a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)

b: =>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

Lời giải:
a. 

$2x(3x-1)=6x^2-13$

$\Leftrightarrow 6x^2-2x=6x^2-13$

$\Leftrightarrow 2x=13$

$\Leftrightarrow x=\frac{13}{2}$

b. 

$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$

$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$

$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$

\(\dfrac{x}{3}=\dfrac{y}{4}\)
Ta có: \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{14}{7}\)=2
* \(\dfrac{x}{3}=2=>x=6\)
*\(\dfrac{y}{4}=2=>y=8\)
Vậy( x, y) ∈{ 6, 8}
Kiểm tra lại nhaa

áp dụng tính chất dãy tỉ số bằng nhau

Ta có:x/3=y/4=x+y/3+4=14/7=2

Vậy x=2.3=6

       y=2.4=8

\(\dfrac{1}{3}< \dfrac{x}{y}< \dfrac{1}{2}\Rightarrow\dfrac{4}{12}< \dfrac{x}{y}< \dfrac{6}{12}\Rightarrow\dfrac{x}{y}=\dfrac{5}{12}\Rightarrow\dfrac{x}{5}=\dfrac{y}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{12}=\dfrac{2x}{10}=\dfrac{3y}{36}=\dfrac{2x+3y}{10+36}=\dfrac{19}{46}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{95}{46}\\y=\dfrac{114}{23}\end{matrix}\right.\)

Mà \(x,y\in Z\)

Vậy ko có x,y nguyên thỏa mãn đề

khó phết

2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0

=> 2x=3y;  5y=2z ;  3z=5x => x/3=y/2; y/2=z/5

=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31

      x/3 = 3y/6=2z/10 = (x-3y+2z)/7

=>  (12x+5y-3z)/ (x-3y+2z)=31/7

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\Rightarrow\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+4}{6}=\dfrac{3y-15}{-12}=\dfrac{z+1}{5}=\dfrac{2x+4-3y+15+z+1}{6-\left(-12\right)+5}=\dfrac{\left(2x-3y+z\right)+\left(4+15+1\right)}{23}=\dfrac{72+20}{23}=\dfrac{92}{23}=4\)

\(\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\\ \dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\\ \dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2x-3y+z+4+15+1}{2\cdot3-3\cdot\left(-4\right)+5}=\dfrac{92}{23}=4\)

Do đó: x=10; y=-11; z=4

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}\text{ và }2x-3y+z=72\)

\(\text{Áp dụng tính chất dãy tỉ số bằng nhau:}\)

\(\dfrac{x+2}{3}=\dfrac{y-5}{-4}=\dfrac{z+1}{5}=\dfrac{2\left(x+2\right)-3\left(y-5\right)+z+1}{2.3-3.\left(-4\right)+5}=\dfrac{92}{23}=4\)

\(\Rightarrow\dfrac{x+2}{3}=4\Rightarrow x+2=12\Rightarrow x=10\)

\(\dfrac{y-5}{-4}=4\Rightarrow y-5=-16\Rightarrow y=-11\)

\(\dfrac{z+1}{5}=4\Rightarrow z+1=20\Rightarrow z=19\)

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

+) Xét \(2x+3y-1=0\) có:

\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)

+) Xét \(2x+3y-1\ne0\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\Rightarrow x=2\)

\(\Rightarrow y=3\)

Vậy...

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+1+3y-2}{5+7}=\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)

\(\Rightarrow6x=12\Rightarrow x=2\)

Thay vào biểu thức, ta có:

\(2.2+\dfrac{1}{5}=\dfrac{3y-2}{7}\Rightarrow1=\dfrac{3y-2}{7}\Rightarrow3y-2=7\)

\(\Rightarrow3y=9\Rightarrow y=3\)

Vậy \(\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có

\(\dfrac{2x+1}{5}\)=\(\dfrac{3y-2}{7}\)=\(\dfrac{2x+3y-1}{6x}\)=\(\dfrac{2x+1+3y-2}{5+7}\)=\(\dfrac{2x+3y-1}{12}\)=

\(\dfrac{2x+3y-1}{6x}\)(1)

TH1: 2x+3y-1≠0

Từ (1) ⇒6x=12⇔x=2

Thầy x=2vào biểu thức trên ta có

\(\dfrac{2\cdot2+3y-1}{12}\)=\(\dfrac{3y-2}{7}\)⇔y=\(\dfrac{1}{5}\)

TH2: 2x+3y-1=0

⇒2x+1=0 và 3y-2=0

⇔x=\(\dfrac{-1}{2}\);y=\(\dfrac{2}{3}\)

Vậy (x;y)∈{(2;\(\dfrac{1}{5}\));(\(\dfrac{-1}{2}\);\(\dfrac{2}{3}\))}