Question

Ngiệm của phương trình \(\frac{6x-1}{3x+2}-\frac{2x-5}{x-3}=0\) là x=

(ghi cách giải luôn nha)

Answer 1

\(\frac{\left(6x-1\right)\left(x-3\right)-\left(2x-5\right)\left(3x+2\right)}{\left(3x+2\right)\left(x-3\right)}=0\) với x\(\ne-\frac{2}{3};3\)

\(\Leftrightarrow6x^2-19x+3-\left(6x^2-11x-10\right)\)= 0

\(\Leftrightarrow-8x+13=0\)

\(\Rightarrow x=\frac{13}{8}\left(TMĐK\right)\)

vậy n₀ là \(\frac{13}{8}\)

Answer 2

Ta có :

\(\frac{6x-1}{3x+2}-\frac{2x-5}{x-3}=0=>\frac{\left(6x-1\right)\left(x-3\right)}{\left(3x+2\right)\left(x-3\right)}-\frac{\left(2x-5\right)\left(3x+2\right)}{\left(x-3\right)\left(3x+2\right)}=0\)

=>\(\frac{\left(6x-1\right)\left(x-3\right)-\left(2x-5\right)\left(3x+2\right)}{\left(x-3\right)\left(3x+2\right)}=0\)

\(=>\frac{6x^2-19x+3-6x^2+11x+10}{\left(x-3\right)\left(3x+2\right)}=0\)

=>\(\frac{-8x+13}{\left(x-3\right)\left(3x+2\right)}=0=>-8x+13=0=>x=1,625=\frac{13}{8}\)

Answer 3