Question

a) \(\dfrac{2x+1}{x-1}\)=\(\dfrac{5\left(x-1\right)}{x+1}\)

b) \(\dfrac{x-3}{x-2}\)+\(\dfrac{x-2}{x-4}\)=-1

Answer 1

a: \(\Leftrightarrow\left(2x+1\right)\left(x+1\right)=5\left(x^2-2x+1\right)\)

\(\Leftrightarrow5x^2-10x+5-2x^2-2x-x-1=0\)

\(\Leftrightarrow3x^2-13x+4=0\)

\(\Leftrightarrow3x^2-12x-x+4=0\)

=>(x-4)(3x-1)=0

=>x=4 hoặc x=1/3

b: \(\Leftrightarrow-\left(x-2\right)\left(x-4\right)=\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2\)

\(\Leftrightarrow-x^2+6x-8=x^2-7x+12+x^2-4x+4\)

\(\Leftrightarrow2x^2-11x+16+x^2-6x+8=0\)

\(\Leftrightarrow3x^2-17x+24=0\)

\(\Delta=17^2-4\cdot3\cdot24=1>0\)

Do đó PT có 2 nghiệm phân biệtlà:

\(\left\{{}\begin{matrix}x_1=\dfrac{17-1}{6}=\dfrac{16}{6}=\dfrac{8}{3}\\x_2=\dfrac{17+1}{6}=\dfrac{18}{6}=3\end{matrix}\right.\)