Tìm x, y biết: (2x+1)/5=(4y-5)/9=(2x+4y-4) /7x Giúp mình nha(^-^)

Đáp án: $(x,y)\in\{(2, \dfrac72), (-\dfrac12,\dfrac54)\}$

Giải thích các bước giải:

Ta có:

$\dfrac{2x+1}{5}=\dfrac{4y-5}{9}$

$\to 2x+1=\dfrac{20y-25}{9}$

$\to 2x=\dfrac{20y-25}{9}-1$

$\to 2x=\dfrac{20y-34}{9}$

$\to x=\dfrac{10y-17}{9}$

Lại có:

$\dfrac{2x+1}{5}=\dfrac{2x+4y-4}{7x}$

$\to \dfrac{2\cdot \dfrac{10y-17}{9}+1}{5}=\dfrac{2\cdot \dfrac{10y-17}{9}+4y-4}{7\cdot \dfrac{10y-17}{9}}$

$\to \dfrac{ \dfrac{20y-34}{9}+1}{5}=\dfrac{ \dfrac{20y-34}{9}+4y-4}{\dfrac{70y-119}{9}}$

$\to \dfrac{ \dfrac{20y-34+9}{9}}{5}=\dfrac{ \dfrac{20y-34+36y-36}{9}}{\dfrac{70y-119}{9}}$

$\to \dfrac{ \dfrac{20y-25}{9}}{5}=\dfrac{ \dfrac{56y-70}{9}}{\dfrac{70y-119}{9}}$

$\to \dfrac{20y-25}{45}=\dfrac{ 56y-70}{70y-119}$

$\to \dfrac{4y-5}{9}=\dfrac{8y-10}{10y-17}$

$\to \dfrac{4y-5}{9}=\dfrac{2(4y-5)}{10y-17}$

$\to \dfrac{4y-5}{9}-\dfrac{2(4y-5)}{10y-17}=0$

$\to (4y-5)(\dfrac19-\dfrac{2}{10y-17})=0$

$\to 4y-5=0\to 4y=5\to y=\dfrac54\to x=\dfrac{10\cdot \dfrac54-17}{9}=-\dfrac12$

Hoặc $\dfrac19-\dfrac{2}{10y-17}=0$

$\to \dfrac19=\dfrac2{10y-17}$

$\to 10y-17=18$

$\to 10y=35$

$\to y=\dfrac72\to x=\dfrac{10\cdot \dfrac72-17}{9}=2$