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第159頁

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解：原式?$=x^2-\frac {1}{4}y^2-(x^2-xy+\frac {1}{4}y^2) $?

? $= x^2-\frac {1}{4}y^2-x^2+xy-\frac {1}{4}y^2$?

? $=xy-\frac {1}{2}y^2，$?

?$ $?當(dāng)?$x = - 2，$??$y=\frac {1}{2}$?時(shí)，

?$ $?原式?$=(-2)×\frac {1}{2}-\frac {1}{2}×(\frac {1}{2})^2=-1-\frac {1}{8}=-\frac {9}{8}。$?

解：原式?$=x^2+4x + 4 - x^2+1-2x^2-3x + 2$?

? $=-2x^2+x + 7$?

?$ $?因?yàn)?$2x^2-x - 2 = 0，$?

所以?$-2x^2+x=-2，$?

?$ $?則原式?$=-2 + 7 = 5。$?

解：原式?$=-3x^2+xy+\frac {2}{3}xy^2-\frac {2}{9}y^3-\frac {1}{3}xy^2+\frac {2}{9}y^3+3x^2$?

? $ =xy+\frac {1}{3}xy^2$?

?$ $?因?yàn)?$\vert xy - 2\vert +(y + 2)^2=0，$?

?$ $?所以?$xy - 2 = 0$?且?$y + 2 = 0，$?

?$ $?解得?$xy = 2，$??$y=-2，$?則?$xy^2=-4，$?

?$ $?原式?$=2-\frac {4}{3}=\frac {2}{3}$?

解：?$ (1)(ax - 3)(2x + 4)-x^2-b$?

?$=2ax^2+4ax-6x-12-x^2-b$?

?$=(2a - 1)x^2+(4a - 6)x+(-12 - b)。$?

?$ $?因?yàn)榇鷶?shù)式?$(ax - 3)(2x + 4)-x^2-b$?化簡后，

不含有?$x^2$?項(xiàng)和常數(shù)項(xiàng)，

?$ $?所以?$2a - 1 = 0，$??$-12 - b = 0，$?

?$ $?所以?$a=\frac {1}{2}，$??$b=-12。$?

?$ (2)$?因?yàn)?$a=\frac {1}{2}，$??$b=-12，$?

?$ $?所以?$(b - a)(-a - b)+(-a - b)^2-a(2a + b)$?

?$ =a^2-b^2+a^2+2ab + b^2-2a^2-ab$?

?$ =ab$?

?$ =\frac {1}{2}×(-12)$?

?$=-6$?

$-22$

解：?$(3a + 1，a - 2)\otimes (a + 2，a - 3)$?

?$=(3a + 1)(a - 3)-(a - 2)(a + 2)$?

?$ =3a^2-9a+a-3-(a^2-4)$?

?$ =3a^2-9a+a-3-a^2+4$?

?$ =2a^2-8a + 1。$?

?$ $?因?yàn)?$a^2-4a + 1 = 0，$?

所以?$a^2=4a - 1。$?

?$ $?則?$(3a + 1，a - 2)\otimes (a + 2，a - 3)$?

?$=2(4a - 1)-8a + 1$?

?$ =8a-2-8a + 1$?

?$=-1$?

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