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\(\left(a+b+c\right)\left(ab+ac+bc\right)=\left(a+b+c\right)\left(ab+ac+bc+c^2-c^2\right)\)
\(=\left(a+b+c\right)\left(\left(a+c\right)\left(b+c\right)-c^2\right)\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2\left(a+b\right)+c\left(a+c\right)\left(b+c\right)-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)-c^2a-c^2b+abc+c^2a+c^2b+c^3-c^3\)
\(=\left(a+b\right)\left(a+c\right)\left(b+c\right)+abc=\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)+2018=2018\)
\(\Rightarrow\left(a+b\right)\left(a+c\right)\left(b+c\right)=0\)
Ta có:
\(A=\left(b^2c+2018\right)\left(c^2a+2018\right)\left(a^2b+2018\right)\)
\(A=\left(b^2c+abc\right)\left(c^2a+abc\right)\left(a^2b+abc\right)\)
\(A=bc\left(a+b\right)ac\left(b+c\right)ab\left(a+c\right)\)
\(A=\left(abc\right)^2\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(A=2018^2.0=0\)
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\Rightarrow6^2=a^2+b^2+c^2+2.12\Rightarrow a^2+b^2+c^2=12\)
Ta có:
\(a^2+b^2+c^2=ab+bc+ca\left(=12\right)\)
\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ca\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow M=0}\)
Chúc bạn học tốt.
Ta có: \(x^2+y^2+z^2\ge xy+yz+zx\)
\(\Rightarrow a^{2018}+b^{2018}+c^{2018}\ge\left(ab\right)^{1009}+\left(bc\right)^{1009}+\left(ca\right)^{1009}\)
Dấu = xảy ra \(\Leftrightarrow a=b=c\)
Mà đẳng thức trên xảy ra dấu =
\(\Leftrightarrow a=b=c\Leftrightarrow P=0\)
Bài kia tí nghĩ nốt, khó v
Sửa đề em nhé: \(\frac{2}{ab}-\frac{1}{c^2}=4\) và tính \(a+b+2c\)
Có: \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ca}+4=4\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{a}=\frac{-1}{c}\\\frac{1}{b}=\frac{-1}{c}\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}a=-c\\b=-c\end{cases}}\)\(\Leftrightarrow a+b+2c=0\)