Question

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

Answer 1

\(\Leftrightarrow\frac{2x}{x-1}+\frac{1}{x-1}=\frac{5x}{x+1}-\frac{5}{x+1}\)

\(\Rightarrow-\frac{5x}{x+1}+\frac{5}{x+1}+\frac{2x}{x-1}+\frac{1}{x-1}=0\)

\(\Rightarrow-\frac{3x^2-13x+4}{\left(x-1\right)\left(x+1\right)}=0\)

=>3x²-13x+4=0

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

=>x-4=0 hoặc 3x-1=0

=>x=4 hoặc 1/3

Answer 2

suy ra \(\frac{\left(2x+1\right)\cdot\left(x+1\right)}{\left(x-1\right)\cdot\left(x+1\right)}=\frac{5\left(x-1\right)\cdot\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

=>    (2x+1)(x+1)=5*(x-1)(x-1)

<=> 2x²+3x+1=5x²-10x+5

=> 13x=3x²+4

=> 3x²+4-13x=0

=> (3x²-x)-(12x-4)=0

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

=> x=4 hoặc x=1/3

Answer 3

<=>(2x+1)(x+1)/(x+1)(x-1)=5(x-1)(x+1)/(x+1)(x-1)

=>2x^2+2x+x+1=5x-5(x+1)

=>2x^2+2x+x+1=5x-5x-5

<=>2x^2+2x+x-5x+5x=-5-1

=>2x^2+3=-6

=>x^2+3=-3

=>2x=-15

=>x=-15/2

Answer 4

mình làm sai rùi xl bạn nha