Quy đồng mẫu các phân thức: a) x/2x^2+7x-15;x+2/x^2+3x-10; 1/x+5 b)1/-x^2+3x-2;1/x^2+5x-6; 1/-x^2+4x-3 03/09/2021 Bởi Allison Quy đồng mẫu các phân thức: a) x/2x^2+7x-15;x+2/x^2+3x-10; 1/x+5 b)1/-x^2+3x-2;1/x^2+5x-6; 1/-x^2+4x-3
\(\begin{array}{l}a)\,\dfrac{1}{{2{x^2} + 7x – 15}} = \dfrac{1}{{2{x^2} + 10x – 3x – 15}} = \dfrac{1}{{2x\left( {x + 5} \right) – 3\left( {x + 5} \right)}} = \dfrac{1}{{\left( {2x – 3} \right)\left( {x + 5} \right)}}\\\dfrac{{x + 2}}{{{x^2} + 3x – 10}} = \dfrac{{x + 2}}{{{x^2} + 5x – 2x – 10}} = \dfrac{{x + 2}}{{x\left( {x + 5} \right) – 2\left( {x + 5} \right)}} = \dfrac{{x + 2}}{{\left( {x – 2} \right)\left( {x + 5} \right)}}\\MTC:\,\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)\\ \Rightarrow \dfrac{1}{{2{x^2} + 7x – 15}} = \dfrac{{x – 2}}{{\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)}}\\\dfrac{{x + 2}}{{{x^2} + 3x – 10}} = \dfrac{{\left( {x + 2} \right)\left( {2x – 3} \right)}}{{\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)}}\\\dfrac{1}{{x + 5}} = \dfrac{{\left( {x – 2} \right)\left( {2x – 3} \right)}}{{\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)}}\\b)\,\dfrac{1}{{ – {x^2} + 3x – 2}} = \dfrac{{ – 1}}{{{x^2} – 2x – x + 2}} = \dfrac{{ – 1}}{{x\left( {x – 2} \right) – \left( {x – 2} \right)}} = \dfrac{{ – 1}}{{\left( {x – 2} \right)\left( {x – 1} \right)}}\\\dfrac{1}{{{x^2} + 5x – 6}} = \dfrac{1}{{{x^2} + 6x – x – 6}} = \dfrac{1}{{x\left( {x + 6} \right) – \left( {x + 6} \right)}} = \dfrac{1}{{\left( {x – 1} \right)\left( {x + 6} \right)}}\\\dfrac{1}{{ – {x^2} + 4x – 3}} = \dfrac{{ – 1}}{{{x^2} – 3x – x + 3}} = \dfrac{{ – 1}}{{\left( {x – 1} \right)\left( {x – 3} \right)}}\\MTC:\,\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)\\ \Rightarrow \dfrac{1}{{ – {x^2} + 3x – 2}} = \dfrac{{ – \left( {x – 3} \right)\left( {x + 6} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)}}\\\dfrac{1}{{{x^2} + 5x – 6}} = \dfrac{{\left( {x – 2} \right)\left( {x – 3} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)}}\\\dfrac{1}{{ – {x^2} + 4x – 3}} = \dfrac{{ – \left( {x – 2} \right)\left( {x + 6} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)}}\end{array}\) Bình luận
\(\begin{array}{l}
a)\,\dfrac{1}{{2{x^2} + 7x – 15}} = \dfrac{1}{{2{x^2} + 10x – 3x – 15}} = \dfrac{1}{{2x\left( {x + 5} \right) – 3\left( {x + 5} \right)}} = \dfrac{1}{{\left( {2x – 3} \right)\left( {x + 5} \right)}}\\
\dfrac{{x + 2}}{{{x^2} + 3x – 10}} = \dfrac{{x + 2}}{{{x^2} + 5x – 2x – 10}} = \dfrac{{x + 2}}{{x\left( {x + 5} \right) – 2\left( {x + 5} \right)}} = \dfrac{{x + 2}}{{\left( {x – 2} \right)\left( {x + 5} \right)}}\\
MTC:\,\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)\\
\Rightarrow \dfrac{1}{{2{x^2} + 7x – 15}} = \dfrac{{x – 2}}{{\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)}}\\
\dfrac{{x + 2}}{{{x^2} + 3x – 10}} = \dfrac{{\left( {x + 2} \right)\left( {2x – 3} \right)}}{{\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)}}\\
\dfrac{1}{{x + 5}} = \dfrac{{\left( {x – 2} \right)\left( {2x – 3} \right)}}{{\left( {x + 5} \right)\left( {x – 2} \right)\left( {2x – 3} \right)}}\\
b)\,\dfrac{1}{{ – {x^2} + 3x – 2}} = \dfrac{{ – 1}}{{{x^2} – 2x – x + 2}} = \dfrac{{ – 1}}{{x\left( {x – 2} \right) – \left( {x – 2} \right)}} = \dfrac{{ – 1}}{{\left( {x – 2} \right)\left( {x – 1} \right)}}\\
\dfrac{1}{{{x^2} + 5x – 6}} = \dfrac{1}{{{x^2} + 6x – x – 6}} = \dfrac{1}{{x\left( {x + 6} \right) – \left( {x + 6} \right)}} = \dfrac{1}{{\left( {x – 1} \right)\left( {x + 6} \right)}}\\
\dfrac{1}{{ – {x^2} + 4x – 3}} = \dfrac{{ – 1}}{{{x^2} – 3x – x + 3}} = \dfrac{{ – 1}}{{\left( {x – 1} \right)\left( {x – 3} \right)}}\\
MTC:\,\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)\\
\Rightarrow \dfrac{1}{{ – {x^2} + 3x – 2}} = \dfrac{{ – \left( {x – 3} \right)\left( {x + 6} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)}}\\
\dfrac{1}{{{x^2} + 5x – 6}} = \dfrac{{\left( {x – 2} \right)\left( {x – 3} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)}}\\
\dfrac{1}{{ – {x^2} + 4x – 3}} = \dfrac{{ – \left( {x – 2} \right)\left( {x + 6} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)\left( {x – 3} \right)\left( {x + 6} \right)}}
\end{array}\)