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`(a+b+c)^2=3(ab+bc+ca)`
`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`
`<=>a^2+b^2+c^2=ab+bc+ca`
`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`
`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`
`VT>=0`
Dấu "=" xảy ra khi `a=b=c`
`a^3+b^3+c^3=3abc`
`<=>a^3+b^3+c^3-3abc=0`
`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`
`<=>(a+b)^3+c^3-3ab(a+b+c)=0`
`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`
`**a+b+c=0`
`**a^2+b^2+c^2=ab+bc+ca`
`<=>a=b=c`
Đặt \(a+b-c=x;b+c-a=y;c+a-b=z\)
\(\Rightarrow x+y+z=a+b-c+b+c-a+c+a-b\)
\(=a+b+c\)
Thay \(x;y;z;x+y+z\) vào M, ta được:
\(M=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-x^3-y^3-z^3\)
\(=x^3+y^3+z^3-x^3-y^3-z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\)\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+zy+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(a+b-c+c+a-b\right)\)
\(=3.2b.2c.2a=24abc\)
Vì \(24abc⋮24\forall a,b,c\) nên \(M⋮24\)
Vậy...
Ta có: \(a^3+b^3+c^3=3abc\Rightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\Rightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)\cdot c+c^2\right]-3ab\left(a+b+c\right)=0\)\(\Rightarrow\left(a+b+c\right)\left[a^2+b^2+2ab-ac-bc+c^2\right]-3ab\left(a+b+c\right)=0\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\left(1\right)\\a^2+b^2+c^2-ab-bc-ca=0\left(2\right)\end{matrix}\right.\)
Từ (1) \(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{c}\right)\cdot\left(\dfrac{c+a}{a}\right)=\left(\dfrac{-c}{b}\right)\cdot\left(-\dfrac{a}{c}\right)\cdot\left(-\dfrac{b}{a}\right)=-1\)
Từ (2) \(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=0\) \(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}a=b\\c=b\\a=c\end{matrix}\right.\) \(\Rightarrow a=b=c\) \(\Rightarrow P=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{c}\right)\cdot\left(\dfrac{c+a}{a}\right)=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=8\)
Vậy...
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\cdot\left(2a^2+2b^2+2c^2-2ab-2ac-2bc\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\2a^2+2b^2+2c^2-2ab-2ac-2bc=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a^2-2ab+b^2+a^2-2ac+c^2+b^2-2bc+c^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\end{matrix}\right.\)
Ta có: \(B=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(\Leftrightarrow B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
Trường hợp 1: a+b+c=0
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
Thay a+b=-c; b+c=-a và c+a=-b vào biểu thức \(B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\), ta được:
\(B=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=\dfrac{-\left(a\cdot b\cdot c\right)}{abc}=-1\)
Trường hợp 2: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow a=b=c\)
Ta có: \(B=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}\)
mà a=b=c(cmt)
nên \(B=\dfrac{b+b}{b}\cdot\dfrac{c+c}{c}\cdot\dfrac{a+a}{a}=\dfrac{2b}{b}\cdot\dfrac{2c}{c}\cdot\dfrac{2a}{a}=2\cdot2\cdot2=8\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
\(\Leftrightarrow\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(\frac{ab+bc+ca+c^2}{abc\left(a+b+c\right)}\right)=0\)
\(\Leftrightarrow\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc\left(a+b+c\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a+b=0\\b+c=0\\c+a=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\) \(\Rightarrow B=0\)
\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\frac{\left(a+b\right)^3}{c^3}+\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}\)
\(\frac{\left(a^2b+ab^2\right)^3+\left(b^2c+c^2b\right)^3+\left(c^2a+a^2c\right)^3}{\left(abc\right)^3}\)
\(\left(\frac{a}{c}+\frac{b}{c}\right)^3+\left(\frac{b}{a}+\frac{c}{a}\right)^3+\left(\frac{c}{b}+\frac{a}{b}\right)^3\)
\(\left(\frac{a+b}{c}\right)^3+\left(\frac{b+c}{a}\right)^3+\left(\frac{c+a}{b}\right)^3\)
dễ thấy \(\frac{c}{a+b}=\frac{1}{2}< =>\frac{a+b}{c}=2\)
làm tương tự với 3 cái còn lại ta đc:
\(2^3+2^3+2^3=24\)