Question

Bài 1: Tìm x biết 

a)\(\frac{x-3}{x+5}=\frac{5}{7}\)

b)\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

Answer 1

\(a,7\left(x-3\right)=5\left(x+5\right)\)

\(\Leftrightarrow2x=46\Leftrightarrow x=23\)

\(b,\left(x^2+2x-3\right)=x^2-4\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)

Answer 2

a ĐKXĐ x khác -5

ta có 7(x-3)=5(x+5)

     7x-21=5x+5

   => 2x=26

=> x=13

b, ĐkxĐ x khác -2  x khác -3

ta có :(x-1)(x+3)=(x-2)(x+2)

     x^2+2x-3-x^2 +2 = 0

=>2x+1=0

=>x=1/2

Answer 3

a)Từ \(\frac{x-3}{x+5}=\frac{5}{7}\)\(\Rightarrow\)7(x-3)=5(x+5)

5x+25=7x-21

5x-7x=(-21)-25

(-2x)=(-46)

x=23

b)Từ \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)\(\Rightarrow\)(x-1).(x+3)=(x+2).(x-2)

(x-1).x+(x-1).x=(x+2).x-(x+2).2

x²-x+3x-3=x²+2x-2x-4

Đưa về 2x=(-1)   => x=\(-\frac{1}{2}\)