〔x+y+z〕的平方-〔x-y-z〕的平方,分解因式是什么? (x+y)的平方+(x+z)的平方+(y+z)的平方等于因式...
\u62ec\u53f7x+y+z\u62ec\u53f7\u7684\u5e73\u65b9\u51cf\u53bb\u62ec\u53f7x-y-z\u62ec\u53f7\u7684\u5e73\u65b9\u7b49\u4e8e\u56e0\u5f0f\u5206\u89e3\u7b49\u4e8e\u591a\u5c11( x + y + z )\u201d - ( x - y - z )\u201d
= [ (x+y+z) + (x-y-z) ][ (x+y+z) - (x-y-z) ]
= ( x +y +z +x -y -z )( x +y +z -x +y +z )
= ( 2x )( 2y + 2z )
= 4x( y + z )
=x²+2xy+y²+x²+2xz+z²+y²+2yz+z²
=2x²+2(y+z)x+2(y²+yz+z²)
=(x+y+z+x-y-z)(x+y+z-x+y+z)
=2x(2y+2z)
=4x(y+z)
4x(y+z)
绛旓細(x+y+z)^2-(x-y-z)^2 =(x+y+z+x-y-z)(x+y+z-x+y+z)=2x(2y+2z)=4x(y+z)
绛旓細閫塂銆 锛坸-z)=4锛坸-z)(y-z)鈮4*锛堬紙x-y+y-z)/2)=(x-z) 鈭村彧鏈夊綋绛夊彿鎴愮珛鏃跺師寮忔墠鎴愮珛鏃㈣婊¤冻鏉′欢x-y=y-z锛岃繖鏄熀鏈笉绛夊紡鐨勬潯浠 鈭撮墄+z-2y=0锛岄塂
绛旓細= ( x +y +z +x -y -z )( x +y +z -x +y +z )= ( 2x )( 2y + 2z )= 4x( y + z )
绛旓細(x+y+z)鐨勫钩鏂-(x-y-z)鐨勫钩鏂 =(x+y+z+x-y-z)(x+y+z-x+y+z)=2x(2y+2z)=4x(y+z)绁濆涔犺繘姝,鏈涢噰绾硚~
绛旓細锛坸-z)鐨勫钩鏂-4锛坹-z)(x-y)=[(x-y)+(y-z)]^2-4(y-z)(x-y)=(x-y)^2+2(x-y)(y-z)+(y-z)^2-4(y-z)(x-y)=(x-y)^2-2(x-y)(y-z)+(y-z)^2 =[(x-y)-(y-z)]^2 =(x-2y+z)^2
绛旓細鎴愮珛銆(x-z)²-4(x-y)(y-z)=x²+4y²+z²-4xy+2xz-4yz=(x-2y+z)²=0
绛旓細鐢骞虫柟宸叕寮忔妸瀹冩墦寮锛(x+y+z+x-y-z)(x+y+z-x+y+z)锛2x(2y+2z)锛4x(y+z)
绛旓細濡傛灉浣犵殑鎰忔濇槸 (x+y-z)鐨勫钩鏂 閭ｄ箞灞曞紑灏卞緱鍒皒^2+y^2+z^2+2xy-2xz-2yz 鑰屽鏋滄槸(xy-z)鐨勫钩鏂 寰楀埌x^2 y^2 +z^2 -2xyz
绛旓細(x-y-z)^2 =(x-y-z)(x-y-z)=x^2+y^2+z^2-2xy-2xz+2yz
绛旓細鍘熷紡=[(x-y)+z]²=(x-y)²+2(x-y)z+z²=x²-2xy+y²+2xz-2yz+z²
