Biến đổi các phân thức hữu tỉ

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

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 \(C = \dfrac{3}{{2 - 3}} =  - 3\) .

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 $P = \dfrac{{ - 1}}{{x + 2}}$ .

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 \(x =  - 1\) .

Ta có \(M = \left( {\dfrac{{x + 1}}{{x - 1}} - \dfrac{{x - 1}}{{x + 1}}} \right):\dfrac{{4x}}{{3x - 3}}\)   ĐK: \(x \ne  \pm 1\)

\( = \left[ {\dfrac{{{{\left( {x + 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} - \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right]:\dfrac{{4x}}{{3\left( {x - 1} \right)}}\) \( = \dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x - 1} \right)}}{{4x}}\) \( = \dfrac{{4x}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\dfrac{{3\left( {x - 1} \right)}}{{4x}} = \dfrac{3}{{x + 1}}\) .

Vậy \(M = \dfrac{3}{{x + 1}}\) .

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 \(M = 2\) .

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: \(x \ne  \pm 1\)

\(M\) 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: \(x \pm 1\)

Ta có \(E = \dfrac{{x{{\left( {1 - x} \right)}^2}}}{{1 + {x^2}}}:\left[ {\left( {\dfrac{{1 - {x^3}}}{{1 - x}} + x} \right).\left( {\dfrac{{1 + {x^3}}}{{1 + x}} - x} \right)} \right]\)

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

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

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

Suy ra \(E = \dfrac{x}{{\left( {1 + {x^2}} \right){{\left( {1 +x} \right)}^2}}}\)

Ta thấy với \(x \pm 1\) thì \(1 + {x^2} \ge 1 > 0\) và \({\left( {1 + x} \right)^2} > 0\) nên \(\left( {1 + {x^2}} \right){\left( {1 + x} \right)^2} > 0\) .

Suy ra \(E = \dfrac{x}{{\left( {1 + {x^2}} \right){{\left( {1 + x} \right)}^2}}} > 0 \Rightarrow x > 0\) nên B đúng, A, C sai.

\(E = \dfrac{x}{{\left( {1 + {x^2}} \right){{\left( {1 + x} \right)}^2}}} < 0 \Rightarrow x < 0\) nên D sai.

ĐKXĐ: \(x \ne 2.\)

Ta có: \(B = \dfrac{{x - 1}}{{x - 2}} = 1 + \dfrac{1}{{x - 2}}\)

\(B = 1 + \dfrac{1}{{x - 2}} \in Z \Leftrightarrow \dfrac{1}{{x - 2}} \in Z \Leftrightarrow x - 2 \in Ư(1) = \left\{ { \pm 1} \right\}\).

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

\(Q = \left[ {\dfrac{{{{(x - 1)}^2}}}{{3x + {{(x - 1)}^2}}} - \dfrac{{1 - 2{x^2} + 4x}}{{{x^3} - 1}} + \dfrac{1}{{x - 1}}} \right]:\dfrac{{3x}}{{{x^3} + x}}\)

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

\( = \left[ {\dfrac{{{{\left( {x - 1} \right)}^2}}}{{{x^2} + x + 1}} + \dfrac{{2{x^2} - 4x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{1}{{x - 1}}} \right].\dfrac{{x({x^2} + 1)}}{{3x}}\)

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

Vậy \(Q = \dfrac{{{x^2} + 1}}{3}\)  với \(x \ne  \pm 1;x \ne 0\).

Ta có: Q = \(\dfrac{{{x^2} + 1}}{3}\)  với \(x \ne 0;x \ne  \pm 1\).

Ta có: \({x^2} \ge 4\,\,\forall x \ge 2 \Rightarrow {x^2} + 1 \ge 5\,\,\forall x \ge 2\)\( \Rightarrow \dfrac{{{x^2} + 1}}{3} \ge \dfrac{5}{3} \,\,\forall x \ge 2\).

Dấu “=” xảy ra khi \(x = 2\left( {tm} \right)\).

Vậy  \(Min\,\,Q = \dfrac{5}{3} \Leftrightarrow x = 2\).

Từ \(x + y + z = 0 \Rightarrow x + y =  - z \Rightarrow {x^2} + 2xy + {y^2} = {z^2} \Rightarrow {x^2} + {y^2} - {z^2} =  - 2xy\).

Tương tự ta có: \(\left\{ \begin{array}{l}{y^2} + {z^2} - {x^2} =  - 2yz\\{z^2} + {x^2} - {y^2} =  - 2zx\end{array} \right.\)                           

Do đó: \(A = \dfrac{{xy}}{{ - 2xy}} + \dfrac{{yz}}{{ - 2yz}} + \dfrac{{zx}}{{ - 2zx}} =  - \dfrac{1}{2} - \dfrac{1}{2} - \dfrac{1}{2} =  - \dfrac{3}{2}\).

Vậy \(A =  - \dfrac{3}{2}.\)