-7x²+28xy-28y²
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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
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+x^2+2xy+y^2+x+y=42(x+y)^2+(x+y)=42(x+y)(x+y+1)=0则x+y=0或x+y=-1
x²+xy+y²-(x²-xy+y²)=2xy=14-28=-14,xy=-714=x²+xy+y²+xy-xy=x²+2xy+y²+7=(x+y)²
x^2+xy+y=14,y^2+xy+x=28两式相加得x²+2xy+y²+(x+y)=42(x+y)²+(x+y)-42=0(x+y+7)(x+y-6)=0所以x+y+7=0或x+y-6=0即x+y=-7或x
x²+xy+y²=14,x²-xy+y²=28相加2(x²+y²)=42x²+y²=21相减2xy=-14所以(x+y)²=x²+2xy+y
题目好像有问题,是y=√(x-7)+√(7-x)+28吧.根式有意义,x-7≥07-x≥0,要两不等式均有解,只有x-7=7-x=0x=7y=0+0+28±√xy=±√(28×7)=±√(2×2×7×7)=±2×7=±14
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
x^3y(-4y)^2+(-7xy)^2*(-xy)-5xy^3*(-3x)^2=16x^3y^3-49x^3y^3-45^3y^3=-78x^3y^3如果本题有什么不明白可以追问,
1、(-7x^y)(2x^y-3xy^3+xy)=-14x^(2y)+21x^(y+1)y^3-7x^(y+1)y;2、((x-y)^6)/((y-x)^3)/(x-y)=-(x-y)^3/(x-y)=-(x-y)^2;很高兴为您解答,sk
8(x平方-2y平方)-x(7x+y)+xy=8x平方-16y平方-7x平方-xy+xy=x平方-4y平方=(x-2y)(x+2y)手机提问的朋友在客户端右上角评价点【满意】即可.再问:太棒了!连我这么不懂数学的人,都看懂了!
6或-7x2+xy+y+y2+xy+x=14+28=42(x+y)*(x+y)+(x+y)-42=0设(x+y)=TT*T+T-42=0T1=6T2=-7(x+y)=6或-7
x0.5是什么意思呀?再问:是x的平方和y的平方再答:那么y0.5是什么意思?你干脆把题再打一遍,打清楚我给你解,好吗?x²+y²+xy+y=14;x²+y²+xy+x=28吗?再问:x²+
(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
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+y=0,x+13y=1,求x²+12xy
题目是不是x^2y-7xy+6yx^2y-7xy+6y=y(x^2-7x+6)=y(x-1)(x-6)
因为,x-y=3xy所以:-3x+6xy+3y/[(7x-5y)-9xy-(3x-y)]=[-3(x-y)+6xy]/[(7x-5y)-9xy-(3x-y)]=[-3*3xy+6xy]/[7x-5y-9xy-3x+y]=-3xy/[4x-4
-4x^3y^2+28x^2y-2xy=-2xy(2X^2y-14x+1)题目没错吗?
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?
题目应是x^2+xy-2y^2-x+7y-6=(x^2-xy+2x)+(2xy-2y^2+4y)-(3x-3y+6)=x(x-y+2)+2y(x-y+2)-3(x-y+2)=(x-y+2)(x+2y-3)