若2x=1-4x,3y=2t-2-搜答案-智慧作业帮

x=1-2t,2t=1-xy=(5-2t)/(7-4t)=[5-(1-x)]/[7-2(1-x)]=(4-x)/(5-2x)x=(5y-4)/(2y-1)

左边的式子乘以2,右面的式子乘以3,相加.最后y=（5-14x）/3

y用x代,得t=-((x-2)*2)/2+5t

x-1=(t+1)/(t-1)-1=2/(t-1)t-1=2/(x-1)t=(x+1)/(x-1)t^2+t+1=(x+1)^2/(x-1)^2+(x+1)/(x-1)+1=(3x^2+1)/(x-1

(2x-y)(2x+y)+(2x-y)(y-4x)+2y(y-3x)=4x^2-y^2+2xy-8x^2-y^2+4xy+2y^2-6xy=-4x^2=-4(-1/4)^2=-1/4

由x=1-t/2+t得t=2(x-1)将其代入y=5+4t/3-2t得y=5-4(x-1)/3

答案：y=2x+3x=2t,y=4t+3因为2x=4t4t+3=y

2x（x+y）=2x(2t)=4t所以2x+2y=x-yx=-3y把x=-3y代入3x+2y=7中y=-1y=-1代入3x+2y=7中x=33-(-1)=4tt=1

∵x+y=1，∴y=1-x＞0，∴t=2+x-14(1−x)=3−[(1−x)+14(1−x)]≤3−2(1−x)•14(1−x)=2，当且仅当x=12时取等号．故选A．

2分之x+y=3分之t+x=4分之y+t(x+y)/2=(x+t)/3=(y+t)/4设(x+y)/2=(x+t)/3=(y+t)/4=kx+y=2kx+t=3ky+t=4k三式相加得2(x+y+t)

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

这种方程组解法如下,设:(x+y)÷2=(t+x)÷3=(y+t)÷4=k,由此可知x+y=2k,t+x=3k,y+t=4k解得x=k/2,y=3k/2,t=5k/2,带入x+y+t=27,解得k=6

3x+4y=2t.[1]2x+3y=t-1.[2][1]*3:9x+12y=6t[2]*4:8x+12y=4t-4二式相减得：x=2t+4y=(2t-3x)/4=(2t-6t-12)/4=-t-3x

x（x*x-1)+x(5-x*x)-6y+7=-x+5x-6y+7=2(2x-3y)+7=2*(-4)+7=-8+7=-1再问：能在写详一点吗-x+5x-6y+7再答：x（x*x-1)+x(5-x*x

有所谓的拉格朗日乘数法可以解决这个多元函数求极值问题设L(x,y,t,u)=5(x-y)^2+4y^2+3t^2+u(2x+y+t-8)分别对x,y,t,u求偏导再令其等于0,得4个方程:L'x=10

①-②得：3y=-2t则y=-2t/3将y=-2t/3代入②中得：x=10t/3将x=10t/3,y=-2t/3代入3x+2y=7中得：10t-4t/3=726t/3=7t=21/26在我回答的右上角

x=3t+1t=(x-1)/3y=2t-1=(2x-5)/3

y=∫(t-1)^3(t-2)dt,dy/dx=(x-1)^3(x-2).

[x-(t+3)]²+[y+(1-4t²)]²=-16t^4-9+(t+3)²+(1-4t²)²是圆则r²=-16t^4-9+(t

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009）y=0+(1×1004-2009)y=-1005y=1005/2