Đáp án:
b)x=-4
3)(x;y)=\(
(0;0);(3;3);(\frac{{5 + \sqrt 5 }}{2};\frac{{5 - \sqrt 5 }}{2});(\frac{{5 - \sqrt 5 }}{2};\frac{{5 + \sqrt 5 }}{2})
\)
Giải thích các bước giải:
\(
\begin{array}{l}
b)\frac{{13}}{{2x^2 + x - 21}} + \frac{1}{{2x + 7}} = \frac{6}{{x^2 - 9}} \\
đkxđ:\left\{ {\begin{array}{*{20}c}
{x \ne \pm 3} \\
{x \ne \frac{{ - 7}}{2}} \\
\end{array}} \right. \\
\Leftrightarrow \frac{{13}}{{(x - 3)(2x + 7)}} + \frac{1}{{2x + 7}} = \frac{6}{{(x - 3)(x + 3)}} \\
\Leftrightarrow 13(x + 3) + (x - 3)(x + 3) = 6(2x + 7) \\
\Leftrightarrow x^2 - 9 + 13x + 39 = 12x + 42 \\
\Leftrightarrow x^2 + x - 12 = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = 3(l)} \\
{x = - 4(tm)} \\
\end{array}} \right. = > x = - 4 \\
3)\left\{ {\begin{array}{*{20}c}
{x^2 + y = 4x} \\
{y^2 + x = 4y} \\
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}c}
{x^2 - y^2 + y - x = 4x - 4y} \\
{x^2 + y = 4x} \\
\end{array}} \right. \\
\Leftrightarrow \left\{ {\begin{array}{*{20}c}
{(x - y)(x + y) - 5(x - y) = 0} \\
{x^2 + y = 4x} \\
\end{array}} \right. \\
\Leftrightarrow \left\{ \begin{array}{l}
(x - y)(x + y - 5) = 0 \\
x^2 + y = 4x \\
\end{array} \right. \\
\Leftrightarrow \left\{ {\begin{array}{*{20}c}
{\left[ {\begin{array}{*{20}c}
{x = y} \\
{y = 5 - x} \\
\end{array}} \right.} \\
{x^2 + y = 4x(1)} \\
\end{array}} \right. \\
+ )x = y = > (1) \Leftrightarrow x^2 + x = 4x \\
\Leftrightarrow x^2 - 3x = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = y = 0} \\
{x = y = 3} \\
\end{array}} \right. = > (x;y) = (0;0);(3;3) \\
+ )y = 5 - x = > (1) \Leftrightarrow x^2 + 5 - x = 4x \\
\Leftrightarrow x^2 - 5x + 5 = 0 \\
\Leftrightarrow \left[ {\begin{array}{*{20}c}
{x = \frac{{5 + \sqrt 5 }}{2}} \\
{x = \frac{{5 - \sqrt 5 }}{2}} \\
\end{array} = > \left[ {\begin{array}{*{20}c}
{y = \frac{{5 - \sqrt 5 }}{2}} \\
{y = \frac{{5 + \sqrt 5 }}{2}} \\
\end{array}} \right.} \right. = > (x;y) = (\frac{{5 + \sqrt 5 }}{2};\frac{{5 - \sqrt 5 }}{2});(\frac{{5 - \sqrt 5 }}{2};\frac{{5 + \sqrt 5 }}{2}) \\
\end{array}
\)
