Toán Lớp 8: Cho P= $(x+y)^{2}$ + $(y+z)^{2}$ + $(z+x)^{2}$ và Q= (x+y).(y+z) + (y+z).(z+x) + (z+x).(x+y). Chứng minh rằng nếu P=Q thì x=y=z

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Toán Lớp 8:  Cho P= $(x+y)^{2}$ + $(y+z)^{2}$ + $(z+x)^{2}$ và Q= (x+y).(y+z) + (y+z).(z+x) + (z+x).(x+y). Chứng minh rằng nếu P=Q thì x=y=z

thaolanphuobg · Answer

Giải đáp:   nếu P=Q th&igrave; x=y=z   Lời giải và giải thích chi tiết:   ta c&oacute; :   P=Q   &hArr;(x+y)&sup2;+(y+z)&sup2;+(z+x)&sup2;=(x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)   &hArr;2x&sup2;+2y&sup2;+2z&sup2;+2xy+2yz+2xz=3xy+3yz+3xz+y&sup2;+x&sup2;+z&sup2;   &hArr;x&sup2;+y&sup2;+z&sup2;-xy-yz-xz=0   &hArr;2x&sup2;+2y&sup2;+2z&sup2;-2xy-2yz-2xz=0   &hArr;(x-y)&sup2;+(y-z)&sup2;+(z-x)&sup2;=0   M&agrave; (x-y)&sup2;+(y-z)&sup2;+(z-x)&sup2;&ge;0(&forall;x,y,z)   dấu '=' xảy ra &hArr;x=y=z   Vậy nếu P=Q th&igrave; x=y=z

ngocbichchau · Answer

Ta c&oacute; : Q= (x+y)(y+z)+(y+z)(z+x)+(z+x)(x+y)   =x&sup2; +y&sup2; +z&sup2;+ 3xy +3yz+3xz   P= (x+y)&sup2;+(y+z)&sup2;+(z+x)&sup2;   = x&sup2;+2xy+y&sup2;+y&sup2;+2yz+z&sup2;+z&sup2;+2zx+x&sup2;   =2x&sup2;+2y&sup2;+2z&sup2;+2xy+2yz+2xz   Lại c&oacute; P - Q = 2x&sup2; + 2y&sup2; + 2z&sup2; + 2xy + 2yz + 2xz - x&sup2; - y&sup2; -z&sup2;- 3xy -3yz-3xz   = x&sup2;+y&sup2;+z&sup2; - xy - yz - xz=0   => x&sup2;+y&sup2;+z&sup2; = xy + yz +xz   m&agrave; x&sup2;+y&sup2;+z&sup2; $ geq$ xy + yz +xz   => dấu bằng xảy ra khi x = y = z khi P=Q