已知(x+y)²=28,(x-y)²=8,求x²+xy+y²
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x(x+y)(x-y)-x(x+y)²=x(x+y)[(x-y)-(x+y)]=x(x+y)(x-y-x-y)=-2xy(x+y)=-2×(-1/2)×1=1
y=3x/5原式=x/(x+3x/5)+(3x/5)/[x-3x/5]-(9x^3/25)/(x^3-9x^3/25)=8/3-3/2-9/16=29/48
#includeusingnamespacestd;intmain(){doublex,y;cout再问:谢谢大虾~有点小问题想问下:#includeusingnamespacestd;voidmai
你的题是什么意思?“^/”是什么意思?
|x|+x+y=10|y|+x-y=12两式相加得|x|+|y|+2x=22.(1)两式相减得|x|-|y|+2y=-2.(2)所以-|y|+2y<0若y>0,则显然-y+2y<0,即y<0,矛盾若y
X+Y分之X-Y等于3x=-2yX+Y分之2(x-y)减X+Y分之3X+Y=(-x-3y)/(x+y)=1
因为(x-y)/(x+y)=3,则(x+y)/(x-y)=1/3则5(x-y)(x+y)-(x+y)/2(x-y)=5*3-1/(3*2)=15-1/6=89/6
∵x-y=3,∴(x+1)2-2x+y(y-2x)=x2+2x+1-2x+y2-2xy=x2+y2-2xy+1=(x-y)2+1=(3)2+1=3+1=4.
这就是一直线,再空间中把直线也叫曲线,因为再未知的情况下都叫曲线,即使结果是直线,就象我们在写东西的时候,不知道他是男的还是女的,就写成"他"一样
X^2表示平方X^2+4X+4+Y^2-2Y+1=0(X+2)^2+(Y-1)^2=0因为平方大于=0所以X+2=0Y-1=0X=-2Y=1X^2+Y^2=5
1.1/x+1/y=(x+y)/xy=-1/(x+y)去分母可得,(x+y)^2+xy=0即x^2+y^2=-3xy所以y/x+x/y=(x^2+y^2)/xy=-32.由1/x-1/y=5可得y-x
4x=5y,x/y=5/4(x+y)/y=x/y+1=5/4+1=9/4
x²+y²=19(x+y)²=x²+y²+2xy=25xy=3(x-y)*(x-y)=(x-y)²=x²+y²-2xy=
∵y=9x^2-6x+6=9[x-(1/3)]^2+5∴当x=1/3时,y有最小值且ymin=5.∵y=-4x^2+28x+1=-4[x-(7/2)]^2+50∴当x=7/2时,y有最大值且ymax=
4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/
y=|x2−1|x−1=|x−1||x+1|x−1=−|x+1| x<1x+1 &n
原式分母为:x2+2x-c=x2+2x+1-c-1=(x+1)2-c-1,∵(x+1)2≥0,无论x取任何实数,这个式子都有意义,∴-c-1>0,解得:c<-1.故答案为:c<-1
原式化简=x+2y=2011
绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=03x-y=0(1)x+y=0(2)(1)+(2)4x=0x=0,代入(2)y=-x=0(x-y)/(xy)无意义,因此题目错了.