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Ta có:\(a^2-b=b^2-c\)
\(\Leftrightarrow a^2-b^2=b-c\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\)
\(\Leftrightarrow a+b=\frac{b-c}{a-b}\)
\(\Leftrightarrow a+b+1=\frac{b-c}{a-b}+1\)
\(\Leftrightarrow a+b+1=\frac{a-c}{a-b}\)
Cmtt ta có:
\(\hept{\begin{cases}b^2-c=c^2-a\Leftrightarrow b+c+1=\frac{b-a}{b-c}\\c^2-a=a^2-b\Leftrightarrow c+a+1=\frac{c-b}{c-a}\end{cases}}\)
\(\Rightarrow\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)=\frac{a-c}{a-b}.\frac{b-c}{b-a}.\frac{c-b}{c-a}=-1\)
Cre:mạng
GT không hợp lí
Theo định lí cosi 3 số
a^3+b^3+c^3>=3*canbacba(a^3*b^3*c^3)
<=> a^3+b^3+c^3>=3abc
dấu"=" khi a=b=c
trái Gt a,b,c đôi một khác nhau
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{bc+ca+ab}{abc}=0\)
\(\Rightarrow bc+ca+ab=0\)
\(\Rightarrow\hept{\begin{cases}bc=-ac-ab\\ca=-bc-ab\\ab=-bc-ca\end{cases}}\)
\(A=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ba}\)
\(A=\frac{a^2}{a^2+bc-ac-ab}+\frac{b^2}{b^2+ca-bc-ab}+\frac{c^2}{c^2+ab-bc-ca}\)
\(A=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
Mình tiếp tục nhé
\(A=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)=\left(a-b\right)\left(b-c\right)\left[\left(a+b\right)-\left(b+c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Vậy A = 1
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:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
