1.(ax+by)²+(bx-ay)²
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1.(ax+by)²+(bx-ay)²
2.(x²-x)²-4(x²-x-1)
3.4(x+y)²-9(x-y)²
4.(a+b)(a-b)+4(b-1)
5.x²+4xy+4y²-2x-4y-3
6.(x+1)(x+2)(x+3)(x+4)+1
7,3x²-11x+10
2.(x²-x)²-4(x²-x-1)
3.4(x+y)²-9(x-y)²
4.(a+b)(a-b)+4(b-1)
5.x²+4xy+4y²-2x-4y-3
6.(x+1)(x+2)(x+3)(x+4)+1
7,3x²-11x+10
打了好久,你一分都没有给,好吝啬哦,开个玩笑.
1、平方展开=a^2x^2+b^2y^2+b^2x^2+a^2y^2
=(x^2+y^2)(a^2+b^2)
2、原式=x^4-2x^3+x^2-4x^2+4x+4= x^4-2x^3-3x^2+4x+4= (x^4-2x^3)-(3x^2-4x-4)=x^3(x-2)-(x-2)( 3x+2)=(x-2)( x^3-3x-2)
3、原式=(2(x+y)-3(x-y))( 2(x+y)+3(x-y))=(5x-y)(y-x)
4、原式=(a+b)(a-b)+4(b-1)
=a^2-b^2+4b-4
=a^2-(b^2-4b+4)
=a^2-(b-2)^2
=(a+b-2)(a-b+2)
5、原式=x^2+4xy+4y^2-2x-4y-3
=(x+2y)^2-2(x+2y)-3
=(x+2y-3)(x+2y+1)
6、原式=(x+1)(x+4)*(x+2)(x+3)+1
=(x^2+5x+4)(x^2+5x+6)+1
=(x^2+5x+4)[(x^2+5x+4)+2]+1
=(x^+5x+4)^2+2(x^2+5x+4)+1
=[(x^2+5x+4)+1]^2
=(x^2+5x+5)^2
7、原式=(3x-5)(x-2)
1、平方展开=a^2x^2+b^2y^2+b^2x^2+a^2y^2
=(x^2+y^2)(a^2+b^2)
2、原式=x^4-2x^3+x^2-4x^2+4x+4= x^4-2x^3-3x^2+4x+4= (x^4-2x^3)-(3x^2-4x-4)=x^3(x-2)-(x-2)( 3x+2)=(x-2)( x^3-3x-2)
3、原式=(2(x+y)-3(x-y))( 2(x+y)+3(x-y))=(5x-y)(y-x)
4、原式=(a+b)(a-b)+4(b-1)
=a^2-b^2+4b-4
=a^2-(b^2-4b+4)
=a^2-(b-2)^2
=(a+b-2)(a-b+2)
5、原式=x^2+4xy+4y^2-2x-4y-3
=(x+2y)^2-2(x+2y)-3
=(x+2y-3)(x+2y+1)
6、原式=(x+1)(x+4)*(x+2)(x+3)+1
=(x^2+5x+4)(x^2+5x+6)+1
=(x^2+5x+4)[(x^2+5x+4)+2]+1
=(x^+5x+4)^2+2(x^2+5x+4)+1
=[(x^2+5x+4)+1]^2
=(x^2+5x+5)^2
7、原式=(3x-5)(x-2)
1.(ax+by)²+(bx-ay)²
如果x=1.y=2是方程(ax-bx-12)²+绝对值ay-by+1=0的一个解试求a.b
bX-bY=aX+aY
求两圆X²+Y²+2aX+2aY+2a²-1=0与X²+Y²+2bX+
证明:|ax+by ay+bz az+bx||ay+bz az+bx ax+by||az+bx ax+by ay+bz|
1.方程ax²+bx+C=0,a、b、c都是实数,且满足(2-a)²+|C+8|+√(a²
(ax+by)^2+(ay-bx)^2+2(ax+by)(ay-by)
已知方程组ax+by=5bx+ay=2
已知方程组ax+by=4bx+ay=5
证明|by+az bz+ax bx+ay| |x y z|
线性代数证明题证明行列式 ax+by ay+bz az+bx ay+bz az+bx ax+by az+bx ax+by
1.若x0是方程ax²+bx+c=0(a≠0)的根,△=b²-4ac,M=(2ax0+b)²