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\(a^2+b^2+c^2+ab+bc+ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2+2ab+2bc+2ca=0\)
\(\Leftrightarrow a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2=0\)
\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a+b=0\\b+c=0\\c+a=0\end{cases}}\Leftrightarrow a=b=c=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có 3(a2 + b2 + c2) = (a + b + c)2
<=> 3a2 + 3b2 + 3c2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
<=> 2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca = 0
<=> (a - b)2 + (b - c)2 + (c - a)2 = 0
<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)(đpcm)
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\(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(< \frac{2}{3^2-1}+\frac{2}{5^2-1}+\frac{2}{7^2-1}+...+\frac{2}{2017^2-1}\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
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\(x\div y=-2\div5\Leftrightarrow\frac{x}{-2}=\frac{y}{5}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{-2}=\frac{y}{5}=\frac{2x+5y}{2.\left(-2\right)+5.5}=\frac{21}{21}=1\)
\(\Leftrightarrow\hept{\begin{cases}x=1.\left(-2\right)=-2\\y=1.5=5\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)