(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/25 06:05:10
![(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2](/uploads/image/z/580468-4-8.jpg?t=%2810y%2Bx%29%5E2-%2810x%2By%29%3D138+%E7%94%A8%E4%B8%80%E5%85%832%E6%AC%A1%E6%96%B9%E7%A8%8B%E8%A7%A3y%3Dx%2B2)
x)04ԮЌ3*+5m
-OYdGfgk>WVh$铡KΆ\'?m]݀;=|]=+⌴M,l
F=7VHLV@BV$ف\ lq
(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2
(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2
(10y+x)^2-(10x+y)=138 用一元2次方程解y=x+2
y=x+2代入(10y+x)^2-(10x+y)=138 并化简:121x^2+429x+260=0解得x1=-2.7696,x2=-0.7758,y1=-0.7696,y2=1.2304
x,y 满足x(x-2)-y(y-6)=-10,求x,y的值
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuaihuajian
3x+x+4y+(90+10)x=3y-(2x+y)+510y=34x
{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
已知x²+4x+y²+6y+13=0,求x-y+(2y²/x+y) 请在10;20分前回答
若x-y不等于0,2x-3y=0,则分式(10x-11y)/(x-y)的值是()?(10x-11y)/(x-y) =[(8x-8y)+(2x-3y)]/(x-y) =8为什么8X-8Y=8?(10x-11y)/(x-y)=[(8x-8y)+(2x-3y)]/(x-y)=8+(2x-3y)]/(x-y)=8+0=8
x+(9-8-y)=10 y+y-(9-8-y)=2 x-y=xx-9.9x xy=11y-1 y=? x=? x+y-xy=?x+(9-8-y)=10y+y-(9-8-y)=2x-y=xx-9.9xxy=11y-1 y=?x=?x+y-xy=?
已知2x-y=10,求式子[(x^2+y^2)-(x-y)^2+2y(x-y)]/4y (4xy-2y²)/4y是怎么变成=2y(2x-y)÷4y
x^2+4y^2-6x+4y+10=0,求x^4-y^4/(x+2y)(x-2y)*x+2y/xy^2+y^3除以(x^2+y^2/y)^2的值
(x+y)^2/y^2-xy÷[-y^2+xy/(x-y)^2,x=0,y=根号10
x+2y=5 2x+y=10
已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y 就是已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y /4y就是除以4y
x+5y=4,2x+10y=?
2x+4y=10 X*y=3
2x-y=0与x+y=10
25(x+y)^2+10(y-x)+1=?因式分解
25(x-y)^2+10(y-x)+1=