[2x+4y-1=0,3x+2y-5=0]用代入法解方程,不要用加减消元法,

{x+y=1 ,xy=-6{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0 3x+4y-5=0 x^2+y^2=1求x.y 解下列方程组：{x(2x-3)=0,y=x²-1{(3x+4y-3)(3x+4y+3)=0,3x+2y=5{(x-y+2)(x+y)=0,x²+y²=8{(x+y)((x+y-1)=0,(x-y)(x-y-1)=0 x*x+2x+y*y-4y+5=0 x= y= {4x-3y-10=0 3x-2y=0,{3x-4(x-y)=2 2x-3y=1,{2(x+y)-(x-y)=3 (x+y)-2(x-y)=1 若(x*x+y*y)(x*x+y*y)-4x*x*y*y=0,求代数式(x*x+5xy+y*y)/(x*x+2xy+y*y)的值 若2x-3y+4=0则x（x*x-1)+x(5-x*x)-6y+7 已知(3x+1)(3x-3)-3x+2)^2=-7,(y+7)(y-7)-y(y-7)=0,求8y^2-5y(-y+3x)+4y(-4y-5/2x)的值 已知x*x+4x+y*y-2y+5=0,则x*x+y*y=? [(x+2y)²-（x+y)(3x-y)-5y²]÷(2x),其中x,y满足x²+4x+y²-y+17/4=0已知、x+y=6且xy=4(1)x²+y² (2)（x-y）² (3)x{四次方}+y{四次方}第一题没错 书上就这样写的！ 已知X,Y,满足(X+3Y)(X-3Y)=-10(Y^2-6/5)和2X(Y-1)+4(1/2X-1)=0 （X+Y)^2 X^2Y+XY^2注：*/* 是分数 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1 若x^5+x^4y+x^4+x+y+1=0,且3x+2y=1,求x,y的值 x^2-x+y^2+2y+5/4=0 求x^-1+y^0 |x-2y+1|+(x+y-5)²=0,则5x+4y= 若2x-3y=0,求5x+4y/3x-2y 已知x,y>0 2x+y+3=xy 求5x+4y最小值 常微分方程的通解dy/dx=(x-y+1)/(x+y-3)y^4=2y^n+y=0y''+6y'+9y=e^(-3x)y''+y'-2y=4e^(2x)