies 2(x+1)=3(x-1), so that 2(a+1+1)=3a , a = 4.

With a = 4, the equation to solve become  2[3(2+x) - 2(4-x)] = 16 , then x = 2.

Replace x = 3 into the equation,we have :

m−33+34=1712⇔m3−1=23⇔m3=53⇔m=5

Replace x = 3 into the equation,we have :

\(\dfrac{m-3}{3}+\dfrac{3}{4}=\dfrac{17}{12}\Leftrightarrow\dfrac{m}{3}-1=\dfrac{2}{3}\Leftrightarrow\dfrac{m}{3}=\dfrac{5}{3}\Leftrightarrow m=5\)

3 - a = 4 - b = 2 => a = 1 ; b = 2 => a + b + c = 502

=> m =\(\dfrac{502}{2008}=\dfrac{1}{4}\)

a = \(\dfrac{2}{3}\)

2/3 nhé

tk nhé

Replacing x = 1/2 we have: 4x - 3a = 0  

Have 4.1/2 - 3a = 0

 Have 2 - 3a = 0

 Have 3a = 2

 Has a = 2/3

Dress a = 2/3 good

Replacing x = 1/2 we have: 4x - 3a = 0  

<=> 4.1/2 - 3a = 0

=> 2 - 3a = 0

=> 3a = 2

=> a = 2/3

Dress a = 2/3 good