tìm x,y biết: (5x-1)/3=(7y-6)/5=(5x-7y-7)/4x

Đáp án: $(x,y)\in\{(2,3), (\dfrac15, \dfrac67)\}$

Giải thích các bước giải:

ĐKXĐ: $x\ne 0$

Ta có:

$\dfrac{5x-1}{3}=\dfrac{7y-6}{5}=\dfrac{5x-7y-7}{4x}=\dfrac{(5x-1)+(7y-6)}{3+5}$

$\to \dfrac{5x-1}{3}=\dfrac{7y-6}{5}=\dfrac{5x-7y-7}{4x}=\dfrac{5x-7y-7}{8}$

$\to \dfrac{5x-7y-7}{4x}=\dfrac{5x-7y-7}{8}$

$\to 5x-7y-7=0$ hoặc $4x=8$

Nếu $5x-7y-7=0$

$\to \dfrac{5x-1}{3}=\dfrac{7y-6}{5}=0$

$\to 5x-1=7y-6=0$

$\to x=\dfrac15, y=\dfrac67$

Nếu $4x=8\to x=2$

$\to \dfrac{5\cdot 2-1}{3}=\dfrac{7y-6}{5}$

$\to3=\dfrac{7y-6}{5}$

$\to 7y-6=15$

$\to 7y=21$

$\to y=3$