Câu 20:
Ta có:
$(a-b)^2 +(b-c)^2 +(c-a)^2 =(a+b-2c)^2 +(b+c-2a)^2 +(c+a-2b)^2$
$\Rightarrow(a-b)^2 +(b-c)^2 +(c-a)^2 - (a+b-2c)^2 - (b+c-2a)^2 - (c+a-2b)^2=0$
$\Rightarrow[(a-b)^2 - (a+b-2c)^2] + [(b-c)^2 - (b+c-2a)^2] + [(c-a)^2 - (c+a-2b)^2]=0$
$\Rightarrow[(a-b-a-b+2c)(a-b+a+b-2c)] + [(b-c-b-c+2a)(b-c+b+c-2a)]+[(c-a-c-a+2b)(c-a+c+a-2b)]=0$
$\Rightarrow[(2c-2b)(2a-2c)]+[(2a-2c)(2b-2a)]+[(2b-2a)(2c- 2b)]=0$
$\Rightarrow 4(c-b)(a-c) + 4(a-c)(b-a) + 4(b-a)(c-b)=0$
$\Rightarrow 4[(c-b)(a-c)+(a-c)(b-a) + (b-a)(c-b) =0$
$\Rightarrow (c-b)(a-c) + (a-c)(b-a)+(b-a)(c-b)=0$
$\Rightarrow ac-c^2 - ab + bc +ab - a^2 - bc + ac + bc -b^2 - ac+ab=0 $
$\Rightarrow (ac+ ac-ac) - c^2 - (ab - ab -ab) - a^2 + (bc -bc+ bc) - b^2=0$
$\Rightarrow ac - c^2 + ab - a^2 +bc =0$
$\Rightarrow a^2 + b^2 + c^2 - ab - bc - ac=0$
$\Rightarrow 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ac=0$
$\Rightarrow a^2 + a^2 + b^2+b^2+ c^2 +c^2 - 2ab - 2bc - 2ac=0$
$\Rightarrow (a^2 - 2ab + b^2) +( a^2 - 2ac + c^2) + ( b^2 - 2bc + c^2)=0$
$\Rightarrow ( a-b)^2 + (a-c)^2 +(b-c)^2 =0$
Với mọi a,b,c thì `(a-b)^2 ge 0 ; (a-c)^2 ge 0 ;(b-c)^2 ge 0`
Dấu bằng xảy ra khi:
$\begin{cases} (a-b)^2=0 \\ (a-c)^2 =0 \\ (b-c)^2 =0 \end{cases}$
`=>`$\begin{cases} a-b=0 \\ a-c =0\\ b-c=0 \end{cases}$
`=> a=b=c`
Vậy$(a-b)^2 +(b-c)^2 +(c-a)^2 =(a+b-2c)^2 +(b+c-2a)^2 +(c+a-2b)^2$ thì a=b=c
Câu 21: Ta có:
$(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)$
$\Rightarrow [(a+d)+(b+c)][(a+d)-(b+c)] =[(a-d)-(b-c)][(a-d)+(b-c)]$
$\Rightarrow(a+d)^2 - (b+c)^2 - [(a-d)-(b-c)][(a-d)+(b-c)]=0$
$ \Rightarrow ( a+d)^2 - (b+c)^2 - (a-d)^2 + (b-c)^2=0$
$\Rightarrow a^2 + 2ad + d^2 - b^2 - 2bc - c^2 - a^2 + 2ad - d^2 + b^2- 2bc+c^2=0$
$\Rightarrow (a^2 - a^2) + (2ad - 2ad) + (d^2 - d^2) - (2bc+2bc) - (c^2 - c^2) + (2ad + 2ad)=0$
$\Rightarrow 4ad - 4bc =0$
$\Rightarrow 4ad = 4bc$
$ \Rightarrow ad = bc$
$ \Rightarrow\dfrac{a}{c } = \dfrac{b}{d}$
Vậy $(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)$ thì $\dfrac{a}{c } = \dfrac{b}{d}$
