Đáp án:
3)
a) \(\left( {x + y} \right)\left( {5x - 1} \right)\)
b) \(= \left( {x - 1} \right)\left( {x + 6} \right)\)
4b) \(\dfrac{{{x^2} - 3x}}{{10\left( {{x^2} - 1} \right)}}\)
Giải thích các bước giải:
3)
a) \(5{x^2} + 5xy - x - y\)
\(\begin{array}{l} = \left( {5{x^2} + 5xy} \right) - \left( {x + y} \right)\\ = 5x\left( {x + y} \right) - \left( {x + y} \right)\\ = \left( {x + y} \right)\left( {5x - 1} \right)\end{array}\)
b) \({x^2} + 5x - 6\)
\(\begin{array}{l} = {x^2} - x + 6x - 6\\ = \left( {{x^2} - x} \right) + \left( {6x - 6} \right)\\ = x\left( {x - 1} \right) + 6\left( {x - 1} \right)\\ = \left( {x - 1} \right)\left( {x + 6} \right)\end{array}\)
4)
b) \(\dfrac{x}{{5x + 5}} - \dfrac{x}{{10x - 10}}\)
\(\begin{array}{l} = \dfrac{x}{{5\left( {x + 1} \right)}} - \dfrac{x}{{10\left( {x - 1} \right)}}\\ = \dfrac{{2x\left( {x - 1} \right) - x\left( {x + 1} \right)}}{{10\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \dfrac{{2{x^2} - 2x - {x^2} - x}}{{10\left( {x + 1} \right)\left( {x - 1} \right)}}\\ = \dfrac{{{x^2} - 3x}}{{10\left( {{x^2} - 1} \right)}}\end{array}\)
