Rút gọn phân thức đại số

Với $2 < x < 6 \Rightarrow x - 2 > 0$ và $x - 6 < 0.$

$\Rightarrow |x - 2| = x - 2$ và  $|x - 6| = 6 - x.$

$A = \dfrac{{3\left| {x - 2} \right| - 5\left| {x - 6} \right|}}{{4{x^2} - 36{\rm{x}} + 81}} = \dfrac{{3(x - 2) - 5(6 - x)}}{{{{(2x - 9)}^2}}} = \dfrac{{3x - 6 - 30 + 5x}}{{{{(2x - 9)}^2}}} = \dfrac{{8x - 36}}{{{{(2x - 9)}^2}}} = \dfrac{{4(2x - 9)}}{{{{(2x - 9)}^2}}} = \dfrac{4}{{2x - 9}}.$

Ta có: $N = \dfrac{{\left( {{x^2} - 4{y^2}} \right)\left( {x - 2y} \right)}}{{{x^2} - 4xy + 4{y^2}}}$$= \dfrac{{\left( {x + 2y} \right)\left( {x - 2y} \right)\left( {x - 2y} \right)}}{{{{\left( {x - 2y} \right)}^2}}} = \dfrac{{\left( {x + 2y} \right){{\left( {x - 2y} \right)}^2}}}{{{{\left( {x - 2y} \right)}^2}}} = x + 2y$

Vậy $N = x + 2y$.

Thay $x =  - 9998$ và $y =  - 1$ vào $N = x + 2y$ ta được $N =  - 9998 + 2.\left( { - 1} \right) =  - 10000$.

Điều kiện: $x + 2 \ne 0 \Leftrightarrow x \ne  - 2$.

Ta có: $\dfrac{{{x^3} + 2{x^2} + 6x + 15}}{{x + 2}} = \dfrac{{{x^2}\left( {x + 2} \right) + 6x + 12 + 3}}{{x + 2}} = \dfrac{{{x^2}\left( {x + 2} \right) + 6\left( {x + 2} \right)}}{{x + 2}} + \dfrac{3}{{x + 2}}$$= \dfrac{{\left( {{x^2} + 6} \right)\left( {x + 2} \right)}}{{x + 2}} + \dfrac{3}{{x + 2}}$

$= {x^2} + 6 + \dfrac{3}{{x + 2}}$

Vì $x \in \mathbb{Z} \Rightarrow {x^2} + 6 \in \mathbb{Z}$ nên để phân thức trên đạt giá trị nguyên thì $\dfrac{3}{{x + 2}} \in \mathbb{Z} \Rightarrow x + 2 \in$ Ư$\left( 3 \right) = \left\{ { - 3; - 1;1;3} \right\}$

+ $x + 2 = 1 \Leftrightarrow x =  - 1\,\left( {TM} \right)$

+ $x + 2 =  - 1 \Leftrightarrow x =  - 3\,\left( {TM} \right)$

+ $x + 2 = 3 \Leftrightarrow x = 1\,\left( {TM} \right)$

+ $x + 2 =  - 3 \Leftrightarrow x =  - 5\,\left( {TM} \right)$

Vậy có $4$ giá trị của $x$ thỏa mãn đề bài là $- 1;\, - 3;\,1;\, - 5$.

Điều kiện: $2x + 1 \ne 0 \Leftrightarrow 2x \ne  - 1 \Leftrightarrow x \ne  - \dfrac{1}{2}$ .

Ta có: $\dfrac{5}{{2x + 1}} \in \mathbb{Z} \Rightarrow 2x + 1 \in$Ư$\left( 5 \right)$ $= \left\{ { - 1;1; - 5;5} \right\}$.

+ $2x + 1 =  - 1 \Leftrightarrow 2x =  - 2 \Leftrightarrow x =  - 1\,\,\left( {TM} \right)$

+ $2x + 1 = 1 \Leftrightarrow 2x = 0 \Leftrightarrow x = 0\,\,\left( {TM} \right)$

+ $2x + 1 =  - 5 \Leftrightarrow 2x =  - 6\, \Leftrightarrow x =  - 3\left( {TM} \right)$

+ $2x + 1 = 5 \Leftrightarrow 2x = 4\, \Leftrightarrow x = 2\,\left( {TM} \right)$

Vậy $x \in \left\{ { - 1;0;2; - 3} \right\}$.

Ta có: $Q = \dfrac{{{x^4} - {x^3} - x + 1}}{{{x^4} + {x^3} + 3{x^2} + 2x + 2}}$$= \dfrac{{{x^3}\left( {x - 1} \right) - \left( {x - 1} \right)}}{{{x^4} + {x^3} + {x^2} + 2{x^2} + 2x + 2}}$ $= \dfrac{{\left( {x - 1} \right)\left( {{x^3} - 1} \right)}}{{{x^2}\left( {{x^2} + x + 1} \right) + 2\left( {{x^2} + x + 1} \right)}}$

$= \dfrac{{\left( {x - 1} \right)\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}{{\left( {{x^2} + 2} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + 2}}$

Nhận thấy ${\left( {x - 1} \right)^2} \ge 0;\,{x^2} + 2 \ge 2 > 0;\,\forall x$  nên $\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + 2}} \ge 0;\,\forall x$ hay $Q \ge 0;\,\forall x$

Vậy $Q$ luôn nhận giá trị không âm với mọi $x$.

Với ${x^2} + 4x + 4 \ne 0 \Leftrightarrow {\left( {x + 2} \right)^2} \ne 0 \Leftrightarrow x \ne  - 2$. Ta có:

 $M = \dfrac{{{x^2} + 5x + 5}}{{{x^2} + 4{\rm{x}} + 4}}$$= \dfrac{{{x^2} + 4x + 4 + x + 1}}{{{x^2} + 4x + 4}} = \dfrac{{{x^2} + 4x + 4}}{{{x^2} + 4x + 4}} + \dfrac{{x + 1}}{{{{\left( {x + 2} \right)}^2}}}$ $= 1 + \dfrac{{x + 2}}{{{{\left( {x + 2} \right)}^2}}} - \dfrac{1}{{{{\left( {x + 2} \right)}^2}}}$

$= 1 + \dfrac{1}{{x + 2}} - \dfrac{1}{{{{\left( {x + 2} \right)}^2}}} =  - \left[ {\dfrac{1}{{{{\left( {x + 2} \right)}^2}}} - \dfrac{1}{{x + 2}} + \dfrac{1}{4}} \right] + \dfrac{5}{4}$ $= \dfrac{5}{4} - {\left( {\dfrac{1}{{x + 2}} - \dfrac{1}{2}} \right)^2}$

Ta có: ${\left( {\dfrac{1}{{x + 2}} - \dfrac{1}{2}} \right)^2} \ge 0$ với mọi $x \ne  - 2$. Suy ra $\dfrac{5}{4} - {\left( {\dfrac{1}{{x + 2}} - \dfrac{1}{2}} \right)^2} \le \dfrac{5}{4}$

Dấu “=” xảy ra khi $\dfrac{1}{{x + 2}} - \dfrac{1}{2} = 0 \Leftrightarrow \dfrac{1}{{x + 2}} = \dfrac{1}{2} \Rightarrow x + 2 = 2 \Leftrightarrow x = 0\,\,\left( {TM} \right)$.

Nên GTLN của $Q$ là $\dfrac{5}{4} \Leftrightarrow x = 0$.

Ta có: $Q = \dfrac{{10}}{{{x^2} - 4{\rm{x}} + 9}}$$= \dfrac{{10}}{{{x^2} - 4x + 4 + 5}} = \dfrac{{10}}{{{{\left( {x - 2} \right)}^2} + 5}}$.

Mà ${\left( {x - 2} \right)^2} \ge 0 \Leftrightarrow {\left( {x - 2} \right)^2} + 5 \ge 5,\,\forall x$. Dấu “=” xảy ra khi $x - 2 = 0 \Leftrightarrow x = 2$

nên GTNN của ${\left( {x - 2} \right)^2} + 5$ là $5$ khi $x = 2$.

Ta có: $Q$ đạt GTLN $\Leftrightarrow {\left( {x - 2} \right)^2} + 5$ đạt GTNN.

Hay GTLN của $Q$ là $\dfrac{{10}}{5} = 2 \Leftrightarrow x = 2$.

Ta có: $B = \dfrac{{{x^3} - {x^2}y + x{y^2}}}{{{x^3} + {y^3}}} = \dfrac{{x({x^2} - xy + {y^2})}}{{(x + y)({x^2} - xy + {y^2})}} = \dfrac{x}{{x + y}}.$

Thay $x =  - 5;\,\,y = 10.$ vào biểu thức B, ta được:

$B = \dfrac{{ - 5}}{{ - 5 + 10}} = \dfrac{{ - 5}}{5} =  - 1.$

Ta có: $\dfrac{{2{x^3} - 7{x^2} - 12x + 45}}{{3{x^3} - 19{x^2} + 33x - 9}} = \dfrac{{2{x^3} + 5{x^2} - 12{x^2} - 30x + 18x + 45}}{{3{x^3} - {x^2} - 18{x^2} + 6x + 27x - 9}}$$= \dfrac{{{x^2}(2x + 5) - 6x(2x + 5) + 9(2x + 5)}}{{{x^2}(3x - 1) - 6x(3x - 1) + 9(3x - 1)}}$  

$= \dfrac{{(2x + 5)({x^2} - 6x + 9)}}{{(3x - 1)({x^2} - 6x + 9)}} = \dfrac{{2x + 5}}{{3x - 1}}.$

Ta có: ${a^2}x + 3ax + 9 = {a^2}$$\Leftrightarrow {a^2}x + 3ax = {a^2} - 9 \Leftrightarrow x\left( {{a^2} + 3a} \right) = {a^2} - 9$ $\Leftrightarrow x = \dfrac{{{a^2} - 9}}{{{a^2} + 3a}} \Leftrightarrow x = \dfrac{{\left( {a - 3} \right)\left( {a + 3} \right)}}{{a\left( {a + 3} \right)}} \Leftrightarrow x = \dfrac{{a - 3}}{a}$ (vì $a \ne 0;a \ne  - 3$).

Vậy $x = \dfrac{{a - 3}}{a}$.

Ta có: $\dfrac{{{y^2} - {x^2}}}{{{x^2} - 3xy + 2{y^2}}} = \dfrac{{(y - x)(y + x)}}{{{x^2} - xy - 2xy + 2{y^2}}} = \dfrac{{(x + y)(y - x)}}{{x(x - y) - 2y(x - y)}} = \dfrac{{ - (x + y)(x - y)}}{{(x - y)(x - 2y)}} = \dfrac{{ - (x + y)}}{{x - 2y}}.$

Vậy mẫu thức của phân thức đã rút gọn là $x - 2y$.

Ta có: $\dfrac{{{x^3} + 2{x^2} - 3x - 6}}{{{x^2} + x - 2}} = \dfrac{{({x^3} + 2{x^2}) - 3x - 6}}{{{x^2} - x + 2x - 2}} = \dfrac{{{x^2}(x + 2) - 3(x + 2)}}{{x(x - 1) + 2(x - 1)}} = \dfrac{{(x + 2)({x^2} - 3)}}{{(x - 1)(x + 2)}} = \dfrac{{{x^2} - 3}}{{x - 1}}$

Ta có: $\dfrac{{{x^3} - 1}}{{{x^2} - 2x + 1}} = \dfrac{{(x - 1)({x^2} + x + 1)}}{{{{(x - 1)}^2}}} = \dfrac{{{x^2} + x + 1}}{{x - 1}}$ nên A đúng.

+ $\dfrac{{5(x - y) - 3(y - x)}}{{10(x - y)}} = \dfrac{{5(x - y) + 3(x - y)}}{{10(x - y)}} = \dfrac{{8(x - y)}}{{10(x - y)}} = \dfrac{4}{5}$ nên B đúng.

+ $\dfrac{{{x^2} + 4x + 3}}{{2x + 6}} = \dfrac{{{x^2} + x + 3x + 3}}{{2(x + 3)}} = \dfrac{{x(x + 1) + 3(x + 1)}}{{2(x + 3)}} = \dfrac{{(x + 1)(x + 3)}}{{2(x + 3)}} = \dfrac{{x + 1}}{2}$ nên C đúng.

+ $\dfrac{{16{x^2}y{{(x + y)}^2}}}{{12xy(x + y)}} = \dfrac{{16x.xy{{(x + y)}^2}}}{{12xy(x + y)}} = \dfrac{{4x\left( {x + y} \right)}}{3}$ nên D sai.

+ $\dfrac{{{x^2} - 3xy}}{{21{y^2} - 7xy}} = \dfrac{{x(x - 3y)}}{{7y(3y - x)}} = \dfrac{{x(x - 3y)}}{{ - 7y(x - 3y)}} = \dfrac{{ - x}}{{7y}}$ nên A sai

+ $\dfrac{{2x + 4}}{{{x^2} - x - 6}} = \dfrac{{2\left( {x + 2} \right)}}{{{x^2} - 3x + 2x - 6}} = \dfrac{{2\left( {x + 2} \right)}}{{x\left( {x - 3} \right) + 2\left( {x - 3} \right)}}$ $= \dfrac{{2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 3} \right)}} = \dfrac{2}{{x - 3}}$ nên B sai.

+ $\dfrac{{2x - 6y}}{{{x^2} - 9{y^2}}} = \dfrac{{2\left( {x - 3y} \right)}}{{\left( {x - 3y} \right)\left( {x + 3y} \right)}} = \dfrac{2}{{x + 3y}}$ nên C sai.

+  $\dfrac{{{x^2} - 2x}}{{{x^3} - 4x}} = \dfrac{{x\left( {x - 2} \right)}}{{x\left( {{x^2} - 4} \right)}} = \dfrac{{x\left( {x - 2} \right)}}{{x\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{1}{{x + 2}}$ nên D đúng.

Ta có: $T = \dfrac{{3{a^2} + 6ab + 3{b^2}}}{{a + b}} = \dfrac{{3\left( {{a^2} + 2ab + {b^2}} \right)}}{{a + b}} = \dfrac{{3{{\left( {a + b} \right)}^2}}}{{a + b}} = 3\left( {a + b} \right)$.

Mà $a + b = 3 \Rightarrow T = 3.3 = 9$.

Ta có: $\dfrac{{5{x^2} - 10xy + 5{y^2}}}{{{x^2} - {y^2}}} = \dfrac{{5({x^2} - 2xy + {y^2})}}{{(x - y)(x + y)}} = \dfrac{{5{{(x - y)}^2}}}{{(x - y)(x + y)}} = \dfrac{{5(x - y)}}{{x + y}}$.

Ta có: $\dfrac{{54{{(x - 3)}^3}}}{{63{{(3 - x)}^2}}} = \dfrac{{54{{(x - 3)}^3}}}{{63{{(x - 3)}^2}}} = \dfrac{6}{7}(x - 3)$.

Ta có $\dfrac{{6{x^2}{y^3}\left( {x + 3y} \right)}}{{18{x^2}y{{\left( {x + 3y} \right)}^2}}} = \dfrac{{6{x^2}y.\left( {x + 3y} \right).{y^2}}}{{6{x^2}y\left( {x + 3y} \right).3\left( {x + 3y} \right)}} = \dfrac{{{y^2}}}{{3\left( {x + 3y} \right)}}$.

Ta có $\dfrac{{{a^2} - 2a - 8}}{{{a^2} + 2a}} = \dfrac{{{a^2} - 4a + 2a - 8}}{{a\left( {a + 2} \right)}} = \dfrac{{a\left( {a - 4} \right) + 2\left( {a - 4} \right)}}{{a\left( {a + 2} \right)}}$$= \dfrac{{\left( {a + 2} \right)\left( {a - 4} \right)}}{{a\left( {a + 2} \right)}} = \dfrac{{a - 4}}{a}$ .

Ta có  $A = \dfrac{{2{x^2} - 4x + 2}}{{{{\left( {x - 1} \right)}^2}}} = \dfrac{{2\left( {{x^2} - 2x + 1} \right)}}{{{{\left( {x - 1} \right)}^2}}} = \dfrac{{2{{\left( {x - 1} \right)}^2}}}{{{{\left( {x - 1} \right)}^2}}} = 2$.