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Ta có : \(2016a+bc=\left(a+b+c\right).a+bc=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
\(2016b+ac=\left(a+b+c\right).b+ac=ab+b^2+bc+ac=b\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(b+c\right)\)
\(2016c+ab=\left(a+b+c\right)c+ab=ac+bc+c^2+ab=a\left(b+c\right)+c\left(b+c\right)=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(2016a+bc\right)\left(2016b+ac\right)\left(2016c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\) (đpcm)
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Xét : \(a^3+b^3+c^3=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-bc-ac\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
Suy ra : \(\frac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ac}=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{a^2+b^2+c^2-ab-bc-ac}=a+b+c=2016\)
Vậy ta có điều phải chứng minh.
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\(\frac{a}{ab+a+2016}+\frac{b}{bc+b+1}+\frac{2016c}{ac+2016c+2016}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{abc^2}{ac+abc^2+abc}\)
\(=\frac{a}{a.\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{abc^2}{ac.\left(1+bc+b\right)}\)
\(=\frac{1}{b+bc+1}+\frac{b}{b+bc+1}+\frac{bc}{b+bc+1}\)
\(=\frac{1+b+bc}{b+bc+1}=1\)
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a2+b2+c2=ab+bc+ca
<=>2a2+2b2+2c2=2ab+2bc+2ca
<=>(a2-2ab+b2)+(b2-2bc+c2)+(c2-2ca+a2)=0
<=>(a-b)2+(b-c)2+(c-a)2=0
<=>a=b=c
mà a+b+c=3<=>a=b=c=1
=>P=0
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a) \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Leftrightarrow a=b=c=1\)
b) \(\left(a+b+c\right)^2=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=3\left(ab+bc+ac\right)\)
\(\Leftrightarrow a^2+b^2+c^2=ab+bc+ac\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\Leftrightarrow\left(a^2+b^2-2ab\right)+\left(b^2+c^2-2bc\right)+\left(c^2+a^2-2ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
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\(a^2+b^2+c^2=ab+bc+ac\)
\(a^2+b^2+c^2-ab-bc-ac=0\)
\(2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
mà \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\forall a;b;c\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Rightarrow}a=b=c}\)
\(\Rightarrow\left(a-b+1\right)^{2018}+\left(b-c-1\right)^{2017}+\left(a-c\right)^{2016}\)
\(=\left(a-a+1\right)^{2018}+\left(c-c-1\right)^{2017}+\left(a-a\right)^{2016}\)
\(=1^{2018}+\left(-1\right)^{2017}+0^{2016}\)
\(=1+\left(-1\right)+0\)
\(=0\)
Vậy......
P.s: các phần thay a=b=c vào biểu thức có thể thay toàn bộ bằng a hoặc bằng b hoặc bằng c đều được nha
giúp với m.n
Làm đơn giản thế này thôi nhé An Kì :
Ta có : \(2016a+bc=\left(a+b+c\right)a+bc=a^2+ab+ac+bc=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)Tương tự : \(2016b+ac=\left(a+b\right)\left(b+c\right)\)
\(2016c+ab=\left(a+c\right)\left(b+c\right)\)
\(\Rightarrow\left(2016a+bc\right)\left(2016b+ac\right)\left(2016c+ab\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)