Question

[Phạm Tú Uyên]

Giải các phương trình sau:

a. \(\frac{4}{2x+3}-\frac{7}{3x-5}=0\)

b. \(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\)

c. \(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\)

d. \(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)

Answer 1

[Phan Nghĩa]

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

Answer 2

[Phan Nghĩa]

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)