x加y加z等于5，xy加yz加xz等于七，那么x平方加y的平方z的平方的值为是多少 若x加上y加上z加上等于a,xy加上yz加上xz等于b,求x...
x\u7684\u5e73\u65b9\u52a0y\u7684\u5e73\u65b9\u52a0z\u7684\u5e73\u65b9\u7b49\u4e8e\u4e00\u6c42YZ\u9664x\u52a0xz\u9664y\u52a0xy\u9664z \u7684\u6700\u5c0f\u503cx^2+y^2+z^2=1\uff0c\u6c42yz/x+xz/y+xy/z\u6700\u5c0f\u503c
\u8bbem=y/x,y=mx,\u5219m\u4e3a\u6b63\u5b9e\u6570
x^+m^x^+z^=1
x^=(1-z^)/(m^+1)
\u8bbek=yz/x+xz/y+xy/z\uff0ck\u4e3a\u6b63\u5b9e\u6570,\u5219
k=mz+z/m+mx^/z
=z(m+1/m)+m(1-z^)/(z(m^+1))
kz=z^(m+1/m)+m/(m^+1)-mz^/(m^+1)
z^(m+1/m-m/(m^+1))-kz+m/(m^+1)=0
\u56e0\u4e3a\u6b64\u65b9\u7a0b\u5f0fz\u6709\u89e3\u5219\u6709
k^-4[m+1/m-m/(m^+1)][m/(m^+1)]>=0
k^>=4[m+1/m-m/(m^+1)]m/(m^+1)
k^>=4(m^4+m^+1)/(m^+1)^
k^>=4[3/4(m^+1)^+1/4(m^-1)^]/(m^+1)^
k^>=3+[(m^-1)/(m^+1)]^
k^>=3
k>=\u68393\u6216k<=-\u68393
\u4f46\u7531\u4e8ek \u4e3a\u6b63\u5b9e\u6570\uff0c\u6240\u4ee5k>=3\u5373
yz/x+xz/y+xy/z>=\u68393
(\u6ce8:^\u8868\u793a\u5e73\u65b9\uff0c^4\u8868\u793a4\u6b21\u65b9)
\u7c7b\u4f3c\u4e8e\u4e24\u4e2a\u672a\u77e5\u6570\u7684\u5b8c\u5168\u5e73\u65b9\uff0c\u8bb0\u4f4f\u5c31\u597d\u505a\u4e86\u2026x*x+y*y+z*z=(x+y+z)*(x+y+z)-3(xy+xz+yz)=a*a-3b
解x+y+z=5
两边平方得：
(x+y+z)平方=25
即x平方+y平方+z平方+2xy+2yz+2zx=25
∵xy+yz+xz=7
∴x平方+y平方+z平方+7×2=25
即x平方+y平方+z平方+14=25
∴x平方+y平方+z平方=11
绛旓細瑙ｏ細x+y+z=5 (x+y+z)²=25 x^2+y^2+z^2+2xy+2xz+2yz=25 鈭xy鍔爕z鍔爔z绛変簬涓 鈭磝²+y²+z²=11
绛旓細瑙 x+y+z=5 涓よ竟骞虫柟寰楋細(x+y+z)骞虫柟=25 鍗硏骞虫柟+y骞虫柟+z骞虫柟+2xy+2yz+2zx=25 鈭祒y+yz+xz=7 鈭磝骞虫柟+y骞虫柟+z骞虫柟+7脳2=25 鍗硏骞虫柟+y骞虫柟+z骞虫柟+14=25 鈭磝骞虫柟+y骞虫柟+z骞虫柟=11
绛旓細xy+yz+xz=3鈫(x+z)(5-x-z)+zx=3 5(z+x)-(z+x)²+zx=3 涓よ竟瀵箈姹傚锛5(z'+1)-2(z+x)(z'+1)+z+xz'=0 z'(5-2z-x)+5-z-2x=0 z'=(z+2x-5)/(5-2z-x)z'=0鈫抸=5-2x鈫抷=5-x-(5-2x)=x 椹荤偣(⅓,⅓,4⅓)銆(3,3,-1)鍗硓鐨...
绛旓細瀹炴暟x,y,z 婊¤冻 x+y+z=5, xy+yz+zx=3 ,鍒檢鐨勬渶澶у兼槸锛13/3 瑙ｏ細鈭祒y+yz+zx=3,x+y+z=5 鈭磝+y=5-z 鈭2xy=6-2(y+x)z=6-2(5-z)z=2z^2-10z+6 鈭2*(xy+yz+zx)=6 鈭祒+y+z=5 鈭达紙x+y+z锛塣2=25 x^2+y^2+z^2+2*(xy+xz+yz)=25 x^2+y^2+z^2...
绛旓細寰 Y锛5-X-Z 灏嗘浠ｅ叆XY+YZ+ZX锛3 寰 X锛2-X-Z锛+锛5-X-Z锛塟+ZX锛3 鏁寸悊寰 X锛2+锛圸-5锛塜+锛圸锛2 -5Z+3锛夛紳0 鍥犱负X鏄疄鏁帮紝閭ｄ箞 鍏充簬X鐨勪竴鍏冧簩娆℃柟绋嬬殑 鍒ゅ埆寮忥紙Z-5锛夛季2 -4锛圸锛2 -5Z+3锛夆墺0 瑙ｈ繖涓竴鍏冧簩娆′笉绛夊紡锛屽緱 -1鈮 Z鈮 13/3 鍙傝冭祫鏂欙細http://z...
绛旓細鑻+y+z=5,xy+yz+xz=7,鍒 x^2+y^2+z^2 =(x+y+z)²-2(xy+yz+xz)=25-14 =11
绛旓細瑙ｏ細鈭祒+y+z=5 鈭磝+y=5-z锛寉+z=5-x锛寈+z=5-y 鍒(x+y)(y+z)(z+x)+xyz =(5-x)(5-y)(5-z)+xyz =(25-5x-5y+xy)(5-z)+xyz =125-25x-25y+5xy-25z+5xz+5xy-xyz+xyz =125-25(x+y+z)+5(xy+xz+yx)=125-25脳5+5脳12 =60 ...
绛旓細x+y+z=5 鈫抷=5-x-z 浠ｅ叆锛xy+yz+xz=3鈫(x+z)(5-x-z)+zx=3 5(z+x)-(z+x)²+zx=3 涓よ竟瀵箈姹傚锛5(z'+1)-2(z+x)(z'+1)+z+xz'=0 z'(5-2z-x)+5-z-2x=0 z'=(z+2x-5)/(5-2z-x)z'=0鈫抸=5-2x鈫抷=5-x-(5-2x)=x 椹荤偣(⅓,⅓,...
绛旓細x+y+z=5 涓よ竟骞虫柟寰 x^2+y^2+z^2+2(xy+yz+xz)=25 3+2(xy+yz+xz)=25 2(xy+yz+xz)=22 xy+yz+xz=11
绛旓細鈭祒+y+z=5锛鈭(x+y+z)²=25锛屽嵆x²+y²+z²+2(xy+yz+zx)=25 鍙堚埖xy+yz+zx=3锛屸埓x²+y²+z²=19锛(5-y-z)²+y²+z²=19 y²+z²-5y-5z+yz+3=0 y²+(z-5)y+z²-5z+3=0 鈭祔鏄疄鏁...
