xyz不等于0
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(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z
x+3y+2z=0.(1)3x-y+z=0.(2)(1)-2*(2)得-5x+5y=0x=yx:y=1:1
设(y+z)/x=(x+z)/y=(x+y)/z=k;y+z=kx;x+z=ky;y+z=kx;2(x+y+z)=k(x+y+z);k=2或x+y+z=0;所以,(y+z)(x+z)(x+y)/xyz
由|3x-2y+z|≥0,|2x+y+2z|≥0,且|3x-2y+z|+|2x+y+2z|=0,得|3x-2y+z|=|2x+y+2z|=0∴3x-2y+z=2x+y+2z=0由3x-2y+z=2x+
2^x=10^z(2^x)^y=(10^z)^y2^xy=10^yz5^y=10^z(5^y)^x=(10^z)^x5^xy=10^xz相乘2^xy*5^xy=10^yz*10^xz(2*5)^xy=
是X+Y/5=Y+X/6=Z+X/7吧由X+Y/5=Y+X/6解得,X=24Y/25把上式代入:Y+X/6=Z+X/7解得Z=179Y/175所以X:Y:Z=(24Y/25):Y:179Y/175=1
把z看成已知数,解3x-4y-z=02x+y-8z=0,得x=3z,y=2z原式=(9z^2+4z^2+z^2)/(6z^2+2z^2+3z^2)=14/11
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
x-2y-3z=0(1)2x+y-2z=0(2)(1)+(2)*2x-2y-3z+4x+2y-4z=05x-7z=05x=7zx:z=7:5(2)-(1)*22x+y-2z-2x+4y+6z=05y+
令(y+z)/x=(z+x)/y=(x+y)/z=t∴y+z=xt,z+x=yt,x+y=zt三式相加得:2(x+y+z)=(x+y+z)t∴(2-t)(x+y+z)=0∴2-t=0或x+y+z=0若
4x-5y+2z=0①x+4y-3z=0②有②得4x+16y-12z=0③①-③得21y-14z=o即Z=1.5Y④④带入①得4X-5Y+3Y=0既X=0.5Y∴x:y:z=0.5Y:y:1.5Y=1
2x-3y+z=0①3x-2y-6z=0②由①×6得12x-18y+6z=0③由②+③得15x-20y=015x=20yx=4y/3把x=4y/3代入①得8y/3-3y+z=0-y/3+z=0z=y/
4x-5y+2z=0①x+4y-3z=0②由①+②得:5x-(y+z)=05x=y+z由②*4-①得:21y-14z=0y=2z/35x=2z/3+z=5z/3x=z/3x:y=(z/3):(2z/3
4x-5y+2z=0①x+4y-3z=0②有②得4x+16y-12z=0③①-③得21y-14z=o即Z=1.5Y④④带入①得4X-5Y+3Y=0既X=0.5Y∴x:y:z=0.5Y:y:1.5Y=1
2x-y+z=0上式两边同时除以x可得y/x-z/x=2标注为①x-2y+3z=0上式两边同时除以x刻碟2y/x-3z/x=1标注为②联解①与②可得y/x=5,z/x=3所以(x²+3y
31x+19y-23应该是:31x+19y-23z,少了一个z就不好解答了4x-5y+2z=0①x+4y-3z=0②由①+②得:5x-(y+z)=0所以5x=y+z由②*4-①得:21y-14z=0所
(x+y)/z=(x+z)/y=(z+y)/xx,y,z等价x=y=z(x+y)(x+z)(z+x)/xyz=8
x+y-z/z=x-y+z/y=-x+y+z/x=kx+y=z(k+1)y+z=x(k+1)x+z=y(k+1)2x+2y+2z=z(k+1)+x(k+1)+y(k+1)(k-1)(x+y+z)=0x
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,
(1/2)x=(1/3)y=(1/4)z=kx=2k,y=3k,z=4kx:y:z=2:3:4