Đáp án:
B3:
b) \(x = 3\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)x\left( {x - 3} \right)\left( {x + 3} \right)\\
b)x\left( {{x^2} - 2x + 1} \right)\\
= x{\left( {x - 1} \right)^2}\\
B2:\\
a){x^3} + {x^2} + 2x + 2\\
b)\dfrac{{2{x^4} + {x^3} - 3{x^2} + 5x - 2}}{{{x^2} - x + 1}}\\
= \dfrac{{2{x^4} - {x^3} + 2{x^3} - {x^2} - 2{x^2} + x + 4x - 2}}{{{x^2} - x + 1}}\\
= \dfrac{{{x^3}\left( {2x - 1} \right) + {x^2}\left( {2x - 1} \right) - x\left( {2x - 1} \right) + 2\left( {2x - 1} \right)}}{{{x^2} - x + 1}}\\
= \dfrac{{\left( {2x - 1} \right)\left( {{x^3} + {x^2} - x + 2} \right)}}{{{x^2} - x + 1}}\\
= \dfrac{{\left( {2x - 1} \right)\left( {{x^3} + 2{x^2} - {x^2} - 2x + x + 2} \right)}}{{{x^2} - x + 1}}\\
= \dfrac{{\left( {2x - 1} \right)\left[ {{x^2}\left( {x + 2} \right) - x\left( {x + 2} \right) + \left( {x + 2} \right)} \right]}}{{{x^2} - x + 1}}\\
= \dfrac{{\left( {2x - 1} \right)\left( {x + 2} \right)\left( {{x^2} - x + 1} \right)}}{{{x^2} - x + 1}}\\
= \left( {2x - 1} \right)\left( {x + 2} \right)\\
B3:\\
a)DK:x \ne \pm 1\\
b)A = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{x - 1}}{{x + 1}}\\
A = \dfrac{1}{2} \to \dfrac{{x - 1}}{{x + 1}} = \dfrac{1}{2}\\
\to 2x - 2 = x + 1\\
\to x = 3
\end{array}\)
