求大侠.①(4XY+8Y^2)-( )=X^2+2XY+8Y^2.②已知X-Y=2,则3-2X+2Y=?③当X=1时,P

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  • ①(4XY+8Y^2)-(2XY- X^2 )=X^2+2XY+8Y^2.

    ②已知X-Y=2,则3-2X+2Y= -1

    ③当X=1时,PX^3+QX+6的值为2012,则当X=-1时,PX^3+QX+6的值为( )?

    ④若(X^2-3XY+Y+X-5)^2+|3X+1|=0,则X^2-3XY+Y-2=( 3 ).

    ⑤已知a^2-ab=2,则3a^2-3ab+4=(10 )

    讲解 过程:

    ①省略

    ②已知X-Y=2,则3-2X+2Y=3- 2(X-Y) =3-4= -1

    ③ X=1时,PX^3+QX+6的值为2012 ②则 P+Q=2006.当X=-1时,PX^3+QX+6=-P-Q+6=-(PX+Q)+6=-2000

    ④若(X^2-3XY+Y+X-5)^2+|3X+1|=0,即 (X^2-3XY+Y+X-5)^2= -|3X+1| 因为|3X+1|不可能负数, (X^2-3XY+Y+X-5)^2也大于等于0,所以 X^2-3XY+Y+X-5 =0 ,则X^2-3XY+Y-5=X^2-3XY+Y-3-2=0所以X^2-3XY+Y-2=3

    ⑤已知a^2-ab=2,则两边都乘以3 则3a^2-3ab=6 即3a^2-3ab+4=6+4=10