(1 3x y)(1 3x-y)(1 9x² y²)运用乘法公式计算
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(x+1)(y+1)
把13X^2-6XY+Y^2-4x+1=0配方,得13X^2-6XY+Y^2-4x+1=09x^2-6xy+y^2+4x^2-4x+1=0(3x-y)^2+(2x-1)^2=0又因为(2x-1)^2>
13x^2-6xy+y^2-4x+1=09x²-6xy+y²+4x²-4x+1=0(3x-y)²+(2x-1)²=0∵(3x-y)²≥0&n
第一题设x+y=axy=b则x^2+y^2=a^2-2b原方程变为a^2-2b+b=13①b+a=7②这就变成了简单的二元二次方程由②得b=7-a带入①得a^2+a-20=0(a-4)(a+5)=0得
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
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?
13x^2-6xy+y^2-4x+1=0(9x^2-6xy+y^2)+4x^2-4x+1=0(3x-y)^2+(2x-1)^2=0加数均非负,和为0.则均为0所以有:3x=y2x-1=0x=1/2y=
1-x-y+xy=(1-x)-(y-xy)=(1-x)-y(1-x)=(1-x)(1-y)
13x^2-6xy+y^2-4x+1=9X^-6XY+Y^+4X^-4X+1=(3X-Y)^+(2X-1)^=0则(3X-Y)=0且(2X-1)=0(两个非负数的和是0则这两个数同时为0)所以x=1/
xy-x-y+1=x(y-1)-y+1=x(y-1)-(y-1)=(y-1)(x-1)
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)^
x²+y²-4x-6y+13=0x²-4x+4+y²-6y+9=0(x-2)²+(y-3)²=0x=2y=3(-y/x^3)^3÷(-1/x
2和3再问:要完整过程!!!~~再答:=。=∵x平方-2xy+y平方=1∴(x-y)平方=1∴x-y=1或x-y=-1。再代入第一个方程里面。算呗
令6x^2+13xy+6y^2+5x+5y+1=(ax+by+c)(dx+ey+f)6x^2+13xy+6y^2+5x+5y+1=(ax+by+c)(dx+ey+f)=adx^2+(ae+bd)xy+
(x-1)(y-1)
13x^2-6xy+y^2-4x+1=09x^2-6xy+y^2+4x^2-4x+1=0(3x-y)^2+(2x-1)^2=03x-y=0,2x-1=0,x=0.5y=3*0.5=1.5(xy-x^2
13x²-6xy+y²-4x+1=0即(9x^2-6xy+y^2)+(4x^2-4x+1)=0∴(3x-y)^2+(2x-1)^2=0x,y在实数范围内(3x-y)^2(2x-1)
题目想必是漏了平方吧3X²+12XY+13Y²=3(x+3y)(x+y)+4y²=4y²x=-y则有:-y+3y=1y=1/24y²=4*1/4=13
解:原式=xy+y-x-1=y(x+1)-(x+1)=(x+1)(y-1)