cho ab(x²+y²)+xy(a²+b²)=ab.trong đó x,y khác 0,x+y=1.CMR:a=b Giúp mjnhf vs mính cần gấpp ạ 16/07/2021 Bởi Kinsley cho ab(x²+y²)+xy(a²+b²)=ab.trong đó x,y khác 0,x+y=1.CMR:a=b Giúp mjnhf vs mính cần gấpp ạ
`ab(x²+y²)+xy(a²+b²)=ab` `⇔ab((x+y)^2-2xy)+xy(a^2+b^2)=ab` `⇔ ab(1-2xy)+xy(a²+b²)=ab` `⇔ab-2axby+xy(a²+b²)=ab` `⇔xy(a^2-2ab+b^2)=0` `⇔xy(a-b)^2=0` `⇔a-b=0(x;y`khác `0)` `⇔a=b` Bình luận
$\begin{array}{l} ab\left( {{x^2} + {y^2}} \right) + xy\left( {{a^2} + {b^2}} \right) = ab\\ \Leftrightarrow ab\left[ {{{\left( {x + y} \right)}^2} – 2xy} \right] + xy\left( {{a^2} + {b^2}} \right) = ab\\ \Leftrightarrow ab{\left( {x + y} \right)^2} – 2xyab + xy\left( {{a^2} + {b^2}} \right) = ab\\ \Leftrightarrow ab + xy\left( {{a^2} + {b^2} – 2ab} \right) = ab\\ \Leftrightarrow xy{\left( {a – b} \right)^2} = 0\\ \Leftrightarrow a = b\left( {x,y \ne 0} \right) \end{array}$ Bình luận
`ab(x²+y²)+xy(a²+b²)=ab`
`⇔ab((x+y)^2-2xy)+xy(a^2+b^2)=ab`
`⇔ ab(1-2xy)+xy(a²+b²)=ab`
`⇔ab-2axby+xy(a²+b²)=ab`
`⇔xy(a^2-2ab+b^2)=0`
`⇔xy(a-b)^2=0`
`⇔a-b=0(x;y`khác `0)`
`⇔a=b`
$\begin{array}{l} ab\left( {{x^2} + {y^2}} \right) + xy\left( {{a^2} + {b^2}} \right) = ab\\ \Leftrightarrow ab\left[ {{{\left( {x + y} \right)}^2} – 2xy} \right] + xy\left( {{a^2} + {b^2}} \right) = ab\\ \Leftrightarrow ab{\left( {x + y} \right)^2} – 2xyab + xy\left( {{a^2} + {b^2}} \right) = ab\\ \Leftrightarrow ab + xy\left( {{a^2} + {b^2} – 2ab} \right) = ab\\ \Leftrightarrow xy{\left( {a – b} \right)^2} = 0\\ \Leftrightarrow a = b\left( {x,y \ne 0} \right) \end{array}$