Question

Đã trả lời bởi chuyên gia

Answer 1

Lời giải

Đáp án đúng là: D

\(\frac{{x + y}}{{{x^3} + {x^2}y + x{y^2} + {y^3}}}:\frac{{{x^2} + xy - 2{y^2}}}{{{x^4} - {y^4}}}\)

\( = \frac{{x + y}}{{{x^2}\left( {x + y} \right) + {y^2}\left( {x + y} \right)}}:\frac{{{x^2} + 2xy - xy - 2{y^2}}}{{\left( {{x^2} - {y^2}} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{{x + y}}{{\left( {{x^2} + {y^2}} \right)\left( {x + y} \right)}}:\frac{{x\left( {x + 2y} \right) - y\left( {x + 2y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{1}{{\left( {{x^2} + {y^2}} \right)}}:\frac{{\left( {x - y} \right)\left( {x + 2y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{1}{{\left( {{x^2} + {y^2}} \right)}}:\frac{{x + 2y}}{{\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}\)

\( = \frac{1}{{\left( {{x^2} + {y^2}} \right)}} \cdot \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2}} \right)}}{{x + 2y}} = \frac{{x + y}}{{x + 2y}}\).

Ta có \(\frac{{x + y}}{{{x^3} + {x^2}y + x{y^2} + {y^3}}}:\frac{{{x^2} + xy - 2{y^2}}}{{{x^4} - {y^4}}} = 2\)

\(\frac{{x + y}}{{x + 2y}} = 2\)

\[x + y = 2x + 4y\]

\[x = - 3y\]

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Answer 2