a) Ta có:
\(a^2+b^2+c^2=ab+bc+ca\)
\(\Leftrightarrow\) \(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\) \(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\) \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)
Ta có: (a-b)2 \(\geq\) 0; (b-c)2 \(\geq\) 0; (a-c)2 \(\geq\) 0 (2)
(1)(2) \(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases} \) \(\Leftrightarrow\) a=b=c
b) Ta có: \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab+2ac+2bc=3a^2+3b^2+3c^2\)
\(\Leftrightarrow\) \(3a^2+3b^2+3c^2-a^2-b^2-c^2-2ac-2bc-2ab=0\)
\(\Leftrightarrow\) \(2a^2+2b^2+2c^2-2ab-2ac-2bc=0\)
\(\Leftrightarrow\) \(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Ta có: (a-b)2 \(\geq\) 0; (b-c)2 \(\geq\) 0; (a-c)2 \(\geq\) 0 (2)
(1)(2) \(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases} \) \(\Leftrightarrow\) a=b=c
c. Ta có: \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab+2ac+2bc=3ab+3bc+3ac\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac=0\)
\(\Leftrightarrow\) \(a^2+b^2+c^2-ab-bc-ac=0\)
\(\Leftrightarrow\) \(2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\) \(\left(a^2-2bc+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
\(\Leftrightarrow\) \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Ta có: (a-b)2 \(\geq\) 0; (b-c)2 \(\geq\) 0; (a-c)2 \(\geq\) 0 (2)
(1)(2) \(\Rightarrow\) \(\begin{cases} (a-b)^{2}=0\\ (b-c)^{2}=0\\ (a-c)^{2}=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a-b=0\\ b-c=0\\ a-c=0 \end{cases} \) \(\Leftrightarrow\) \(\begin{cases} a=b\\ b=c\\ a=c \end{cases} \) \(\Leftrightarrow\) a=b=c
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Học tại nhà - Toán - Bài 7: CMR: a = b = c nếu có 1 trong các điều kiện sau:1/ a2 + b2 + c2 = ab + bc + ca.2/ (a + b + c)2 = 3(a2 + b2 + c2)3/ (a + b + c)2 = 3 (ab + bc + ca).