Toán Lớp 8: cho a, b, c là những hằng số khác 0 không Chứng minh rằng nếu giá trị của hai biểu thức (a^2+b^2+c^2)(x^2+y^2+z^2) và (ã+by+cz)^2) bằng

Question

Toán Lớp 8:  cho a, b, c là những hằng số khác 0 không Chứng minh rằng nếu giá trị của hai biểu thức (a^2+b^2+c^2)(x^2+y^2+z^2) và (ã+by+cz)^2) bằng nhau thì x/a=y/b=z/c CÁC BẠN GIẢI KỸ TRONG MÌNH NHÉ ???? TẠI TRONG SÁCH CÓ LỜI GIẢI MÀ MÌNH KHÔNG HIỂU ????????

cattien · Answer

Giải đáp+Lời giải và giải thích chi tiết:    (a&sup2;+b&sup2;+c&sup2;)(x&sup2;+y&sup2;+z&sup2;)=(ax+by+cz)&sup2;     =>(a&sup2;+b&sup2;+c&sup2;)x&sup2;+(a&sup2;+b&sup2;+c&sup2;)y&sup2;+(a&sup2;+b&sup2;+c&sup2;)z&sup2;=[(ax+by)+cz]&sup2;      =>a&sup2;x&sup2;+b&sup2;x&sup2;+c&sup2;x&sup2;+a&sup2;y&sup2;+b&sup2;y&sup2;+c&sup2;y&sup2;+a&sup2;z&sup2;+b&sup2;z&sup2;+c&sup2;z&sup2;=(ax+by)&sup2;+2(ax+by)cz+c&sup2;z&sup2;                                                                                        (Áp dụng (A+B)&sup2;=A&sup2;+2AB+B&sup2;)     =>a&sup2;x&sup2;+b&sup2;x&sup2;+c&sup2;x&sup2;+a&sup2;y&sup2;+b&sup2;y&sup2;+c&sup2;y&sup2;+a&sup2;z&sup2;+b&sup2;z&sup2;+c&sup2;z&sup2;=a&sup2;x&sup2;+2axby+b&sup2;y&sup2;+2axcz+2bycz+c&sup2;z&sup2;     =>a&sup2;x&sup2;+b&sup2;x&sup2;+c&sup2;x&sup2;+a&sup2;y&sup2;+b&sup2;y&sup2;+c&sup2;y&sup2;+a&sup2;z&sup2;+b&sup2;z&sup2;+c&sup2;z&sup2;-a&sup2;x&sup2;-2axby-b&sup2;y&sup2;-2axcz-2bycz-c&sup2;z&sup2;=0     =>b&sup2;x&sup2;+c&sup2;x&sup2;+a&sup2;y&sup2;+c&sup2;y&sup2;+a&sup2;z&sup2;+b&sup2;z&sup2;-2axby-2axcz-2bycz=0     Có 2axby=2.ay.bx     2axcz=2.az.cx     2bycz=2.bz.cy     =>(a&sup2;y&sup2;-2.ay.bx+b&sup2;x&sup2;)+(a&sup2;z&sup2;-2.az.cx+c&sup2;x&sup2;)+(b&sup2;z&sup2;-2.bz.cy+c&sup2;y&sup2;)=0     =>(ay-bx)&sup2;+(az-cx)&sup2;+(bz-cy)&sup2;=0     (Áp dụng (A-B)&sup2;=A&sup2;-2AB+B&sup2;)     Có (ay-bx)&sup2;&ge;0     (az-cx)&sup2;&ge;0     (bz-cy)&sup2;&ge;0     mà (ay-bx)&sup2;+(az-cx)&sup2;+(bz-cy)&sup2;=0     =>(ay-bx)&sup2;=(az-cx)&sup2;=(bz-cy)&sup2;=0     =>ay-bx=az-cx=bz-cy=0     =>ay=bx=az=cx=bz=cy     =>a/x=b/y=a/x=c/z=b/y=c/z     =>a/x=b/y=c/z hay x/a=y/b=z/c(đpcm)