已知2x减3y减z=0,x 3y减14z=0
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∵|2x-3y+1|+(x+3y+5)的二次方=0∴2x-3y+1=0x+3y+5=0x=-2y=-1∴(-2x*y)的二次方(-y的二次方)×6xy平方的值=4x⁴y*(-y²
由x+y-z=0,2x-y+2z=0可得:z=-3xy=-4x则3x-2y+5z/5x-3y+2z=3x+8x-15x/5x+12x-6x=-4x/11x=-4/11
x+2y-3z=0⑴2x+3y+5z=0⑵⑵-⑴得x+y-2z=0⑶⑶-⑴得y=z代入⑶得x=y=z所以(x+y+z)除以(x-y+z)=3
4x-y+3z=0(1)2x+y+6z=0(2)()+(2)6x+9z=06x=-9zz/x=-2/3(1)*2-(2)8x-2y-2x-y=06x-3y=06x=3yx/y=1/2z/x=-2/3x
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)=(
2x+5y+3z=0,3x-5y-8z=0,两式相加:5x-5z=0,所以,x:z=1(1)式乘以3:6x+15y+9z=0(2)式乘以2:6x-10y-16z=0两式相减:25y+25z=0,y:z
两个方程相减得到,6z=2x,然后z:x=3:1,方程乘上3,得到3x-9y+6z=0,减去方程2得到6y=6z,y:z=1:1,所以x:y:z=1:3:3
x+4y+3z=3x-2y-5z=0则x+4y+3z=0①3x-2y-5z=0,则6x-4y-10z=0②①②两式相加,得7x-7z=0,所以x=z代入①,得z+4y+3z=0,所以y=-z所以x+2
4x-5y+2z=0(1)x+2y=3z(2)(2)×4-(1)得:13y=14zy=14/13z(1)×2+(2)×5得:13x=11zx=11/13z所以:x:y:z=11/13:14/13:1=
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+2y-4z=0(1)3x-y-z=0,(2)(2)*2+(1)得6x-2y-2z+x+2y-4z=07x=6zx=6z/7y=11z/7则3x+2y+z/x+2y+z=(3*6z/7+2*11z/
∵|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
即(x²-2x+1)+(y²+4y+4)+(z²+6z+9)=0(x-1)²+(y+2)²+(z+3)²=0平方相加则都等于0所以x-1=0
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
4x-3y+z=0(1)x+2y-8z=0(2)(1)-(2)×4得-11y+33z=0∴y=3z把y=3z代入(2)得x=2z把x=2z,y=3z代入x+y-z/x-y+2z得原式=(2z+3z-z
∵x-y=l,xy=2,∴x3y-2x2y2+xy3=xy(x2-2xy+y2)=xy(x-y)2=2×1=2.
两式相加,得3x-z=0可得z/x=5将z=5x代入1式13x-y=0得y/x=13所以x:y:z=1:13:5