若2x-3y z=0,x-2y-3y=0
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首先,显然x,y,z均不为0.然后分开看xy:yz=3:2,两边除以y,得x:z=3:2yz:zx=2:1,除以z,得y:x=2:1,两边同时乘以3,得x:y=3:6所以:x:y:z=3:6:2,不能
x²+y²+xy=x²+y²-2xycos120度同理y²+z²+yz=y²+z²-2yzcoa120度x²+
f(x,y,z)=yz+xz使得,y^2+z^2=1,yz=3令F(x,y,z)=yz+xz+a(y²+z²-1)+b(yz-3)Fx=z=0Fy=z+2ay+bz=0Fz=y+x
x^2-yz-8x+7=0……(1),y^2+z^2+yz-6x+6=0……(2);(1)×3+(2)得到:(y-z)^2=-3x^2+30x-27=-3(x-1)(x-9)>=0所以:1
(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz>3(xy+yz+zx)所以只要求证x^2+y^2+z^2>xy+yz+zx2(x^2+y^2+z^2)>2(xy+yz+zx)(x^
应该是设X/2=Y/1=Z/3=K则X=2KY=KZ=3K则有xy+xz+yz=992K^2+6K^2+3K^2=99==>K^2=9所以4x^2-2xz+3yz-9y^2=2X(2X-Z)+3Y(Z
若3/x=2/y=5/z,则x/3=y/2=z/5,设x/3=y/2=z/5=K,则x=3K,y=2K,z=5K将其代入上式,得6k2+10k2+15k231k2-------------------
证明:(x-(yz/x))/(1-yz)=(y-(xz/y))/(1-xz),十字相乘得:(x-(yz/x))×(1-xz)=(y-(xz/y))×(1-yz),化简:x-(yz/x)-x²
左边=x^2y+xy^2+y^2z+yz^2+z^2x+zx^2+3xyz=(x^2y+xyz+zx^2)+(y^2x+xyz+zy^2)+(z^2y+xyz+xz^2)=x*(xy+yz+zx)+y
令(y+z)/(1+yz)=X1,(y-z)/(1-yz)=X2,因为f(x)=lg((1+x)/(1-x))所以f(X1)=lg((1+X1)/(1-X1)=1,f(X2)=lg((1+X2)/(1
x+y+z=2√x+2√(y-1)+2√(z-2)[x-2√x+1]+[(y-1)-2√(y-1)+1]+[(z-2)+2√(z-2)+1]=0(√x-1)^2+[√(y-1)-1]^2+[√(z-2
由(x+y)/xy=5/6(y+z)/yz=-2/3得1/x+1/y=5/6①1/y+1/z=-2/3②1/x-1/z=3/21/z=1/x-3/2③z=2x/(2-3x)代入2(x+3)+xz=0化
是指所构造的方程存在实数解时,其判别式△不小于0.再问::t^2-(y+z)t+yz=0这个是什么意思再答:题目抄错了,应当是证明x²≥3.利用韦达定理啊!依条件式知:yz=x²,
答案是:(2*X)/((X-Z)*(X+Z))再问:解题过程给我写下1再答:=(2X+Z-Y)/[(x-y)(x+z)]-(y-z)/[(x-z)(x-y)]=[(2x+z-y)(x-z)-(y-z)
1.=x^2-(y+z)^2=(x+y+z)(x-y-z)2.a^2-b^2+c^2-2ac=(a-c)^2-b^2=(a-c-b)(a-c+b)ac-b可知原式
(x+2y-z)^2+(z-x)^2=0所以x+2y-z=0,z-x=0x=z所以2y=0,y=0代入xz^2+yz-5√(xz^2+yz+9)+3=0x^3-5√(x^3+9)+3=0(x^3+9)
x+3y+7z=02x+5y+11z=0x=2z,y=-3zz=0,x=y=0分式无意义(x^2+y^2+z^2)/(xy+2yz+3xz),z≠0=[(2z)^2+(-3z)^2+z^2]/[(-2
解题思路:本题的关键是将三个方程两边取倒数,化简后分别将方程等号左边和右边相加,得到1/x+1/y+1/z的值,最后将要求的分式化简,把1/x+1/y+1/z的值带入即可。解题过程:
|x-3|+|y+z|+|2z+1|=0则|x-3|=0x=3|y+z|=0y=-z=1/2|2z+1|=0z=-1/2xy-yz=3x1/2-1/2x(-1/2)=7/4
①x:y:z因为xy:yz:zx=3:2:1所以xy:yz=3:2所以x:z=3:2同理yz:zx=2:1所以y:x=2:1=6:3所以x:y:z=3:6:2②x/yz:y/zx=x^2:y^2=(x