1) pp: biến đổi tương đương

ta có: VT= \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+x^2\right).\)

        = \(\left(ax\right)^2+\left(ay\right)^2+\left(az\right)^2+\left(bx\right)^2+\left(by\right)^2+\left(bz\right)^2+\left(cx\right)^2+\left(cy\right)^2+\left(cz\right)^2\)     (*)

VP=\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)+\left(bz-cy\right)^2+\left(cx-az\right)^2+\left(ay-bx\right)^2\)

=\(\: \left(ax\right)^2+\left(by\right)^2+\left(cz\right)^2+2\left(axby+bycz+czax\right)+\left(bz\right)^2+\left(cy\right)^2+\left(cx\right)^2+\left(az\right)^2\)

\(+\left(ay\right)^2+\left(bx\right)^2-2\left(bzcy+cxaz+aybx\right)\)   (**)

Từ (*),(**)=> VT-VP=0=> VT=VP=> \(\left(a^2+b^2+c^2\right)\left(x^2+y^2+x^2\right).\)=\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)+\left(bz-cy\right)^2+\left(cx-az\right)^2+\left(ay-bx\right)^2\)   (đpcm)

2) áp dụng BĐT Schwartz ta có: 

\(\left(a+b+c\right)^2\le\left(1+1+1\right)\left(a^2+b^2+c^2\right)\)

=>\(2010^2\le3\left(a^2+b^2+c^2\right)\)  (vì a+b+c=2010)

=>\(a^2+b^2+c^2\ge\frac{2010^2}{3}=1346700\)

Dấu '=' xảy ra khi: a=b=c

Vậy GTNN của a^2 +b^2 +c^2 là 1346700 khi a=b=c

\(\frac{bz-cy}{a}=\frac{cx-az}{b}=\frac{ay-bx}{c}=\frac{bxz-cyx}{ax}=\frac{cxy-azy}{by}=\frac{ayz-bxz}{cz}=\frac{bxz-cxy+cyz-azy+ayz-bxz}{ax+by+cz}=0\)

\(\frac{bz-cy}{a}=0\Rightarrow bz=cy\Rightarrow\frac{b}{y}=\frac{c}{z}\)

\(\frac{cx-az}{b}=0\Rightarrow cx=az\Rightarrow\frac{c}{z}=\frac{a}{x}\)

\(\frac{ay-bx}{c}=0\Rightarrow ay=bx\Rightarrow\frac{a}{x}=\frac{b}{y}\)

\(\Rightarrow\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\left(đpcm\right)\)

Học hành thế này! Tớ mách cô Hiền nhé!

\(1.\)

Theo đề ra, ta có:

\(ax+by=c\)

\(bx+cy=a\Leftrightarrow ax+by+bx+cy+cx+ay=c+a+b\)

\(cx+by=b\)

\(\Leftrightarrow x\left(a+b+c\right)+y\left(a+b+c\right)=a+b+c\)

\(\Leftrightarrow\left(x+y-1\right)\left(a+b+c\right)=0\)

Ta có: \(x,y\)thỏa mãn \(\Rightarrow a+b+c=0\Rightarrow a+b=\left(-c\right)\)

Khi đó ta có:

\(a^3+b^3+c^3=a^3+3ab\left(a+b\right)+b^3-3ab\left(a+b\right)+c^3\)

\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3=\left(-c\right)^3-3ab\left(-c\right)+c^3=3abc\)\(\left(đpcm\right)\)

Đặt: \(\frac{ay-bx}{c}=\frac{cx-az}{b}=\frac{bz-cy}{a}=G\)

\(\Rightarrow G=\frac{cay-cbx}{c^2}=\frac{bcx-baz}{b^2}=\frac{abz-acy}{a^2}\)

\(\Rightarrow G=\frac{cay-cbx+bcx-baz+abz-acy}{c^2+b^2+a^2}\)

\(\Rightarrow G=0\)

\(\Rightarrow\left(ay-bx\right)^2=\left(cx-az\right)^2=\left(bz-cy\right)^2=0\)

\(\Rightarrow\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)

Vì bz-cy/a=cx-az/b=ay-bx/c 

=> a(bz-cy)/a^2=b(cx-az)/b^2=c(ay-bx)/c^2 

=> abz-acy/a^2=bcx=baz/b^2=cay-cbx/c^2 

theo tính chất của dãy tỉ số bằng nhau : 

=> abz-acy/a^2=bcx=baz/b^2=cay-cbx/c^2=a^2+... 

= 0/a^2+b^2+c^2=0 

vì bz-cy/a=0=>bz=cy=>y/b=z/c (1) 

vì cx-az/b=0=>cx=az=>x/a=z/c (2) 

từ (1) và (2) => x/a=y/b=z/c

t i c k nhé!! 4645767856875897696890806895789568467856

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

Ta có: bx−cyabx−cya = cx−axbcx−azb = ay−bxcay−bxc

⇒ bx−cyabx−cya = a(bx−cy)a²a(bx−cy)a²  = abx−acya²abx-acya²

    cx−azbcx−axb = b(cx−az)b²b(cx−az)b² = bcx−baxb²bcx−baxb²

     ay−bxcay−bxc = c(ay−bx)c²c(ay−bx)c² = cay−cbxc²cay−cbxc²

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

bx−cyabx−cya = cx−azbcx−axb = cy−bxccy−bxc = abx−acy+bcx−bax+cay−cbxa²+b²+c²abx−acy+bcx−bax+cay−cbxa²+b²+c²  = 0

\(\Rightarrow\) bx - cy = 0

    cx - ax = 0

    ay - bx = 0

\(\Rightarrow\) bx = cy

    cx = ax

    ay = bx

\(\Rightarrow\) xcxc = ybyb 

    xaxa = xcxc 

    ybyb = xaxa 

\(\Rightarrow\) xaxa = ybyb = xcxc 

cyabx o dau vay