Giải thích các bước giải:
`i, \ (2x+1)/(x-1)=(5(x-1))/(x+1)`
ĐKXĐ : `x ne +-1`
`<=> (2x+1)/(x-1)-(5x-5)/(x+1)=0`
`<=> ((2x+1)(x+1)-(5x-5)(x-1))/((x-1)(x+1))=0`
`<=> 2x^2+2x+x+1-(5x^2-5x-5x+5)=0`
`<=> 2x^2+3x+1-5x^2+10x-5=0`
`<=> -3x^2+13x-4=0`
`<=> -3x^2+13x-4=0`
`<=> 3x^2-13x+4=0`
`<=> 3x^2-x-12x+4=0`
`<=> x(3x-1)-4(3x-1)=0`
`<=> (x-4)(3x-1)=0`
`<=> [(x-4),(3x-1=0):} <=>`\(\left[ \begin{array}{l}x=4 \qquad \text{(tm)}\\x=\dfrac{1}{3} \qquad \text{(tm)}\end{array} \right.\)
Vậy `S={1/3;4}`
`j, \ (x-1)/(x+2)-x/(x-2)=(5x-2)/(4-x^2)`
ĐKXĐ : `x ne +-2`
`<=> (x-1)/(x+2)-x/(x-2)-(5x-2)/(4-x^2)=0`
`<=> ((x-1)(x-2)-x(x+2)+5x-2)/(x^2-4)=0`
`<=> x^2-2x-x+2-x^2-2x+5x-2=0`
`<=> 0x=0 qquad text{(Luôn đúng)}`
Vậy `S={x in RR | x ne +-2}`
`k, \ (x-2)/(2+x)-3/(x-2)=(2(x-11))/(x^2-4)`
ĐKXĐ : `x ne +-2`
`<=> (x-2)/(2+x)-3/(x-2)-(2(x-11))/(x^2-4)=0`
`<=> ((x-2)(x-2)-3(x+2)-2(x-11))/(x^2-4)=0`
`<=> x^2-4x+4-3x-6-2x+22=0`
`<=> x^2-9x+20=0`
`<=> x^2-4x-5x+20=0`
`<=> x(x-4)-5(x-4)=0`
`<=> (x-5)(x-4)=0`
`<=> [(x-5=0),(x-4=0):} <=> [(x=5 qquad text{(tm)}),(x=4 qquad text{(tm)}):}`
Vậy `S={4;5}`
