Question

tìm x trong tỉ lện thức:

a)\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

b)\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

Answer 1

a\(\dfrac{\left(x-1\right)}{x+5}=\dfrac{6}{7}\) => (x-1)7=(x+5)6=7x-7=6x+30

<=> 7x=6x+37 => x=37

Answer 2

b\(\dfrac{\left(x-2\right)}{x-1}=\dfrac{\left(x+4\right)}{x+7}\) => (x-2)(x+7)=(x+4)(x-1) rồi làm tườn tự như câu a để tìm x.

Answer 3

a) Ta có:

\(\dfrac{x-1}{x+5}=\dfrac{6}{7}\Rightarrow\left(x-1\right).7=\left(x+5\right).6\\ \Leftrightarrow7x-7=6x+30\\ \Rightarrow x=37\)

b) Ta có:

\(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\Rightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\\ \Leftrightarrow x^2+7x-2x-14=x^2+4x-x-4\\ \Leftrightarrow x^2+5x-14=x^2+3x-4\\ \Leftrightarrow2x=10\Rightarrow x=5\)