Question

Câu1 : Tìm x biết :

a) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

b)\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

c)\(\dfrac{x-1}{x+2}=\dfrac{x-2}{x+3}\)

Answer 1

\(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)(x\(\ne\)1)

<=>(x-1)(x+1)=7.9

<=>x²-1=63

<=>x²=64

<=>x=+-8

Answer 2

\(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

<=>(x+4)²=100

\(< =>\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

\(< =>\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

Answer 3

ý C từ nghĩ đã

Answer 4

c)ĐKXĐ : x # -2 ; x # - 3

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

<=> \(\dfrac{\left(x-1\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}\)= \(\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}\)

<=> x² + 3x - x - 3 = x² - 4

<=> x² - x² + 3x - x = 3 - 4

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)