已知X,Y,Z是正数,则X Y XY=8
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∵x,y,z为正数∴利用柯西不等式(x+2y+3z)(1/x+2/y+3/z)>=(1+2+3)²所以1/x+2/y+3/z>=(1+2+3)²/(x+2y+3z)=18所以1/X
配凑柯西不等式1/(x+y)+1/(y+z)+1/(z+x)≤[1/2(xy)^0.5]+[1/2(yz)^0.5]+[1/2(zx)^0.5]=(1/2){1*[z/(x+y+z)]^0.5+1*[
∵xyx+y=2∴xy=2(x+y)∴原式=3x−5×2(x+y)+3y−x+3×2(x+y)−y=−7x−7y5x+5y=−75
4^x>0,4^y>0,4^z>0所以4^x+4^y+4^z≥3(4^x*4^y*4^z)的立方根=3*[4^(x+y+z)]的立方根=3*[4^1]的立方根所以最小值=3*(4的立方根)
2^x=3^y=5^z=ax,y,z均为正数所以a>1lga>02^x=3^y=5^z=a取对数xlg2=lgaylg3=lgazlg5=lga所以2x=2lga/lg2=lga/(1/2*lg2)3
解题思路:,解题过程:最终答案:略
这么简单的题目,你们不要老是依靠答案,要自己算出答案来,就算错了,那也是你自己算出来的,就算你骗了老师,但你同事也骗了你自己
x,y,z均为正数,xy+yz+zx=1,求x+y+z的最小值设M=2(x+y+z)² 则M=2x²+2y²+2z²+4xy+4yz+4zx=(x²
解;已知正数x,y满足,x2+y2=1,则1=x2+y2≥2xy,∴xy≤12…① 又xyx+y=11x+1y≤12 1x•1y=xy2…②①②联立得xyx
是2x,3y,5z吧,打错了?lg2^x=lg3^y=lg5^z,x*lg2=y*lg3=z*lg5,x=ylg3/lg2,z=y*lg3/lg5,所以2x=2ylg3/lg2,5z=5y*lg3/l
∵xyx+y=-2,yzy+z=43,zxz+x=-43,∴1x+1y=-12,1y+1z=34,1z+1x=-34,∴2(1x+1y+1z)=-12,即1x+1y+1z=-14,则xyzxy+yz+
柯西【x^2/(y+z)+y^2/(x+z)+z^2/(x+y)】*(y+z+x+z+x+y)≥(x+y+z)^2即x^2/(y+z)+y^2/(x+z)+z^2/(x+y)≥(x+y+z)/2=(3
由(1)、(3)得y=xx−2,z=6xx−3,故x≠0,代入(2)解得x=2710,所以y=277,z=-54.检验知此组解满足原方程组.∴10x+7y+z=0.故选D.
还有什么地方不懂的可以再问哈~
(已知x,y,z为正数,令3^x=4^y=6^z=t3^x=tx=log3(t)1/x=logt(3)4^y=ty=log4(t)1/y=logt(4)6^z=tz=log6(t)1/z=logt(6
lg2^x=lg3^y=lg5^z,x*lg2=y*lg3=z*lg5,x=ylg3/lg2,z=y*lg3/lg5,所以2x=2ylg3/lg2,5z=5y*lg3/lg5,比较2lg3/lg2、5
用几何方法做
(xy+yz)?
×=负3y=1z=2(1)X的平方+y的平方+Z的平方-2xy-2xz-2yz=(-3)×(-3)+1×1+2×2+6+12-4=9+1+4+14=-282)(x-y)(y-z)(z-x)分之xyz=
由柯西不等式可得(x+2y+2y+3z+3z+x)(1x+2y+42y+3z+93z+x)≥(1+2+3)2,∵x+2y+3z=1,∴2(1x+2y+42y+3z+93z+x)≥36,∴1x+2y+4