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\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ac\right)^2\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)
\(\Leftrightarrow a^4+b^4+c^4=2\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(ab+bc+ac\right)\right]\)\(\Leftrightarrow a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\)
Câu hỏi của Khoa Nguyễn Đăng - Toán lớp 8 - Học toán với OnlineMath
+ a + b + c = 0 \(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
+ \(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=\left[-2\left(ab+bc+ca\right)\right]^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=4\left(ab+bc+ca\right)^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+a^2bc+abc^2\right)\right]\)
\(=2\left(ab+bc+ca\right)^2+4\left(ab^2c+abc^2+a^2bc\right)\)
\(=2\left(ab+bc+ca\right)^2+4abc\left(a+b+c\right)\)
\(=2\left(ab+bc+ca\right)^2\)
a+b+c=0\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow ab+bc+ac=\frac{-a^2-b^2-c^2}{2}\)
\(\Rightarrow2\left(ab+bc+ac\right)^2=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)(1)
Lại có \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(=a^4+b^4+c^4+2\left(ab+bc+ac\right)^2-2abc\left(a+b+c\right)\)
\(=a^4+b^4+c^4+2\left(ab+bc+ac\right)^2\)(do a+b+c=0)
Thay vào (1)
\(2\left(ab+bc+ca\right)^2=\frac{a^4+b^4+c^4}{2}+\left(ab+cb+ac\right)^2\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\frac{a^4+b^4+c^4}{2}\)
\(\Rightarrowđpcm\)
Từ a+b+c=6 \(\Rightarrow\)a+b=6-c
Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9
\(\Leftrightarrow\)ab=9-c(a+b)
Mà a+b=6-c (cmt)
\(\Rightarrow\)ab=9-c(6-c)
\(\Rightarrow\)ab=9-6c+c2
Ta có: (b-a)2\(\ge\)0 \(\forall\)b, c
\(\Rightarrow\)b2+a2-2ab\(\ge\)0
\(\Rightarrow\)(b+a)2-4ab\(\ge\)0
\(\Rightarrow\)(a+b)2\(\ge\)4ab
Mà a+b=6-c (cmt)
ab= 9-6c+c2 (cmt)
\(\Rightarrow\)(6-c)2\(\ge\)4(9-6c+c2)
\(\Rightarrow\)36+c2-12c\(\ge\)36-24c+4c2
\(\Rightarrow\)36+c2-12c-36+24c-4c2\(\ge\)0
\(\Rightarrow\)-3c2+12c\(\ge\)0
\(\Rightarrow\)3c2-12c\(\le\)0
\(\Rightarrow\)3c(c-4)\(\le\)0
\(\Rightarrow\)c(c-4)\(\le\)0
\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)
*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)
*
Ta có :
\(\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\left(1\right)\)
\(\Leftrightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\) (vì \(a+b+c=0\))
\(\left(1\right)+\left(2\right)\Rightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)
\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)
\(\Rightarrow dpcm\)
b, Ta có \(m=a+b+c\)
\(\Rightarrow am+bc=a\left(a+b+c\right)+bc=a\left(a+b\right)+ac+bc=\left(a+c\right)\left(a+b\right)\)
CMTT \(bm+ac=\left(b+c\right)\left(b+a\right)\);\(cm+ab=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)