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Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}\Leftrightarrow\left(a+b+c\right)\left(ab+bc+ac\right)-abc=0\)
\(\Leftrightarrow a^2b+a^2c+b^2a+b^2c+abc+abc+bc^2+ac^2=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\Leftrightarrow...\)
\(P=0\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow\frac{ab+bc+ac}{abc}=\frac{1}{abc}\Leftrightarrow ab+bc+ac=1\)
\(A=\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{abc}\Leftrightarrow1=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right).abc\Leftrightarrow1=bc+ac+ab\)
\(A=\left(bc+ac+ab+a^2\right)\left(bc+ac+ab+b^2\right)\left(bc+ac+ab+c^2\right)\)
\(A=\left[c\left(a+b\right)+a\left(a+b\right)\right]\left[c\left(a+b\right)+b\left(a+b\right)\right]\left[c\left(c+b\right)+a\left(c+b\right)\right]\)
\(A=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(A=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)
từ đẳng thức: a^3+b^3+c^3=3abc
suy ra a=b=c hoặc a^2+b^2+c^2+ab+ac+bc=0
thay vào bt M
tìm được M=8 hoặc M=-1
hok tốt
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+3a^2b+3b^2a+c^3-3a^2b-3b^2a-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2=ab+bc+ca\end{cases}}\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a=b=c\end{cases}}\).Với a+b+c=0 thì \(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}\Rightarrow}M=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=-1\)
Với a=b=c thì \(M=8\)
\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=0\)
\(\Leftrightarrow a+b+c=0\)
Xét : \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right).\left(b+c\right).\left(c+a\right)=-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\) luôn chia hết cho 3
xét a + b + c = 0 khi đó a + b = -c ; b + c = -a ; a + c = -b
Ta có : \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{\left(-a\right)\left(-b\right)\left(-c\right)}{abc}=-1\)
xét a + b + c \(\ne\)0 . thì \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow a+b=2c;b+c=2a\)\(\Rightarrow a-c=2\left(c-a\right)\)\(\Rightarrow a=c\)( loại vì a khác c )
Vậy A = -1
Ta có: \(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
Vì \(a+b+c\ne0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(c-a\right)^2\ge0\end{cases}\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0}\)
\(\Rightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Rightarrow a=b=c}\)
\(\Rightarrow A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}=\frac{2a.2a.2a}{a.a.a}=\frac{8a^3}{a^3}=8\)
Ta có: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)
\(\Rightarrow A=\left(a+b\right)\left(a^2-ab+b^2\right)\left(b+c\right)\left(b-c\right)\left(c+a\right)=0\)