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a) Ta có: \(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(Vt\ge0\left(\forall a,b,c\right)\) nên dấu "=" xảy ra khi:
\(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Rightarrow a=b=c\)
Ta có : a2 + b2 + c2 = ab + bc + ca
=> 2a2 + 2b2 + 2c2 = 2ab + 2bc + 2ca
=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
= (a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2) = 0
=> (a - b)2 + (b - c)2 + (c - a)2 = 0
=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\Rightarrow a=b=c\left(\text{đpcm}\right)\)
b) Ta có : 2(x2 + t2) + (y + t)(y - t) = 2x(y + t)
=> 2x2 + 2t2 + y2 - t2 = 2xy + 2t
=> 2x2 + t2 + y2 = 2xt + 2xy
=> 2x2 + t2 + y2 - 2xt - 2xy = 0
=> (x2 - 2xy + y2) + (x2 + t2 - 2xt) = 0
=> (x - y)2 + (x - t)2 = 0
=> \(\hept{\begin{cases}x-y=0\\x-t=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y\\x=t\end{cases}}\Rightarrow x=y=t\left(\text{đpcm}\right)\)
c) Ta có a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
=> a2 + b2 + c2 = 0
=> a = b = c = 0
Khi đó A = (0 - 1)2003 + 02004 + (0 + 1)2005
= - 1 + 0 + 1 = 0
Vậy A = 0
a. \(x^2+y^2+z^2=xy+yz+xz\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}}\Leftrightarrow x=y=z\)( đpcm )
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)\)
\(=2\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Suy ra, \(ĐTBĐ\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\left(đpcm\right)\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-ac-bc\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=4a^2+4b^2+4c^2-4ab-4ac-4bc\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow VT\ge0}\)
Dấu "=" xảy ra <=> a = b = c (đpcm)
đặt A= \(\left(a-b\right)^2\)+(b-c)^2+(c-a)^2=a^2-2ab+b^2+b^2-2bc+c^2+
-2ca+a^2
=2(a^2+b^2+c^2-ab-bc-ca)
=>A=2A
=>A=0
=>(a-b)^2+(b-c)^2+(c-a)^2=0
=>\(\left[{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
=>a=b=c (ĐPCM)