a)x mũ 2 + y mũ 2 phần x-y +2xy phần y-x b)5x-7 phần 2(x-1) – 4x phần x mũ 2 – 1 + 9 – 3x phần 2(x-1) 26/11/2021 Bởi Daisy a)x mũ 2 + y mũ 2 phần x-y +2xy phần y-x b)5x-7 phần 2(x-1) – 4x phần x mũ 2 – 1 + 9 – 3x phần 2(x-1)
Đáp án: b) \(\dfrac{{x – 1}}{{x + 1}}\) Giải thích các bước giải: \(\begin{array}{l}a)\dfrac{{{x^2} + {y^2}}}{{x – y}} + \dfrac{{2xy}}{{y – x}}\\ = \dfrac{{{x^2} + {y^2} – 2xy}}{{x – y}}\\ = \dfrac{{{{\left( {x – y} \right)}^2}}}{{x – y}} = x – y\\b)\dfrac{{5x – 7}}{{2\left( {x – 1} \right)}} – \dfrac{{4x}}{{{x^2} – 1}} + \dfrac{{9 – 3x}}{{2\left( {x – 1} \right)}}\\ = \dfrac{{5x – 7 + 9 – 3x}}{{2\left( {x – 1} \right)}} – \dfrac{{4x}}{{{x^2} – 1}}\\ = \dfrac{{2x + 2}}{{2\left( {x – 1} \right)}} – \dfrac{{4x}}{{{x^2} – 1}}\\ = \dfrac{{x + 1}}{{x – 1}} – \dfrac{{4x}}{{{x^2} – 1}}\\ = \dfrac{{{x^2} + 2x + 1 – 4x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \dfrac{{{x^2} – 2x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\ = \dfrac{{{{\left( {x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \dfrac{{x – 1}}{{x + 1}}\end{array}\) Bình luận
Đáp án:
b) \(\dfrac{{x – 1}}{{x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\dfrac{{{x^2} + {y^2}}}{{x – y}} + \dfrac{{2xy}}{{y – x}}\\
= \dfrac{{{x^2} + {y^2} – 2xy}}{{x – y}}\\
= \dfrac{{{{\left( {x – y} \right)}^2}}}{{x – y}} = x – y\\
b)\dfrac{{5x – 7}}{{2\left( {x – 1} \right)}} – \dfrac{{4x}}{{{x^2} – 1}} + \dfrac{{9 – 3x}}{{2\left( {x – 1} \right)}}\\
= \dfrac{{5x – 7 + 9 – 3x}}{{2\left( {x – 1} \right)}} – \dfrac{{4x}}{{{x^2} – 1}}\\
= \dfrac{{2x + 2}}{{2\left( {x – 1} \right)}} – \dfrac{{4x}}{{{x^2} – 1}}\\
= \dfrac{{x + 1}}{{x – 1}} – \dfrac{{4x}}{{{x^2} – 1}}\\
= \dfrac{{{x^2} + 2x + 1 – 4x}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{x^2} – 2x + 1}}{{\left( {x – 1} \right)\left( {x + 1} \right)}}\\
= \dfrac{{{{\left( {x – 1} \right)}^2}}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \dfrac{{x – 1}}{{x + 1}}
\end{array}\)