Đáp án:
m) x=-4
Giải thích các bước giải:
\(\begin{array}{l}
i){x^4} + 3{x^2} - 4 = 0\\
\to {x^4} - {x^2} + 4{x^2} - 4 = 0\\
\to {x^2}\left( {{x^2} - 1} \right) + 4\left( {{x^2} - 1} \right) = 0\\
\to \left( {{x^2} - 1} \right)\left( {{x^2} + 4} \right) = 0\\
\to {x^2} - 1 = 0\left( {do:{x^2} + 4 > 0\forall x} \right)\\
\to \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
l)DK:x \ne 1\\
\dfrac{{4x\left( {x - 1} \right) + 3 - 11\left( {x - 1} \right)}}{{x - 1}} = 0\\
\to 4{x^2} - 4x + 3 - 11x + 11 = 0\\
\to 4{x^2} - 15x + 14 = 0\\
\to 4{x^2} - 8x - 7x + 14 = 0\\
\to 4x\left( {x - 2} \right) - 7\left( {x - 2} \right) = 0\\
\to \left( {x - 2} \right)\left( {4x - 7} \right) = 0\\
\to \left[ \begin{array}{l}
x = 2\\
x = \dfrac{7}{4}
\end{array} \right.\\
m)DK:x \ne \left\{ { - 3; - 2} \right\}\\
\dfrac{{\left( {6x + 22} \right)\left( {x + 3} \right) - \left( {2x + 7} \right)\left( {x + 2} \right) - x - 4}}{{\left( {x + 2} \right)\left( {x + 3} \right)}} = 0\\
\to 6{x^2} + 40x + 66 - 2{x^2} - 11x - 14 - x - 4 = 0\\
\to 4{x^2} + 28x + 48 = 0\\
\to 4\left( {x + 3} \right)\left( {x + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 3\left( l \right)\\
x = - 4
\end{array} \right.
\end{array}\)
