Đáp án:
h) \(x = - {3}{{26}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {2;4} \right\}\\
{{\left( {x - 6} \right)\left( {x - 2} \right) - x\left( {x - 4} \right)}}{{\left( {x - 4} \right)\left( {x - 2} \right)}} = 0\\
\to {x^2} - 8x + 12 - {x^2} + 4x = 0\\
\to - 4x + 12 = 0\\
\to x = 3\\
b)DK:x \ne \left\{ {2;4} \right\}\\
{{\left( {x - 3} \right)\left( {x - 4} \right) - {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = {{16}}{5}\\
\to 5\left( {{x^2} - 7x + 12} \right) - 5\left( {{x^2} - 4x + 4} \right) = 16\left( {{x^2} - 6x + 8} \right)\\
\to 5{x^2} - 35x + 60 - 5{x^2} + 20x - 20 = 16{x^2} - 96x + 128\\
\to 16{x^2} - 81x + 88 = 0\\
\to \left[ \begin{array}{l}
x = {{81 + \sqrt {929} }}{{32}}\\
x = {{81 - \sqrt {929} }}{{32}}
\end{array} \right.\\
c)DK:x \ne \left\{ { - 7;{3}{2}} \right\}\\
{{\left( {3x - 2} \right)\left( {2x - 3} \right) - \left( {6x + 1} \right)\left( {x + 7} \right)}}{{\left( {x + 7} \right)\left( {2x - 3} \right)}} = 0\\
\to 6{x^2} - 11x + 6 - 6{x^2} - 43x - 7 = 0\\
\to - 54x - 1 = 0\\
\to x = - {1}{{54}}\\
d)DK:x \ne \pm 1\\
{{\left( {2x + 1} \right)\left( {x + 1} \right) - 5{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to 2{x^2} + 3x + 1 - 5\left( {{x^2} - 2x + 1} \right) = 0\\
\to 2{x^2} + 3x + 1 - 5{x^2} + 10x - 5 = 0\\
\to - 3{x^2} + 13x - 4 = 0\\
\to \left( {4 - x} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = {1}{3}
\end{array} \right.\\
e)DK:x \ne \pm 2\\
{{{{\left( {x - 2} \right)}^2} - 3\left( {x + 2} \right) - 2\left( {x - 11} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = 0\\
\to {x^2} - 4x + 4 - 3x - 6 - 2x + 22 = 0\\
\to {x^2} - 9x + 20 = 0\\
\to \left( {x - 4} \right)\left( {x - 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = 4\\
x = 5
\end{array} \right.\\
g)DK:x \ne \pm 1\\
{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2} - 4}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to {x^2} + 2x + 1 - {x^2} + 2x - 1 = 4\\
\to 4x = 4\\
\to x = 1\left( l \right)\\
\to x \in \emptyset \\
h)DK:x \ne \pm {1}{2}\\
{{8{x^2}}}{{3\left( {1 - 2x} \right)\left( {1 + 2x} \right)}} = {{2x}}{{3\left( {2x - 1} \right)}} - {{8x + 1}}{{4\left( {1 + 2x} \right)}}\\
\to {{32{x^2}}}{{12\left( {1 - 2x} \right)\left( {1 + 2x} \right)}} = {{ - 4.2x\left( {1 + 2x} \right) - 3\left( {8x + 1} \right)\left( {1 - 2x} \right)}}{{12\left( {1 - 2x} \right)\left( {1 + 2x} \right)}}\\
\to 32{x^2} = - 8x - 16{x^2} - 3\left( { - 16{x^2} + 6x + 1} \right)\\
\to 32{x^2} = 32{x^2} - 26x - 3\\
\to 26x = - 3\\
\to x = - {3}{{26}}
\end{array}\)
