Question

giải pt sau:

15.\(\dfrac{7}{x-2}+\dfrac{8}{x-5}=3\)

16. \(\dfrac{x^2-3x+6}{x^2-9}=\dfrac{1}{x-3}\)

17.\(\dfrac{3}{x^2-4}=\dfrac{1}{x-2}+\dfrac{1}{x+2}\)

18.\(\dfrac{1}{2\left(x-1\right)}+\dfrac{8}{x^2-1}=\dfrac{1}{4}\)

Answer 1

15

\(\dfrac{7}{x-2}\)+\(\dfrac{8}{x-5}\)=3 (x khác 2 khác 5)

\(\Leftrightarrow\)7*(x-5)+8(x-2)=3(x-2)(x-5)

\(\Leftrightarrow\)15x-51=3x^2-21x+30\(\Leftrightarrow\)3x^2-36x+81=0

\(\Leftrightarrow\)\(\begin{matrix}&\end{matrix}\)\(\left[{}\begin{matrix}9\\3\end{matrix}\right.\) tmđk

16\(\dfrac{x^2-3x+6}{x^2-9}\)=\(\dfrac{1}{x-3}\)(x khác +_3)

\(\Leftrightarrow\)x^2-3x+6=x+3

\(\Leftrightarrow\)x^2-4x+3=0\(\Leftrightarrow\)\(\left[{}\begin{matrix}3loại\\1\end{matrix}\right.\)

vậy x=1 là nghiệm của pt

Answer 2

17 \(\dfrac{3}{x^2-4}\) = \(\dfrac{1}{x-2}+\dfrac{1}{x+2}\)

<=> x + 2 + x - 2 = 3

<=> 2x = 3

<=> x = \(\dfrac{3}{2}\)