Question

[Hàn Vũ Nhi]

Giải các phương trình sau :

a) \(\frac{6x+1}{x^2+7x+10}\)+ \(\frac{5}{x-2}\)= \(\frac{3}{x-5}\)

b) \(\frac{2}{x^2-4}\)- \(\frac{x-1}{x\left(x-2\right)}\)+ \(\frac{x-4}{x\left(x+2\right)}\)= 0

c) \(\frac{1}{3-x}\)- \(\frac{1}{x+1}\)= \(\frac{x}{x-3}\)- \(\frac{\left(x-1\right)^2}{x^2-2x-3}\)

d) \(\frac{1}{x-2}\)- \(\frac{6}{x+3}\)=\(\frac{5}{6-x^2-x}\)

e) \(\frac{2}{x+2}\)- \(\frac{2x^2+16}{x^3+8}\)= \(\frac{5}{x^2-2x+4}\)

f) \(\frac{x+1}{x^2+x+1}\)- \(\frac{x-1}{x^2-x+1}\)= \(\frac{2\left(x+2\right)^2}{x^6-1}\)

Answer 1

[๖²⁴ʱ๖ۣۜTɦủү❄吻༉]

a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)

\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)

\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)

\(8x^3-46x^2-60x-180=0\)

=> vô nghiệm 

Answer 2

[Tran Le Khanh Linh]

b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)

\(\Leftrightarrow\frac{2x}{\left(x-2\right)\left(x+2\right)x}-\frac{\left(x+2\right)\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+x-2}{x\left(x-2\right)\left(x+2\right)}+\frac{x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{2x-x^2-x+2+x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)

=> 3x-6=0

<=> x=2 (ktm)

Vậy pt vô nghiệm