证明2x＾2-4xy+4y＾2+2x+2的值总是不小于1.

证明:存在整数x,y满足x^2+4xy+y^2=2022 证明2x＾2-4xy+4y＾2+2x+2的值总是不小于1. 因式分解:12xy^2(y-x)+xy^3(x-y)+4x^2y(x-y) (x^2+2xy+y^2-4xy)*(x^2-2xy+y^2+4xy) (x+y)(x+y+2xy)+(xy+1)(xy-1)(x+y)(x-y)+4(y-1) 试证明(x^3+3x^2y-2xy^2+1)+(x^3-4x^2y+3xy^2-10)+(-xy^2+x^2y-2x^3+3)的值与x,y无关 证明：代数式[(x+y)(x-y)-(x+y)^2-2y(x-y)-2xy]/xy 的值为x,y的值无关 解方程 (x+1)^5/(-1-x)^4=3x+7一定要100%正确阿. (x^-4y^)/(x^+2xy+y^)/(x+2y)/(2x^+2xy) 因式分解 x-xy+3y-3x 2x+xy-y-4x+5y-6 化简：3x^2-[5xy-(x^y-3xy)+4x^2y-8xy] （-3x^y+2xy)-( )=4x^+xy 2x(x-y)+4xy(y-x) 因式分解2x(x-y)+4xy(y-x) 因式分解 计算:2xy/(x+y)².(x+y)/4y x,y为实数,证明x*2-xy+y*2>=x+y-1 已知x.y属于R,用向量法证明x*x+y*y>=2xy 计算:4xy-(x+y)^2/xy(x+y)(x-y)÷x^2+xy-2y^2/x^2y+2xy^2 一条不等式的证明题证明：x^2+y^>=xy+x+y-1 xy-2{xy-[3x^2y-(4xy^2+0.5xy)]+4x^2y}+6xy^2