Đáp án:

B1:

b) x=6

Giải thích các bước giải:

\(\begin{array}{l}
B1:\\
DK:x \ne  \pm 1\\
\dfrac{{{x^2} + 2x + 1 - {x^2} + 2x - 1 - {x^2} - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
 \to \dfrac{{{x^2} + 4x - 3}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
 \to {x^2} + 4x - 3 = 0\\
 \to {x^2} + 4x + 4 - 7 = 0\\
 \to {\left( {x + 2} \right)^2} = 7\\
 \to \left[ \begin{array}{l}
x + 2 = \sqrt 7 \\
x + 2 =  - \sqrt 7 
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 2 + \sqrt 7 \\
x =  - 2 - \sqrt 7 
\end{array} \right.\\
b)DK:x \ne  \pm 4\\
\dfrac{{96 - \left( {2x + 1} \right)\left( {x - 4} \right)}}{{\left( {x + 4} \right)\left( {x - 4} \right)}} = \dfrac{{ - \left( {1 - 3x} \right)\left( {x + 4} \right) - 5\left( {{x^2} - 16} \right)}}{{\left( {x - 4} \right)\left( {x + 4} \right)}}\\
 \to 96 - 2{x^2} + 7x + 4 = 3{x^2} + 11x - 4 - 5{x^2} + 80\\
 \to 4x = 24\\
 \to x = 6\\
c)DK:x \ne \left\{ { - 5;0;5} \right\}\\
\dfrac{{x\left( {x + 25} \right) - 2{{\left( {x + 5} \right)}^2} - \left( {5 - x} \right)\left( {x - 5} \right)}}{{2x\left( {x + 5} \right)\left( {x - 5} \right)}} = 0\\
 \to {x^2} + 25x - 2{x^2} - 20x - 50 + {x^2} - 10x + 25 = 0\\
 \to  - 5x - 25 = 0\\
 \to x =  - 5\left( l \right)\\
 \to x \in \emptyset 
\end{array}\)

\(\begin{array}{l}
B2:\\
a)DK:x \ne \left\{ { - 3;1} \right\}\\
\dfrac{{4 - \left( {2x - 5} \right)\left( {x - 2} \right) + 2x\left( {x + 3} \right)}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = 0\\
 \to \dfrac{{4 - 2{x^2} + 9x - 10 + 2{x^2} + 6x}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = 0\\
 \to 15x = 6\\
 \to x = \dfrac{2}{5}\\
b)DK:x \ne \left\{ { - 2;1} \right\}\\
\dfrac{{3 - x - 2 + 7\left( {x - 1} \right)}}{{\left( {x + 2} \right)\left( {x - 1} \right)}} = 0\\
 \to 6x - 6 = 0\\
 \to x = 1\left( l \right)\\
 \to x \in \emptyset \\
c)DK:x \ne \left\{ {2;4} \right\}\\
\dfrac{{ - 2 - \left( {x - 1} \right)\left( {x - 4} \right) - \left( {x + 3} \right)\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = 0\\
 \to  - 2 - {x^2} + 5x - 4 - {x^2} - x + 6 = 0\\
 \to  - 2{x^2} + 4x = 0\\
 \to  - 2x\left( {x - 2} \right) = 0\\
 \to \left[ \begin{array}{l}
x = 0\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)