Question

giải các phương trình

a.5x+15=0

b.(2x+3)(x-5)=0

c.\(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\)

Answer 1

a) 5x + 15 = 0

<=> 5x = -15

<=> x = -3

\(S=\left\{-3\right\}\)

b) (2x + 3)(x - 5) = 0

TH1 : 2x + 3 = 0

<=> 2x = -3

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

TH2 : x - 5 = 0

<=> x = 5

\(S=\left\{\dfrac{-3}{2};5\right\}\)

c) \(\dfrac{1}{2x-3}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5}{x}\) đkxđ : \(x\ne0;x\ne\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x}{x\left(2x-3\right)}-\dfrac{3}{x\left(2x-3\right)}=\dfrac{5\left(2x-3\right)}{x\left(2x-3\right)}\)

\(\Rightarrow x-3=5\left(2x-3\right)\)

<=> x - 3 = 10x - 15

<=> x - 10x = -15 + 3

<=> -9x = -12

<=> x = \(\dfrac{4}{3}\)

\(S=\left\{\dfrac{4}{3}\right\}\)