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Đáp án:
$\begin{array}{l}
1)Dkxd:x \ne 0;x \ne - 1;x \ne \dfrac{1}{2}\\
D = \left( {\dfrac{{x + 2}}{{3x}} + \dfrac{2}{{x + 1}} - 3} \right):\dfrac{{2 - 4x}}{{x + 1}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{\left( {x + 2} \right).\left( {x + 1} \right) + 2.3x - 3.3x.\left( {x + 1} \right)}}{{3x\left( {x + 1} \right)}}.\dfrac{{x + 1}}{{2.\left( {1 - 2x} \right)}}\\
- \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{{x^2} + 3x + 2 + 6x - 9{x^2} - 9x}}{{3x}}.\dfrac{1}{{2\left( {1 - 2x} \right)}}\\
- \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{ - 8{x^2} + 2}}{{3x}}.\dfrac{1}{{2\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{2.\left( {1 - 4{x^2}} \right)}}{{3x}}.\dfrac{1}{{2\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}{{3x\left( {1 - 2x} \right)}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{1 + 2x}}{{3x}} - \dfrac{{3x - {x^2} + 1}}{{3x}}\\
= \dfrac{{1 + 2x - 3x + {x^2} - 1}}{{3x}}\\
= \dfrac{{{x^2} - x}}{{3x}}\\
= \dfrac{{x - 1}}{3}\\
2){x^2} - 3x - 4 = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 4\left( {do:x \ne - 1} \right)\\
\Leftrightarrow D = \dfrac{{4 - 1}}{3} = 1\\
3)\dfrac{1}{D} = 1:\dfrac{{x - 1}}{3} = \dfrac{3}{{x - 1}} \in Z\\
\Leftrightarrow \left( {x - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Leftrightarrow x \in \left\{ { - 2;0;2;4} \right\}\\
Do:x \ne 0\\
\Leftrightarrow x \in \left\{ { - 2;2;4} \right\}
\end{array}$
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`1,`
`D=((x+2)/(3x)+2/(x+1)-3):(2-4x)/(x+1) - (3x-x^2+1)/(3x) (x\ne 0, x\ne -1, x\ne 1/2)`
`= (((x+2)(x+1))/(3x(x+1)) + (2.3x)/(3x(x+1)) - (3.3x(x+1))/(3x(x+1)) ) . (x+1)/(2-4x) - (3x-x^2+1)/(3x)`
`= ((x^2+x+2x+2)/(3x(x+1)) + (6x)/(3x(x+1)) - (9x^2 + 9x)/(3x(x+1)) ) . (x+1)/(2-4x) - (3x-x^2+1)/(3x)`
`=(x^2+3x+2 +6x - 9x^2-9x)/(3x(x+1)) . (x+1)/(2-4x)-(3x-x^2+1)/(3x)`
`=(-8x^2 +2)/(3x) . 1/(2(1-2x)) - (3x-x^2+1)/(3x)`
`=(2 (1-2x)(1+2x))/(3x) . 1/(2(1-2x)) - (3x-x^2+1)/(3x)`
`= (1+2x)/(3x) - (3x-x^2+1)/(3x)`
`=(1+2x-3x+x^2-1)/(3x)`
`= (x^2 - x)/(3x)`
`=(x(x-1))/(3x)`
`=(x-1)/3`
Vậy `D=(x-1)/3` với `x\ne 0, x\ne -1, x\ne 1/2`
`2,`
`x^2-3x-4=0`
`<=>x^2-4x+x-4=0`
`<=>x(x-4)+(x-4)=0`
`<=>(x+1)(x-4)=0`
Do `x\ne -1 =>x+1 \ne 0`
`<=>x-4=0`
`<=>x=4`
`D=(x-1)/3 (x\ne 0, x\ne -1, x\ne 1/2)`
Thay `x=4` vào ta được :
`D=(4-1)/3=3/3=1`
Vậy `D=1` khi `x^2-3x-4=0`
`3,`
`D=(x-1)/3(x\ne 0, x\ne -1, x\ne 1/2)`
`=>1/D`$=\dfrac{1}{\dfrac{x-1}{3}}$
`=>1/D=3/(x-1)`
Để `1/D\in ZZ`
`=>3/(x-1)\in ZZ`
`=>3\vdots x-1`
`=>x-1\in Ư (3)={1;-1;3;-3}`
`=>x\in {2;0; 4; -2}` do `x\ne 0`
`=>x\in {2;4;-2}`
Vậy `x\in {2;4;-2}` để `1/D\in ZZ`
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