*\(\dfrac{x-1}{x+2}\)-\(\dfrac{x}{x+2}\)=\(\dfrac{5x-2}{4-x^2}\).ĐKXĐ: x\(\ne\pm2\)
<=>\(\dfrac{\left(x-1\right)\left(2-x\right)}{4-x^2}\)-\(\dfrac{x\left(2-x\right)}{4-x^2}\)=\(\dfrac{5x-2}{4-x^2}\)
=>2x-\(x^2\)-2+x-2x+\(x^2\)=5x-2
<=>x-2=5x-2
<=>x-5x=2-2
<=>-4x=0
<=> x = 0(TM)
Vậy phương trình có tập nghiệm là S={0}
*(x+4)(5x+9)-x-4=0
<=>(x+4)(5x+9)-(x+4)=0
<=>(x+4)(5x+9-1)=0
<=>(x+4)(5x+8)=0
<=>x+4= 0 hoặc 5x+8=0
(+) x+4=0 (+)5x+8=0
<=>x=-4 <=>5x=-8
<=>x=\(\dfrac{-8}{5}\)
Vậy phương trình có tập nghiệm là S={\(-4;\dfrac{-8}{5}\)}
\(\dfrac{-1}{x+2}=\dfrac{5x-2}{4-x^2}< =>\dfrac{\left(2-x\right).\left(-1\right)}{4-x^2}=\dfrac{5x-2}{4-x^2}\)
<=> \(\dfrac{x-2}{4-x^2}=\dfrac{5x-2}{4-x^2}\) <=> x-2=5x-2<=>5x-x=-2+2 <=>4x=0
=> x vô số nghiệm
\(\left(x+4\right)\left(5x+9\right)-x-4=0\)
<=>\(5x^2+29x+36-x-4=0\)
<=>\(5x^2+28x+32=0\)
<=>\(\left(x+\dfrac{8}{5}\right)\left(x+4\right)=0\)
=> x=?