\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ca=\)

\(=\left(a^2+b^2+c^2\right)+2\left(ab+bc+ca\right)=13^2=169\)

\(\Rightarrow85+2\left(ab+bc+ca\right)=169\Rightarrow ab+bc+ca=42\)

(a +b+c)² = a² + b² + c² + 2ab + 2ac + 2bc ( hằng đẳng thức mở rộng )

13² = 85 + 2 ( ab + bc +ca )

\(\Rightarrow\)ab + bc + ca = (169 - 85) :2 = 42

a: ĐKXĐ: \(\left(a+b+c\right)^2-\left(ab+bc+ca\right)<>0\)

=>\(a^2+b^2+c^2+2\left(ab+ac+bc\right)-\left(ab+ac+bc\right)<>0\)

=>\(a^2+b^2+c^2+ab+ac+bc<>0\)

=>\(2a^2+2b^2+2c^2+2\left(ab+ac+bc\right)<>0\)

=>\(\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)<>0\)

=>\(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2<>0\)

Dấu '=' xảy ra khi \(\begin{cases}a+b=0\\ b+c=0\\ a+c=0\end{cases}\Rightarrow a=b=c=0\)

=>Để M xác định thì \(a^2+b^2+c^2<>0\)

b: \(\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+ac+bc\right)^2\)

\(=\left(a^2+b^2+c^2\right)\left\lbrack a^2+b^2+c^2+2\left(ab+ac+bc\right)\right\rbrack+\left(ab+ac+bc\right)^2\)

\(=\left(a^2+b^2+c^2\right)^2+2\left(a^2+b^2+c^2\right)\left(ab+ac+bc\right)+\left(ab+ac+bc\right)^2\)

\(=\left(a^2+b^2+c^2+ab+ac+bc\right)^2\)

\(\left(a+b+c\right)^2-ab-ac-bc\)

\(=a^2+b^2+c^2+2\left(ab+ac+bc\right)-\left(ab+ac+bc\right)\)

\(=a^2+b^2+c^2+ab+ac+bc\)

Ta có: \(M=\frac{\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+ac+bc\right)^2}{\left(a+b+c\right)^2-ab-ac-bc}\)

\(=\frac{\left(a^2+b^2+c^2+ab+ac+bc\right)^2}{\left(a^2+b^2+c^2+ab+ac+bc\right)}\)

\(=a^2+b^2+c^2+ab+ac+bc\)

\(P\le a^2+b^2+c^2+3\sqrt{3\left(a^2+b^2+c^2\right)}=12\)

\(P_{max}=12\) khi \(a=b=c=1\)

Lại có: \(\left(a+b+c\right)^2=3+2\left(ab+bc+ca\right)\ge3\Rightarrow a+b+c\ge\sqrt{3}\)

\(a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}=3\)

\(\Rightarrow\sqrt{3}\le a+b+c\le3\)

\(P=\dfrac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}+3\left(a+b+c\right)\)

\(P=\dfrac{1}{2}\left(a+b+c\right)^2+3\left(a+b+c\right)-\dfrac{3}{2}\)

Đặt \(a+b+c=x\Rightarrow\sqrt{3}\le x\le3\)

\(P=\dfrac{1}{2}x^2+3x-\dfrac{3}{2}=\dfrac{1}{2}\left(x-\sqrt{3}\right)\left(x+6+\sqrt{3}\right)+3\sqrt{3}\ge3\sqrt{3}\)

\(P_{min}=3\sqrt{3}\) khi \(x=\sqrt{3}\) hay \(\left(a;b;c\right)=\left(0;0;\sqrt{3}\right)\) và hoán vị

thế bạn bt hok

\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}=\dfrac{1+1+1}{a+b+c}=\dfrac{3}{a+b+c}=\dfrac{3}{1}=3\)

\(\Rightarrow a=b=c=\dfrac{1}{3}\)

\(\Rightarrow A=\dfrac{a^3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}=a^3=\left(\dfrac{1}{3}\right)^3=\dfrac{1}{27}\)