Cộng, trừ các phân thức

Quy tắc: Muốn cộng hai phân thức cùng mẫu thức ta cộng các tử thức với nhau và giữ nguyên mẫu thức.

$\dfrac{A}{B} + \dfrac{C}{B} = \dfrac{{A + C}}{B}\,\,\left( {B \ne 0} \right)$

Muốn trừ phân thức $\dfrac{A}{B}$ cho phân thức $\dfrac{C}{D}$ ta cộng $\dfrac{A}{B}$ với phân thức đối của $\dfrac{C}{D}$ :   $\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{A}{B} + \dfrac{{ - C}}{D}$ .

Phân thức đối của phân thức $\dfrac{{ - x}}{{x - 1}}$ là $- \dfrac{{ - x}}{{x - 1}} = \dfrac{x}{{x - 1}}$ .

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

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

* $\dfrac{{{x^2} + 4}}{{x - 2}} + \dfrac{{4x}}{{2 - x}}$$= \dfrac{{{x^2} + 4}}{{x - 2}} - \dfrac{{4x}}{{x - 2}} = \dfrac{{{x^2} - 4x + 4}}{{x - 2}} = \dfrac{{{{\left( {x - 2} \right)}^2}}}{{x - 2}} = x - 2$  nên B đúng.

* $\dfrac{{2x}}{{x - 2}} + \dfrac{4}{{{x^2} - 4}}$$= \dfrac{{2x}}{{x - 2}} + \dfrac{4}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2x\left( {x + 2} \right) + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{2{x^2} + 4x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}$  nên C sai.

* $\dfrac{{{x^2}}}{{x - 2}} - \dfrac{4}{{x - 2}}$$= \dfrac{{{x^2} - 4}}{{x - 2}} = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x - 2}} = x + 2$  nên D sai.

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

$= \dfrac{{{x^2} - 2xy + {y^2}}}{{xy\left( {x - y} \right)}} = \dfrac{{{{\left( {x - y} \right)}^2}}}{{xy\left( {x - y} \right)}} = \dfrac{{x - y}}{{xy}}$.

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

Phân thức cần tìm là

$\dfrac{{2x - 6}}{{x + 3}} - \dfrac{{x + 1}}{2} = \dfrac{{2(2x - 6) - (x + 3)(x + 1)}}{{2(x + 3)}} = \dfrac{{4x - 12 - {x^2} - x - 3x - 3}}{{2(x + 3)}} = \dfrac{{ - {x^2} - 15}}{{2(x + 3)}}.$

* $\dfrac{{3x - 4}}{{4{x^2}{y^5}}} + \dfrac{{9x + 4}}{{4{x^2}{y^5}}} = \dfrac{{3x - 4 + 9x + 4}}{{4{x^2}{y^5}}} = \dfrac{{12x}}{{4{x^2}{y^5}}} = \dfrac{3}{{x{y^5}}}$  nên A sai.

* $\dfrac{{2x + 5}}{3} + \dfrac{{x - 2}}{3} = \dfrac{{2x + 5 + x - 2}}{3} = \dfrac{{3x + 3}}{3} = x + 1$ nên B sai.

*

 $\begin{array}{l}\,\,\,\,\dfrac{{x + 8}}{{x - 1}} - \dfrac{{2x - 1}}{{x - 1}} - \dfrac{{6x + 2}}{{x - 1}}\\ = \dfrac{{x + 8 - (2x - 1) - (6x + 2)}}{{x - 1}}\\ = \dfrac{{x + 8 - 2x + 1 - 6x - 2}}{{x - 1}}\\ = \dfrac{{ - 7x + 7}}{{x - 1}} = \dfrac{{ - 7(x - 1)}}{{x - 1}} =  - 7.\end{array}$

nên C sai.

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

$= \dfrac{{{x^2} + xy + xy - {y^2} + 2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}}$  $= \dfrac{{{x^2} + {y^2} + 2xy}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{x + y}}{{x - y}}$  nên D đúng.

* $\dfrac{{11x + 13}}{{3x - 3}} + \dfrac{{15x + 17}}{{4 - 4x}} = \dfrac{{11x + 13}}{{3\left( {x - 1} \right)}} - \dfrac{{15x + 17}}{{4\left( {x - 1} \right)}} = \dfrac{{4\left( {11x + 13} \right)}}{{12\left( {x - 1} \right)}} - \dfrac{{3\left( {15x + 17} \right)}}{{12\left( {x - 1} \right)}}$$= \dfrac{{44x + 52 - 45x - 51}}{{12\left( {x - 1} \right)}} = \dfrac{{ - x + 1}}{{12\left( {x - 1} \right)}}$

$= \dfrac{{ - \left( {x - 1} \right)}}{{12\left( {x - 1} \right)}} =  - \dfrac{1}{{12}}$ nên A đúng, B sai.

* $\dfrac{{xy}}{{{x^2} - {y^2}}} - \dfrac{{{x^2}}}{{{y^2} - {x^2}}} = \dfrac{{xy}}{{{x^2} - {y^2}}} + \dfrac{{{x^2}}}{{{x^2} - {y^2}}} = \dfrac{{xy + {x^2}}}{{{x^2} - {y^2}}}$$= \dfrac{{x\left( {x + y} \right)}}{{\left( {x + y} \right)\left( {x - y} \right)}} = \dfrac{x}{{x - y}} = \dfrac{{ - x}}{{y - x}}$ nên C, D đúng.

Điều kiện: $x \ne 1.$

$\begin{array}{l}\dfrac{{4{x^2} - 3x + 5}}{{{x^3} - 1}} - \dfrac{{1 - 2x}}{{{x^2} + x + 1}} - \dfrac{6}{{x - 1}}\\ = \dfrac{{4{x^2} - 3x + 5 - (1 - 2x)(x - 1) - 6({x^2} + x + 1)}}{{(x - 1)({x^2} + x + 1)}}\\ = \dfrac{{4{x^2} - 3x + 5 - x + 1 + 2{x^2} - 2x - 6{x^2} - 6x - 6}}{{(x - 1)({x^2} + x + 1)}}\\ = \dfrac{{ - 12x}}{{{x^3} - 1}}.\end{array}$

Điều kiện: $x \ne \dfrac{6}{5}.$

$\begin{array}{l}D = \,\dfrac{{5{x^2}}}{{5x - 6}} + \dfrac{{{x^2} - 1}}{{6 - 5x}} - \dfrac{{7 + x - {x^2}}}{{5x - 6}}\\ = \dfrac{{5{x^2}}}{{5x - 6}} - \dfrac{{{x^2} - 1}}{{5x - 6}} + \dfrac{{{x^2} - x - 7}}{{5x - 6}}\\ = \dfrac{{5{x^2} - {x^2} + 1 + {x^2} - x - 7}}{{5x - 6}}\\ = \dfrac{{5{x^2} - x - 6}}{{5x - 6}}\\ = \dfrac{{(x + 1)(5x - 6)}}{{5x - 6}} = x + 1.\end{array}$.

Ta có  $P = \dfrac{{6{x^2} + 8x + 7}}{{{x^3} - 1}} + \dfrac{x}{{{x^2} + x + 1}} - \dfrac{6}{{x - 1}}$$= \dfrac{{6{x^2} + 8x + 7}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{6\left( {{x^2} + x + 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$$= \dfrac{{6{x^2} + 8x + 7 + {x^2} - x - 6{x^2} - 6x - 6}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}$$= \dfrac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{1}{{x - 1}}$

Thay $x = \dfrac{1}{2}$ vào $P = \dfrac{1}{{x - 1}}$  ta được $P = \dfrac{1}{{\dfrac{1}{2} - 1}} = \dfrac{1}{{\dfrac{{ - 1}}{2}}} =  - 2$.

Ta có $M = \dfrac{{10}}{{\left( {x + 2} \right)\left( {3 - x} \right)}} - \dfrac{{12}}{{\left( {3 - x} \right)\left( {3 + x} \right)}} - \dfrac{1}{{\left( {x + 3} \right)\left( {x + 2} \right)}}$$= \dfrac{{10\left( {x + 3} \right) - 12\left( {x + 2} \right) - \left( {3 - x} \right)}}{{\left( {x + 2} \right)\left( {3 - x} \right)\left( {x + 3} \right)}}$

$= \dfrac{{10x + 30 - 12x - 24 - 3 + x}}{{\left( {x + 2} \right)\left( {3 - x} \right)\left( {x + 3} \right)}}$$= \dfrac{{ - x + 3}}{{\left( {x + 2} \right)\left( {3 - x} \right)\left( {x + 3} \right)}} = \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}$

Thay $x =  - 0,25$ vào $M = \dfrac{1}{{\left( {x + 2} \right)\left( {x + 3} \right)}}$ ta được $M = \dfrac{1}{{\left( { - 0,25 + 2} \right)\left( { - 0,25 + 3} \right)}} = \dfrac{{16}}{{77}}$ .

Ta có  $2a - b = 7$ ta suy ra $b = 2a - 7$. Thay $b = 2a - 7$ vào $A = \dfrac{{5a - b}}{{3a + 7}} + \dfrac{{3b - 2a}}{{2b - 7}}$ ta được

$A = \dfrac{{5a - \left( {2a - 7} \right)}}{{3a + 7}} + \dfrac{{3\left( {2a - 7} \right) - 2a}}{{2\left( {2a - 7} \right) - 7}}$ $= \dfrac{{3a + 7}}{{3a + 7}} + \dfrac{{4a - 21}}{{4a - 21}} = 1 + 1 = 2$.

$\dfrac{{4x - 7}}{{{x^2} - 3x + 2}} = \dfrac{a}{{x - 1}} + \dfrac{b}{{x - 2}}$

Xét $VP = \,\dfrac{a}{{x - 1}} + \dfrac{b}{{x - 2}} = \dfrac{{a(x - 2) + b(x - 1)}}{{(x - 1)(x - 2)}}$$= \dfrac{{ax - 2a + bx - b}}{{{x^2} - 3x + 2}}$ $= \dfrac{{(a + b)x - 2a - b}}{{{x^2} - 3x + 2}} = \dfrac{{4x - 7}}{{{x^2} - 3x + 2}}$

$\Rightarrow \left\{ \begin{array}{l}a + b = 4\\ - 2a - b =  - 7\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 1\end{array} \right.$

Vậy $a = 3,b = 1$

Ta có  $\dfrac{{{x^2} + 5}}{{{x^3} - 3x - 2}} = \dfrac{a}{{x - 2}} + \dfrac{b}{{{{(x + 1)}^2}}}$

Xét

$\begin{array}{l}VP = \dfrac{a}{{x - 2}} + \dfrac{b}{{{{(x + 1)}^2}}}\\ = \dfrac{{a({x^2} + 2x + 1) + b(x - 2)}}{{(x - 2){{(x + 1)}^2}}}\end{array}$

$\begin{array}{l} = \dfrac{{a{x^2} + 2ax + a + bx - 2b}}{{(x - 2){{(x + 1)}^2}}}\\ = \dfrac{{a{x^2} + (2a + b)x + a - 2b}}{{(x - 2){{(x + 1)}^2}}}\\ = \dfrac{{{x^2} + 5}}{{{x^3} - 3x - 2}}\end{array}$

$\Rightarrow \left\{ \begin{array}{l}a = 1\\2a + b = 0\\a - 2b = 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 1\\b =  - 2\end{array} \right.$

Suy ra $a + b = 1 + \left( { - 2} \right) =  - 1$

ĐK: $x \ne {\rm{\{ }} - 2;2;3\}$.

$\begin{array}{l}P + \dfrac{{4x - 12}}{{{x^3} - 3{x^2} - 4x + 12}} = \dfrac{3}{{x - 3}} - \dfrac{{{x^2}}}{{4 - {x^2}}}\\P = \dfrac{3}{{x - 3}} - \dfrac{{{x^2}}}{{4 - {x^2}}} - \dfrac{{4x - 12}}{{{x^3} - 3{x^2} - 4x + 12}}\\\end{array}$

$\begin{array}{l}P = \dfrac{3}{{x - 3}} + \dfrac{{{x^2}}}{{(x - 2)(x + 2)}} - \dfrac{{4x - 12}}{{{x^2}(x - 3) - 4(x - 3)}}\\P = \dfrac{{3\left( {{x^2} - 4} \right)}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}} + \dfrac{{{x^2}\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}} - \dfrac{{4x - 12}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}}\\P = \dfrac{{3{x^2} - 12 + {x^3} - 3{x^2} - 4x + 12}}{{\left( {x - 3} \right)\left( {{x^2} - 4} \right)}}\\\end{array}$

$\begin{array}{l}P = \dfrac{{{x^3} - 4x}}{{(x - 3)(x - 2)(x + 2)}}\\P = \dfrac{{x({x^2} - 4)}}{{(x - 3)(x - 2)(x + 2)}} = \dfrac{x}{{x - 3}}\end{array}$

Quy tắc: Muốn cộng hai phân thức cùng mẫu thức ta cộng các tử thức với nhau và giữ nguyên mẫu thức.

$\dfrac{A}{B} + \dfrac{C}{B} = \dfrac{{A + C}}{B}\,\,\left( {B \ne 0} \right)$

Muốn trừ phân thức $\dfrac{A}{B}$ cho phân thức $\dfrac{C}{D}$ ta cộng $\dfrac{A}{B}$ với phân thức đối của $\dfrac{C}{D}$ :   $\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{A}{B} + \dfrac{{ - C}}{D}$ .