Đáp án:
B4:
2x-3
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)3x + 3 - 3x = 3\\
b){x^2} - 2x + 1 + 2x - {x^2} = 1\\
B2:\\
a)\dfrac{{{x^2} + 9 + 6x}}{{x\left( {x + 3} \right)}} = \dfrac{{{{\left( {x + 3} \right)}^2}}}{{x\left( {x + 3} \right)}}\\
= \dfrac{{x + 3}}{x}\\
b)\dfrac{{\left( {x - 5} \right)\left( {x + 5} \right)}}{{4\left( {x + 5} \right)}}.\dfrac{{ - 4}}{{x - 5}}\\
= - 1\\
B3:\\
a)3x\left( {4y - x} \right)\\
b)x\left( {x + y} \right) - 2\left( {x + y} \right)\\
= \left( {x + y} \right)\left( {x - 2} \right)\\
B4:\\
\dfrac{{2{x^2} + x - 3}}{{x - 1}} = \dfrac{{2{x^2} - 2x + 3x - 3}}{{x - 1}}\\
= \dfrac{{2x\left( {x - 1} \right) + 3\left( {x - 1} \right)}}{{x - 1}}\\
= \dfrac{{\left( {x - 1} \right)\left( {2x - 3} \right)}}{{x - 1}}\\
= 2x - 3
\end{array}\)
