Y等于-2x, y=2x+1,y=2x-1的图像-搜答案-智慧作业帮

由意义,画出线性规划图,可以看出满足条件的解x>0,y>0.则z=（x+2y)/(2x+y)=[(x+y/2)+3y/2]/(2x+y)=1/2+3y/(4x+2y)=1/2+3/(4x/y+2).其

根据3组不等式约束,画出x,y的取值区域z=x+2y看成一条直线,平移直线过x,y的取值区域即可得最大最小值

原式=1/2x-[1/(x+y)][(x+y)/2x-(x+y)]=1/2x-[1/(x+y)](x+y)(1/2x-1)=1/2x-(1/2x-1)=1/2x-1/2x+1=1

Dx^y+x^-y=2根号2===>(x^y+x^-y)^2=8===>x^2y+x^-2y+2=8===>x^2y+x^-2y=6(x^y-x^-y)^2=x^2y+x^-2y-2=6-2=4==>

2X^2-xy-5x+y+4=o可化为：（x^2-4x+4）+(X^2-XY-X+Y)=0(X-2)^2+(x-Y)(x-1)=0因为x>=YY>=1所以X-Y>=0x-1>=0又因为(X-2)^2>

x-y/x+y=1/2设：x=3y=1代入可得：（y-x/x+y）-（2x+2y／x-y）=-4(1/2)

1)x(x-y)(x+y)-x(x+y)^2=x((x-y)(x+y)-(x+y)^2)=x(x^2-y^2-x^2-2xy-y^2)=x(-2xy-2y^2)=-2xy(x+y)2)（2a+b)(2

[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y=[x²-y²-x²-2xy-y²-xy+xy²]/(y/2)=[(x-2)y

对不起说得不太清楚是（y+2）的平方

分解因式,原式=xy-x/y-y/x+1/(xy)=1-x²-y²+1=2-x²-y²选D

(1)(x^2/x)-y-x-y

（1)x^2/x)-y-x-y=x-y-x-y=-2y(2)(a/a-b)-(a/a+b)-(2b^2/a^2-b^2)=a(a+b-a+b)/(a^2-b^2)-(2b^2/a^2-b^2)=2b/

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

-5|x-2y|≥08|y-1\2|≥0而-5|x-2y|+8|y-1\2|=0所以-5|x-2y|=08|y-1\2|=0x-2y=0y-1/2=0解得x=1y=1/2

方法1：∵x+y+2（-x-y+1）=3（1-y-x）-4（y+x-1）∴x+y-2x-2y+2=3-3y-3x-4y-4x+4∴-x-y+2=7-7y-7x∴6x+6y=5∴x+y=56方法2：∵x

若(x-y)/(x+y)=3那么x-y=3(x+y)x-y=3x+3y故x=-2y所以(x+2y)/(6x-7y)=(-2y+2y)/(6*(-2y)-7y)=0如果不懂,请Hi我,祝学习愉快!

x:2=y:4,则x:y等于

x:2=y:4,根据比例关系4x=2y所以x：y=2:4=1:2

∵f(x,y)=ln[x(1+2/y)]=lnx+ln(1+2/y)∴αf(x,y)/αy=(-2/y^2)/(1+2/y)=-2/[y(y+2)]即αf(1,1)/αy=-2/[1*(1+2)]=-

把原方程整理得：2x^4－4x^2y＋y^4＋1＝0则这个关于x^2的一元二次方程有实数解,故得：(－4y)^2－8(y^4＋1)≥0即有：(y^2－1)^2≤0当且仅当y^2－1＝0时,上述方程有实

x=8/3y=15/4就是这个了...