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a: a^3+b^3+c^3-3abc
=(a+b)^3+c^3-3ab(a+b)-3bac
=(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)
=(a+b+c)(a^2+b^2+c^2-ab-ac-bc)
b: Đề sai rồi bạn
c: 2(a+b+c)*(b/2+c/2-a/2)
=(a+b+c)(b+c-a)
=(b+c)^2-a^2
=c^2+2bc+c^2-a^2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2\right)-\left(a+b+c\right)\left(ab+bc+ac\right)\)
\(=a^3+ab^2+ac^2+a^2b+b^3+c^2b+a^2c+b^2c+c^3-a^2b-abc-a^2c-ab^2-b^2c-abc-abc-bc^2-ac^2\)
\(=a^3+b^3+c^3-3abc\left(đpcm\right)\)
b) Bạn chỉ cần nhân bung cả 2 vế ra là được á .
c) \(2\left(a+b+c\right)\left(\dfrac{b}{2}+\dfrac{c}{2}-\dfrac{a}{2}\right)\)
\(=2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=ab+ac-a^2+b^2+bc-ab+bc+c^2-ac\)
\(=2bc+b^2+c^2-a^2\left(đpcm\right)\)
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1.
\(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
Ta có:
\(\dfrac{\left(a+2b\right)^2+\left(b+2c\right)^2+\left(c+2a\right)^2}{\left(a-2b\right)^2+\left(b-2c\right)^2+\left(c-2a\right)^2}\)
\(=\dfrac{a^2+4b^2+4ab+b^2+4c^2+4bc+c^2+4a^2+4ca}{a^2+4b^2-4ab+b^2+4c^2-4bc+c^2+4a^2-4ca}\)
\(=\dfrac{5\left(a^2+b^2+c^2\right)+4\left(ab+bc+ca\right)}{5\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-10\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)}{-10\left(ab+bc+ca\right)-4\left(ab+bc+ca\right)}\)
\(=\dfrac{-6}{-14}=\dfrac{3}{7}\)
b.
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3abc\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\) \(\Leftrightarrow a=b=c\)
\(\Rightarrow\dfrac{ab+2bc+3ca}{3a^2+4b^2+5c^2}=\dfrac{a^2+2a^2+3a^2}{3a^2+4a^2+5a^2}=\dfrac{6}{12}=\dfrac{1}{2}\)
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Đặt \(\left(\dfrac{1}{a};\dfrac{1}{2b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\)
\(M=\dfrac{x^2}{yz}+\dfrac{y^2}{zx}+\dfrac{z^2}{xy}=\dfrac{x^3+y^3+z^3}{xyz}\)
\(=\dfrac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3}{xyz}=\dfrac{-z^3-3xy\left(-z\right)+z^3}{xyz}\)
\(=\dfrac{3xyz}{xyz}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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