Đáp án:
Bài 2:
d. \(\left[ \begin{array}{l}
x = - 1\\
x = 8
\end{array} \right.\)
f. \(\left[ \begin{array}{l}
x = 5\\
x = 4
\end{array} \right.\)
g. x=0
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
d.\left( {x - 6} \right)\left( {x + 1} \right) - 2\left( {x + 1} \right) = 0\\
\to \left( {x + 1} \right)\left( {x - 6 - 2} \right) = 0\\
\to \left[ \begin{array}{l}
x + 1 = 0\\
x - 8 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 8
\end{array} \right.\\
f.{x^2} - 5x - 4x + 20 = 0\\
\to x\left( {x - 5} \right) - 4\left( {x - 5} \right) = 0\\
\to \left( {x - 5} \right)\left( {x - 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = 4
\end{array} \right.\\
g.x\left( {{x^2} - 4x + 5} \right) = 0\\
\to x = 0\left( {do:{x^2} - 4x + 5 > 0\forall x} \right)\\
B3:\\
a.DK:x \ne \left\{ {2;3} \right\}\\
Pt \to {x^2} - 3x = {x^2} - 4x + 4\\
\to x = 4\left( {TM} \right)\\
b.DK:x \ne \left\{ {1;2} \right\}\\
Pt \to \left( {2x - 4} \right)\left( {x - 2} \right) - \left( {x - 3} \right)\left( {x - 1} \right) = \left( {x - 1} \right)\left( {x - 2} \right)\\
\to 2{x^2} - 8x + 8 - {x^2} + 4x - 3 = {x^2} - 3x + 2\\
\to - x = - 3 \to x = 3\left( {TM} \right)\\
c.DK:x \ne 0\\
Pt \to 2{x^2} - 12 = 2{x^2} + 3x\\
\to 3x = - 12\\
\to x = - 4\left( {TM} \right)\\
d.DK:x \ne \left\{ { - 7;\frac{3}{2}} \right\}\\
Pt \to 6{x^2} - 13x + 6 = 6{x^2} + 43x + 7\\
\to 56x = - 1\\
\to x = - \frac{1}{{56}}\left( {TM} \right)
\end{array}\)
