设x-10xy+25y+2x-10y+1=0,求x-5y=0的值
来源:学生作业帮助网 编辑:作业帮 时间:2024/12/02 07:17:32
x){n_AE)ؕچ:66UVlH4Y{狕XH=\5Ov4<[[3y.+
:M
F!mC[F~8-/WOQxַhӆndWϳ5!+L7 <;l,R}8#mC@`fI:^
B"Y y
设x-10xy+25y+2x-10y+1=0,求x-5y=0的值
设x-10xy+25y+2x-10y+1=0,求x-5y=0的值
设x-10xy+25y+2x-10y+1=0,求x-5y=0的值
應該是求x-5y的值吧 第一方程組可化為:(x-5y)的平方 - 2(x-5y) + 1 =0 將x-5y看做為整體 接一個一元二次方程 可得x-5y的值
x-10xy+25y+2x-10y+1 =(x-5y)^2+2(x-5y)+1 =(x-5y+1)^2 =0 所以x-5y+1=0 所以x-5y=-1
设x-10xy+25y+2x-10y+1=0,求x-5y=0的值
设x,y满足10x^2-16xy+8y^2+6x-4y+1=0,则x-y=?
因式分解(x+y-2xy)(x+y-2)+(xy-1)^2设x+y=a,xy=b,
设x,y∈R,比较x^2+y^2+1与x+y+xy
[(xy+2)(xy-2)-2x^2y^2+4]/xy,其中x=10,y=-1/25
[(xy+2)(xy-2)-2x^2y^2+4]/xy,其中x=10,y=-1/25
X*X-10XY+25Y*Y+X-5Y-2 分解因式
x的2次方y+10xy+25y
化简:xy+5x/y^2+10y+25
因式分解x^2y+10xy+25y
约分:(1)-4x^2y^3/10xy^4(2)(x^2-4xy+4y^2)(y-x)/(2y-x)(x^2-y^2)
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=?
因式分解30题(要过程)am+an+bm+bn x^2+5y-xy-5x x2+xy+2xz+x (a+b)x-(b+a)y -12x(x-y)-(y-x)x^2 2x^2+4x+2 x(x+z)-y(y+z) 2(x^2-3ab)+x(4a-3b) 4x^2+12xy+9y^2-25 x^4-16 16x^4-1 x^2+9x+8 x^2-10xy+24y^2 2x^2+15xy-8y^2 3x^2-6xy+3y^2 24x^2+22x-21 -x^2+4xy-
已知:x>0,y>0,且x- 10√xy+25y=0,求2x+√xy+3y/x+√xy-y
实数x,y满足|2分之5x-y|+y²-10y+25=0,求代数式1-x-2y分之x-y÷x平方-y平方分之x²-4xy+4y
若x+y=10 xy=-1 则x²+y²=(x+y)²-2xy=
已知x+y=3,xy=-10,求(1)x平方-xy+y平方;(2)|x-y|
10x(5x²-y)-2x(5y+25x²)-3xy 求化简