Đáp án:
`a)S={3/2;-1/4}`
`b)S={-1/4}`
`c)S=∅`
Giải thích các bước giải:
`a)|3x-1|-x=2`
`|3x-1|=3x-1` nếu `x≥1/3`
`1-3x` nếu `x<1/3`
*Nếu `x≥1/3` ta có:
`3x-1-x=2`
`⇔2x-1=2`
`⇔2x=2+1`
`⇔2x=3`
`⇔x=3/2(TM)`
*Nếu `x<1/3` ta có:
`1-3x-x=2`
`⇔1-4x=2`
`⇔-4x=2-1`
`⇔-4x=1`
`⇔x=-1/4(TM)`
Vậy `S={3/2;-1/4}`
`b)1/(x-1)-(3x²)/(x³-1)=(2x)/(x²+x+1)(ĐKXĐ:x`$\neq$ `1)`
`⇔1/(x-1)-(3x²)/[(x-1)(x²+x+1)]=(2x)/(x²+x+1)`
`⇔(x²+x+1)/[(x-1)(x²+x+1)]-(3x²)/[(x-1)(x²+x+1)]=[2x(x-1)]/[(x-1)(x²+x+1)]`
`⇒x²+x+1-3x²=2x(x-1)`
`⇔-2x²+x+1=2x²-2x`
`⇔-2x²+x+1-2x²+2x=0`
`⇔(-2x²-2x²)+(x+2x)+1=0`
`⇔-4x²+3x+1=0`
`⇔-4x²+4x-x+1=0`
`⇔-4x(x-1)-(x-1)=0`
`⇔(x-1)(-4x-1)=0`
`⇔`\(\left[ \begin{array}{l}x-1=0\\-4x-1=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1(Ko TM ĐKXĐ)\\x=-\dfrac{1}{4}(TM ĐKXĐ)\end{array} \right.\)
Vậy `S={-1/4}`
`c)(x+5)/(x²-5x)-(x-5)/(2x²+10x)=(x+25)/(2x²-50)(ĐKXĐ:x`$\neq$ `0,x`$\neq$ `5,x`$\neq$ `-5)`
`⇔(x+5)/[x(x-5)]-(x-5)/[2x(x+5)]=(x+25)/[2(x²-25)]`
`⇔[2(x+5)²]/[2x(x+5)(x-5)]-[(x-5)²]/[2x(x+5)(x-5)]=[x(x+25)]/[2x(x+5)(x-5)]`
`⇒2(x+5)²-(x-5)²=x(x+25)`
`⇔2(x²+10x+25)-(x²-10x+25)=x²+25x`
`⇔2x²+20x+50-x²+10x-25=x²+25x`
`⇔(2x²-x²)+(20x+10x)+(50-25)=x²+25x`
`⇔x²+30x+25=x²+25x`
`⇔x²+30x-x²-25x=-25`
`⇔5x=-25`
`⇔x=(-25):5`
`⇔x=-5(Ko` `TM` `ĐKXĐ)`
Vậy `S=∅`
