Question

giải phương trình này giúp mình\(\frac{x}{2\left(x+3\right)}+\frac{x}{2x+2}=\frac{4x}{\left(x+1\right)\left(x+3\right)}\)

 

Answer 1

ĐKXĐ

(x+1)(x+3)\(\ne\)0

<=>x+1\(\ne\)0 và x+3\(\ne\)0

<=>x\(\ne\)-1 và x\(\ne\)-3

Phương trình : \(\frac{x}{2\left(x+3\right)}+\frac{x}{2x+2}=\frac{4x}{\left(x+1\right)\left(x+3\right)}\)

<=>\(\frac{x}{2\left(x+3\right)}+\frac{x}{2\left(x+1\right)}=\frac{4x}{\left(x+1\right)\left(x+3\right)}\)

<=>\(\frac{x+1}{2\left(x+1\right)\left(x+3\right)}+\frac{x+3}{2\left(x+1\right)\left(x+3\right)}=\frac{8x}{2\left(x+1\right)\left(x+3\right)}\)

=>x+1+x+3=8x

<=>x+x-8x=-1-3

<=>-6x=-4

<=>x=2/3(thỏa ĐKXĐ)

Vậy S={2/3}