(2x-y+1)(2x+y-1)

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/09 09:53:33

(2x-y+1)(2x+y-1)
(2x-y+1)(2x+y-1)

(2x-y+1)(2x+y-1)
原式=[2x-(y-1)]*[2x+(y-1)] 用平方差公式
=4x²-(y-1)²
=4x²-(y²-2y+1)
=4x²-y²+2y-1

这是方程吗

希望有帮助!呵呵!

(2x-y+1)(2x+y-1)
=[2x-(y-1)][2x+(y-1)] ------ (将题两括号内符号相反的项括起来,一个添 "+"号,一个添 "-" 号)
=(2x)²-(y-1)²
=4x²-(y²-2y+1)
=4x²-y²+2y-1

计算:1/(2x) - 1/(x+y)((x+y)/(2x) - x - y) 化简1/2x-1/x+y(x+y/2x-x-y) 计算2x/1-x+y/1·(2x/x+y-x-y) 1/2x-(x+y/2x -x-y)÷(x+y) 分式的加减:[(x/y-y/x)除以(x+y)+x(1/y-1/x)]除以1+x/y,其中x=5,y=2 对于(y一撇)y`=y/x(y-1)求导,我求的结果是y``=-y^2-y`x+y/x^2(y-1)^2 怎么不对啊,我的求导过程是:y`x(y-1) - [x(y-1)]` y y`xy-y`x-(y-1+xy`)y -y^2-y`x+yy``=-------------------------=----------------------=-----------------x^2 (y-1 (x/x+y +2y/x+y)乘以xy/ x+2y除以(1/x+1/y)(2)(3x^2/4y)^2乘以2y/3x+ x^2/2y^2除以2y^2/x y=2x*(*代表X)y=(lnx)*y=(x/1+X)* 先化简,再求值 (1)[(x-y)的平方+(x+y)(x-y)]÷2x 其中X=2010,y=2009 (2)y(x+y)+(x+y)(x-y)-x的平方,先化简,再求值(1)[(x-y)的平方+(x+y)(x-y)]÷2x 其中X=2010,y=2009(2)y(x+y)+(x+y)(x-y)-x的平方,其中x=-2,y=二分 (2X-Y)²-4(X-Y)*(X+Y)-(X-Y)*(X-1) 若|x+y-1|+(x-y-2)²=0,求代数式(x+2y)(x-2y)-(2x-y)(-y-2x)的值. 若x+y=4,则多项式1/2(x+y)+4(x-y)-3(x-y)-3/2(x+y)-x+y值是() 已知2x-y=0,求分式【1+(3*y*y*y/x*x*x-y*y*y)】/【1+(2y/x-y)】的值注:x*x*x-y*y*y=(x-y)*(x*x+xy+y*y) 2x(2x-y)+y(y-2x) y(2y+1)-y(y-1)+1 x(x+8)-8(x+2) 分2x(2x-y)+y(y-2x)y(2y+1)-y(y-1)+1x(x+8)-8(x+2)分解因式 [(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y (x²-4y²)÷(2x+y)/xy×1/x(2y-x)(2x+y)/xy 1/x(2y-x) 都是一项 把下列各式因式分解 (1)、a^2(x-y)-4a(y-x)+4(x-y) (2)、(x+y)(x-y)-(x+z)(x-z) 计算:1.(x-1)^2+2(1-x) 2.(x-y)/(x+3y)÷(x^2-y^2)/(x^2+6xy+9y^2)-2y/(x+y)1.(x-1)^2+2(1-x)2.(x-y)/(x+3y)÷(x^2-y^2)/(x^2+6xy+9y^2)-2y/(x+y)