Biến đổi các phân thức hữu tỉ

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

\( = \dfrac{1}{{x + 1}} - \dfrac{{x\left( {x - 1} \right)\left( {x + 1} \right)}}{{{x^2} + 1}}\left( {\dfrac{1}{{{{\left( {x + 1} \right)}^2}}} - \dfrac{1}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right)\)   Điều kiện: \(x \ne  \pm 1\)

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

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

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

\( = \dfrac{{{x^2} + 1 + 2x}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x + 1} \right)\left( {{x^2} + 1} \right)}} = \dfrac{{x + 1}}{{{x^2} + 1}}\) . Vậy \(D = \dfrac{{x + 1}}{{{x^2} + 1}}\).

Phân thức \(B = \left( {\dfrac{1}{{x - 2}} - \dfrac{{2x}}{{4 - {x^2}}} + \dfrac{1}{{2 + x}}} \right):\left( {\dfrac{2}{x} - 1} \right)\) xác định khi

\(\left\{ \begin{array}{l}x - 2 \ne 0\\4 - {x^2} \ne 0\\2 + x \ne 0\\x \ne 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}x \ne 2\\x \ne  - 2\\x \ne 0\\{x^2} \ne 4\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}x \ne 2\\x \ne  - 2\\x \ne 0\end{array} \right.\)  .

ĐK: \(x \ne  \pm 3\)

Theo câu trước ta có: \(Q = \dfrac{{ - 3}}{{x - 3}}\).

Để \(Q = x - 1\) \( \Leftrightarrow \dfrac{{ - 3}}{{x - 3}} = x - 1\).

\( \Leftrightarrow \dfrac{{ - 3}}{{x - 3}} = \dfrac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{x - 3}}\)\( \Rightarrow \left( {x - 1} \right)\left( {x - 3} \right) =  - 3\)\( \Leftrightarrow {x^2} - 4x - 3 + 3 = 0 \Leftrightarrow {x^2} - 4x = 0\)

\( \Leftrightarrow x\left( {x - 4} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\left( {tm} \right)\\x = 4\left( {tm} \right)\end{array} \right.\)

Vậy \(x = 0;x = 4\) thỏa mãn yêu cầu đề bài.

Ta có \(\left| {2x + 1} \right| = 5\)\( \Leftrightarrow \left[ \begin{array}{l}2x + 1 = 5\\2x + 1 =  - 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = 4\\2x =  - 6\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 2\left( {TM} \right)\\x =  - 3\,\left( L \right)\end{array} \right.\)

Thay \(x = 2\) vào \(C = \dfrac{3}{{x - 3}}\) ta được \(C = \dfrac{3}{{2 - 3}} =  - 3\) .

ĐK: \(x \ne  \pm 3\)

Ta có:

\(Q = \left( {\dfrac{9}{{{x^3} - 9x}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{{x^2} + 3x}} - \dfrac{x}{{3x + 9}}} \right)\)

\( = \left( {\dfrac{9}{{x({x^2} - 9)}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{x(x + 3)}} - \dfrac{x}{{3(x + 3)}}} \right)\)

\( = \left( {\dfrac{9}{{x(x - 3)(x + 3)}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{x(x + 3)}} - \dfrac{x}{{3(x + 3)}}} \right)\)

\( = \dfrac{{9 + x(x - 3)}}{{x(x - 3)(x + 3)}}:\dfrac{{3(x - 3) - x.x}}{{3x(x + 3)}}\)

\( = \dfrac{{{x^2} - 3x + 9}}{{x(x - 3)(x + 3)}}:\dfrac{{ - {x^2} + 3x - 9}}{{3x(x + 3)}}\)     

 \( = \dfrac{{{x^2} - 3x + 9}}{{x(x - 3)(x + 3)}} \cdot \dfrac{{3x(x + 3)}}{{ - ({x^2} - 3x + 9)}} \)\(= \dfrac{{ - 3}}{{x - 3}}.\)

Vậy \(Q = \dfrac{{ - 3}}{{x - 3}}.\) 

Ta có  \(C = \left( {\dfrac{{21}}{{{x^2} - 9}} - \dfrac{{x - 4}}{{3 - x}} - \dfrac{{x - 1}}{{3 + x}}} \right):\left( {1 - \dfrac{1}{{x + 3}}} \right)\)

\( = \left[ {\dfrac{{21}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{\left( {x - 4} \right)\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right]:\left( {\dfrac{{x + 3 - 1}}{{x + 3}}} \right)\)   Điều kiện: \(x \ne  \pm 3\)

\( = \dfrac{{21 + {x^2} - x - 12 - {x^2} + 4x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\dfrac{{x + 2}}{{x + 3}}\)

\( = \dfrac{{3x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{{x + 2}}\)

\( = \dfrac{3}{{x - 3}}\). Vậy \(C = \dfrac{3}{{x - 3}}\) .

ĐK: \(x \ne  \pm 3\)

Ta có:

\(Q = \left( {\dfrac{9}{{{x^3} - 9x}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{{x^2} + 3x}} - \dfrac{x}{{3x + 9}}} \right)\)

\( = \left( {\dfrac{9}{{x({x^2} - 9)}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{x(x + 3)}} - \dfrac{x}{{3(x + 3)}}} \right)\)

\( = \left( {\dfrac{9}{{x(x - 3)(x + 3)}} + \dfrac{1}{{x + 3}}} \right):\left( {\dfrac{{x - 3}}{{x(x + 3)}} - \dfrac{x}{{3(x + 3)}}} \right)\)

\( = \dfrac{{9 + x(x - 3)}}{{x(x - 3)(x + 3)}}:\dfrac{{3(x - 3) - x.x}}{{3x(x + 3)}}\)

\( = \dfrac{{{x^2} - 3x + 9}}{{x(x - 3)(x + 3)}}:\dfrac{{ - {x^2} + 3x - 9}}{{3x(x + 3)}}\)     

 \( = \dfrac{{{x^2} - 3x + 9}}{{x(x - 3)(x + 3)}} \cdot \dfrac{{3x(x + 3)}}{{ - ({x^2} - 3x + 9)}} \)\(= \dfrac{{ - 3}}{{x - 3}}.\)

Vậy \(Q = \dfrac{{ - 3}}{{x - 3}}.\) 

Ta có  \(C = \left( {\dfrac{{21}}{{{x^2} - 9}} - \dfrac{{x - 4}}{{3 - x}} - \dfrac{{x - 1}}{{3 + x}}} \right):\left( {1 - \dfrac{1}{{x + 3}}} \right)\)

\( = \left[ {\dfrac{{21}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} + \dfrac{{\left( {x - 4} \right)\left( {x + 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \dfrac{{\left( {x - 1} \right)\left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}} \right]:\left( {\dfrac{{x + 3 - 1}}{{x + 3}}} \right)\)   Điều kiện: \(x \ne  \pm 3\)

\( = \dfrac{{21 + {x^2} - x - 12 - {x^2} + 4x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\dfrac{{x + 2}}{{x + 3}}\)

\( = \dfrac{{3x + 6}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\dfrac{{x + 3}}{{x + 2}}\)

\( = \dfrac{3}{{x - 3}}\). Vậy \(C = \dfrac{3}{{x - 3}}\) .

\(P + 1 = \dfrac{{ - 3x + 7}}{{x + 4}} + 1 = \dfrac{{ - 3x + 7 + x + 4}}{{x + 4}} = \dfrac{{ - 2x + 11}}{{x + 4}} =  - 2 + \dfrac{{19}}{{x + 4}}\)

\(x \in Z\) để \(P + 1 \in Z \Rightarrow \left( {x + 4} \right) \in Ư\left( {19} \right) = \left\{ { \pm 1;\, \pm 19} \right\}\)

Vậy \(x \in \left\{ { - 25; - 5; - 3;15} \right\}\) thì \(P + 1 \in Z\).

Theo câu trước ta có  $P = \dfrac{{ - 1}}{{x + 2}}$ với \(\left( {x \ne  \pm 2;\,x \ne 0} \right)\)

Để \(P = \dfrac{1}{x}\)$ \Leftrightarrow \dfrac{{ - 1}}{{x + 2}} = \dfrac{1}{x}$  \(\left( {x \ne  \pm 2;\,x \ne 0} \right)\)

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

Vậy \(x =  - 1\) .

ĐK: \(x \ne 1;x \ne  - 4\)

Theo kết quả các câu trước ta có: \(P = \dfrac{{ - 3x + 7}}{{x + 4}}\)

Để \(P = 2\) thì \(\dfrac{{ - 3x + 7}}{{x + 4}} = 2\) \( \Rightarrow  - 3x + 7 = 2\left( {x + 4} \right)\) \( \Leftrightarrow  - 3x + 7 = 2x + 8 \Leftrightarrow 5x =  - 1 \Leftrightarrow x =  - \dfrac{1}{5}\left( {tm} \right)\)

Vậy \(x =  - \dfrac{1}{5}\).

Ta có  \(P = \left( {\dfrac{x}{{x + 2}} - \dfrac{{{x^3} - 8}}{{{x^3} + 8}}.\dfrac{{{x^2} - 2x + 4}}{{{x^2} - 4}}} \right):\dfrac{4}{{x + 2}}\)

\( = \left( {\dfrac{x}{{x + 2}} - \dfrac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}.\dfrac{{{x^2} - 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right).\dfrac{{x + 2}}{4}\)      ĐK: \(x \ne  \pm 2\)

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

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

Vậy $P = \dfrac{{ - 1}}{{x + 2}}$ .

Ta có: \(P = \dfrac{{ - 3x + 7}}{{x + 4}}\)

Khi \(x =  - 1(t/m) \Rightarrow P = \dfrac{{ - 3.( - 1) + 7}}{{ - 1 + 4}} = \dfrac{{10}}{3}\)

Vậy khi \(x =  - 1\) thì \(P = \dfrac{{10}}{3}.\)

Ta có  \(P = \left( {\dfrac{x}{{x + 2}} - \dfrac{{{x^3} - 8}}{{{x^3} + 8}}.\dfrac{{{x^2} - 2x + 4}}{{{x^2} - 4}}} \right):\dfrac{4}{{x + 2}}\)

\( = \left( {\dfrac{x}{{x + 2}} - \dfrac{{\left( {x - 2} \right)\left( {{x^2} + 2x + 4} \right)}}{{\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)}}.\dfrac{{{x^2} - 2x + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}} \right).\dfrac{{x + 2}}{4}\)      ĐK: \(x \ne  \pm 2\)

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

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

Vậy $P = \dfrac{{ - 1}}{{x + 2}}$ .

ĐK: \(\left\{ \begin{array}{l}{x^2} + 3x - 4 \ne 0\\x + 4 \ne 0\\1 - x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {x - 1} \right)\left( {x + 4} \right) \ne 0\\x \ne 1\\x \ne  - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 1\\x \ne  - 4\end{array} \right..\)

\(P = \dfrac{{10x}}{{{x^2} + 3x - 4}} - \dfrac{{2x - 3}}{{x + 4}} + \dfrac{{x + 1}}{{1 - x}}\)

\( = \dfrac{{10x}}{{\left( {x - 1} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 3}}{{x + 4}} - \dfrac{{x + 1}}{{x - 1}}\)

\(= \dfrac{{10x - (2x - 3)(x - 1) - (x + 1)(x + 4)}}{{(x - 1)(x + 4)}}\)

\(= \dfrac{{10x - 2{x^2} + 2x + 3x - 3 - {x^2} - 4x - x - 4}}{{(x - 1)(x + 4)}}\)

\(= \dfrac{{ - 3{x^2} + 10x - 7}}{{(x - 1)(x + 4)}} \)\(=  - \dfrac{{ - (x - 1)(3x - 7)}}{{(x - 1)(x + 4)}}\)\( = \dfrac{{ - 3x + 7}}{{x + 4}}.\)

ĐK: \(x \ne  \pm 1\)

\(M\) có giá trị nguyên nghĩa là \(\dfrac{3}{{x + 1}}\) có giá trị nguyên

Suy ra \(3 \vdots \left( {x + 1} \right) \Rightarrow \left( {x + 1} \right) \in \) Ư\(\left( 3 \right) = \left\{ { - 1;1; - 3;3} \right\}\) .

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

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

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

+ \(x + 1 =  - 3 \Leftrightarrow x =  - 4\,\left( {TM} \right)\)

Vậy \(x \in \left\{ { - 4; - 2;2;0} \right\}\)

ĐK: \(\left\{ \begin{array}{l}{x^2} + 3x - 4 \ne 0\\x + 4 \ne 0\\1 - x \ne 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left( {x - 1} \right)\left( {x + 4} \right) \ne 0\\x \ne 1\\x \ne  - 4\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne 1\\x \ne  - 4\end{array} \right..\)

\(P = \dfrac{{10x}}{{{x^2} + 3x - 4}} - \dfrac{{2x - 3}}{{x + 4}} + \dfrac{{x + 1}}{{1 - x}}\)

\( = \dfrac{{10x}}{{\left( {x - 1} \right)\left( {x + 4} \right)}} - \dfrac{{2x - 3}}{{x + 4}} - \dfrac{{x + 1}}{{x - 1}}\)

\(= \dfrac{{10x - (2x - 3)(x - 1) - (x + 1)(x + 4)}}{{(x - 1)(x + 4)}}\)

\(= \dfrac{{10x - 2{x^2} + 2x + 3x - 3 - {x^2} - 4x - x - 4}}{{(x - 1)(x + 4)}}\)

\(= \dfrac{{ - 3{x^2} + 10x - 7}}{{(x - 1)(x + 4)}} \)\(=  - \dfrac{{ - (x - 1)(3x - 7)}}{{(x - 1)(x + 4)}}\)\( = \dfrac{{ - 3x + 7}}{{x + 4}}.\)

Sử dụng kết quả câu trước \(M = \dfrac{3}{{x + 1}}\) với $x \ne \pm 1$.

Để \(M =  - 1\) thì  \(\dfrac{3}{{x + 1}} =  - 1 \Leftrightarrow \dfrac{3}{{x + 1}} = \dfrac{{ - x - 1}}{{x + 1}} \Rightarrow  - x - 1 = 3 \Leftrightarrow x =  - 4\,\left( {TM} \right)\)

Vậy \(x =  - 4\) .

ĐK: \(x \ne 2\)

Đặt \(x - 1 = t\) ta có \(x = t + 1;\,x - 2 = t - 1\)

Do đó \(N = \left( {\dfrac{{x - 1}}{{{{\left( {x - 1} \right)}^2} + x}} - \dfrac{2}{{x - 2}}} \right):\left( {\dfrac{{{{\left( {x - 1} \right)}^4} + 2}}{{{{\left( {x - 1} \right)}^3} - 1}} - x + 1} \right)\)

\( = \left( {\dfrac{t}{{{t^2} + t + 1}} - \dfrac{2}{{t - 1}}} \right):\left( {\dfrac{{{t^4} + 2}}{{{t^3} - 1}} - t} \right)\)

\( = \left( {\dfrac{{t\left( {t - 1} \right) - 2\left( {{t^2} + t + 1} \right)}}{{\left( {t - 1} \right)\left( {{t^2} + t + 1} \right)}}} \right):\left( {\dfrac{{{t^4} + 2}}{{{t^3} - 1}} - \dfrac{{t\left( {{t^3} - 1} \right)}}{{{t^3} - 1}}} \right)\)

\( = \dfrac{{{t^2} - t - 2{t^2} - 2t - 2}}{{{t^3} - 1}}:\dfrac{{{t^4} + 2 - {t^4} + t}}{{{t^3} - 1}}\) \( = \dfrac{{ - {t^2} - 3t - 2}}{{{t^3} - 1}}.\dfrac{{{t^3} - 1}}{{t + 2}}\)

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

Thay \(x - 1 = t\) ta được \(N =  - \left( {x - 1} \right) - 1 =  - x\)

Vì \(x\) là số nguyên nên giá trị của \(N\) cũng luôn là số nguyên.

Sử dụng kết quả câu trước \(M = \dfrac{3}{{x + 1}}\) với $x \ne \pm 1$.

Thay \(x = \dfrac{1}{2}\) (TMĐK) vào \(M = \dfrac{3}{{x + 1}}\) ta được \(M = \dfrac{3}{{\dfrac{1}{2} + 1}} = \dfrac{3}{{\dfrac{3}{2}}} = 3:\dfrac{3}{2} = 3.\dfrac{2}{3} = 2\) . Vậy với \(x = \dfrac{1}{2}\) thì \(M = 2\) .