-24x²y-12xy²-28y²
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解题思路:提取公因式进行分解解题过程:附件最终答案:略
求数学达人来解答x²/xy-x/y=(1-x)/y3x/4x+y-x-2y/4x+y=(3x-x-2y)/(4x+y)(1-y/y+x)÷x/y²-x²=(1-y)*(y
原式=(3x²y+12xy)+(7xy²+28y²)=3xy(x+4)+7y²(x+4)=y(x+4)(3x+7y)
x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7
解题思路:本题主要利用完全平方公式进行因式分解即可求出结果解题过程:解:x²-6xy+9y²=(x-3y)2
对.前提是x不等于y
xy-12=4x+y≥2√(4xy)=4√(xy)xy-4√(xy)-12≥0(√(xy)-6)(√(xy)+2)≥0√(xy)≤-2,√(xy)≥6因为√(xy)≥0所以√(xy)≥6xy≥36所以
-12xy*3x^2y-x^2y*(-3xy)=-36x^3y^2+6x^3y^2=-30x^3y^2
xy-1+x-y=XY+X-Y-1(加法交换律)=(XY+X)-(Y+1)=X(Y+1)-(Y+1)(提取公因式X)=(Y+1)(X-1)(提取公因式Y+1)这样可以么?
这题少括号了吧.
16x³-24x²y²+12xy^4-2y^6=(16x^3-16x^2y^2+4xy^4)-8x^2y^2+8xy^4-2y^6=4x(4x^2-4xy^2+y^4)-
解题思路:提取公因式进行分解.解题过程:解:
(4xy+12y)+[7x-(3xy+4y-x)]=4xy+12y+7x-3xy-4y+x=xy+8x+8y=xy+8(x+y)=(-2)+8*3=-2+24=22
原式=(xy)2+2xy+3-2(x+y)-1-(x+y)2+2(x+y)-1,=(xy)2+2xy+1-(x+y)2,=(xy+1)2-(x+y)2,=(xy+1+x+y)(xy+1-x-y),=(
-12XY×3X平方Y-X平方Y×(-3XY)=-12×3X^3Y^2-(-3X^3Y^2)=-36X^3Y^2+3X^3Y^2=-33X^3Y^2
解题思路::∵x+y=0,x+13y=1,解得x=1/12,y=-1/12∴x²+12xy+13y²=1/144-1/12+13/144=14/144-1/12=2/144=1/72解题过程:已知x+
两式相加得到x+y=5,相减得y-x=1/5,故x=12/5,y=13/5xy=156/25,因为要求的都是正数,而且xy同正负,所以只考虑x,y正数即可故x²+y²=(x+y)^
-24x^2y-12xy^2+28y^3解,得:==-4y(6x^2+3xy-7y^2)分解因式(x-y)²-(y-x)也是因式分解,解,得:==(x-y)^2+(x-y)==(x-y)[(
-12xy*3x^2y-x^2y*(-3xy)=-36x^3y^2+3x^3y^2=33x^3y^2