Đáp án:
d) \(x = \dfrac{4}{3}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)4{x^2} + 3 - 14x - 4{x^2} = 14\\
\to 14x = - 11\\
\to x = - \dfrac{{11}}{4}\\
c)\dfrac{{2\left( {2x + 3} \right) + 5\left( {3x - 5} \right)}}{{10}} = 4\\
\to 4x + 6 + 15x - 25 = 40\\
\to 19x = 59\\
\to x = \dfrac{{59}}{{19}}\\
b)2x\left( {x - 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
c)DK:x \ne \pm 3\\
\dfrac{{x\left( {x + 3} \right) + {{\left( {x - 3} \right)}^2} - {x^2} - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = 0\\
\to {x^2} + 3x + {x^2} - 6x + 9 - {x^2} - 9 = 0\\
\to {x^2} - 3x = 0\\
\to x\left( {x - 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 3\left( l \right)
\end{array} \right.\\
B1:\\
a)4{x^2} + 12x + 9 - 4{x^2} + 1 = 5\\
\to 12x = - 5\\
\to x = - \dfrac{5}{{12}}\\
b)\dfrac{{3x + 3 - 2x + 4 - 2x + 3 + 12x}}{6} = 0\\
\to 11x = - 10\\
\to x = - \dfrac{{10}}{{11}}\\
c)3x\left( {x - 2021} \right) - \left( {x - 2021} \right) = 0\\
\to \left( {x - 2021} \right)\left( {3x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2021\\
x = \dfrac{1}{3}
\end{array} \right.\\
d)DK:x \ne \pm 1\\
\dfrac{{5\left( {x + 1} \right) + 3\left( {x - 1} \right) - 2x - 10}}{{\left( {x - 1} \right)\left( {x + 1} \right)}} = 0\\
\to 5x + 5 + 3x - 3 - 2x - 10 = 0\\
\to 6x - 8 = 0\\
\to x = \dfrac{4}{3}
\end{array}\)
