\(\dfrac{x}{3}=\dfrac{y}{4}\)
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

còn 1 câu 

Áp dụng dãy tỉ số bằng nhau là ra mà.

Giải:

Áp dụng dãy tỉ số bằng nhau,ta có:

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

\(\Rightarrow\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)\(\Rightarrow6x=12\Rightarrow x=2\)

Thay vào (1), ta được:

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

Vậy x=2 , y=3

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

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

1.

Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=4k\end{matrix}\right.\)

\(\Rightarrow x^2-y^2=\left(5k\right)^2-\left(4k\right)^2=25k^2-16k^2=9k^2=4\)

\(\Rightarrow k^2=\dfrac{4}{9}\Rightarrow k=\pm\dfrac{2}{3}\)

\(\circledast k=\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{8}{3}\end{matrix}\right.\)

\(\circledast k=-\dfrac{2}{3}\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{3}\\y=-\dfrac{8}{3}\end{matrix}\right.\)

2.

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

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

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

\(\Rightarrow y=\dfrac{\dfrac{2\cdot2+1}{5}\cdot7+2}{3}=3\)

3.

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Leftrightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

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

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-\left(z-3\right)}{4+9-4}=\dfrac{95-8+3}{9}=10\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10\cdot4+2}{2}=21\\y=\dfrac{10\cdot9+6}{3}=32\\z=10\cdot4+3=43\end{matrix}\right.\)

Theo tính chất của dãy tỉ số bằng nhau ta có:

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

Do đó: \(\dfrac{2x+3y-1}{12}=\dfrac{2x+3y-1}{6x}\)

Nếu:

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

Nếu: \(2x+3y-1\ne0\Rightarrow6x=12\Rightarrow x=2\)

Khi đó ta có:

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

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