已知根号2x 3y-7与(3x 2y-3)²互为相反数
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因为x=√3+1所以x﹥0,x-1﹥0原式=√[x²/(1-2x+x²)]=√[x²/(x-1)²]=x/(x-1)=(√3+1)/(√3+1-1)=(√3+1
x=(4-√7)/3x²=(16+7-8√7)/9=(32-8√7)/9-9/9所以x²=(8/3)x-1即x²+1=(8/3)x所以x+1/x=8/3(x+1/x)
[x(x2y2-xy)-y(x2+x3y)]÷3x2y,=(x3y2-x2y-x2y-x3y2)÷3x2y,=-2x2y÷3x2y,=-23.
(1)原式=xy(x2-y2)=xy(x+y)(x-y);(2)原式=(x2+1-2x)(x2+1+2x)=(x-1)2(x+1)2;(3)原式=x−yx÷(x−y)2x=x−yx×x(x−y)2=1
x3y+2x2y2+xy3=xy(x2+2xy+y2)=xy(x+y)2,∵x+y=5,∴(x+y)2=25,x2+y2+2xy=25,∵x2+y2=13,∴xy=6,∴xy(x+y)2=6×25=1
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
原式=4x29y2•27y364x3•4xy=34x2.故答案为34x2.
反应前XY均为0价,反应后化合价有变化,四氧化还原反应.提一句,4X2+Y2=X3Y+Y2去掉Y2的话是4X2=X3Y,这是不可能的,元素本身发生了变化,应该是核反应
∵x+y=4,∴(x+y)2=16,∴x2+y2+2xy=16,而x2+y2=14,∴xy=1,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=14-2=12.
(x2+3/根号x2+1)^2-(2根号2)^2=(x^4-2x^2+1)/8(x^2+1)=(x^2-1)/8(x^2+1)>=0,又因为不等式两边均为正,所以x2+3/根号x2+1≥2根号2
已知x=(根号3+根号2)分之(根号3-根号2)y=(根号3-根号2)分之(根号3+根号2)x=(根号3+根号2)分之(根号3-根号2)x=(根号3-根号2)^2/(根号3+根号2)(根号3-根号2)
原式=(x3y2-x2y-x2y+x3y2)÷3x2y=(2x3y2-2x2y)÷3x2y=23xy-23.
已知x+y=5,xy=3,代数式x3y-2x平方y平方+xy3=xy(x²-2xy+y²)=xy(x-y)²=3×[(x+y)²-4xy]=3×(25-12)=
∵|x+y+1|≥0,|xy-3|≥0|x+y+1|+|xy-3|=0,∴x+y+1=0,即x+y=-1xy=3xy3+x3y=xy(x²+y²)=yx[(x+y)²-2
x+y=4,xy=2后者平方后二式相加再加后者平方
用点到直线的距离公式,可求出圆心(0,0)到此直线的距离小于半径,位置关系是相交
x3y+xy3=xy(x^2+y^2)=(√3-√2)(√3+√2)((√3-√2)^2)+(√3-√2)^2)=1*(3-2√6+2+3+2√6+2)=10
(x-y)2=x2-2xy+y2=9,当x2+y2=13时,13-2xy=9,解得xy=2.当xy=2,x2+y2=13时,x3y-8x2y2+xy3=xy(x2-8xy+y2)=2×(13-8×2)
∵x+y=3,∴(x+y)2=9,即x2+y2+2xy=9①,又x2+y2-3xy=4②,①-②,得5xy=5,xy=1.∴x2+y2=4+3xy=7.∴x3y+xy3=xy(x2+y2)=7.故答案
∵x-y=l,xy=2,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=xy(x-y)2=2×1=2.