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Xét \(A=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\)
\(=a.\frac{a}{b+c}+b.\frac{b}{c+a}+c.\frac{c}{a+b}\)
\(=a.\left(\frac{a}{b+c}+1-1\right)+b.\left(\frac{b}{c+a}+1-1\right)+c.\left(\frac{c}{a+b}+1-1\right)\)
\(=a.\frac{a+b+c}{b+c}-a+b.\frac{a+b+c}{c+a}-b+c.\frac{a+b+c}{a+b}-c\)
\(=\left(a+b+c\right).\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)-\left(a+b+c\right)\)
\(=\left(a+b+c\right).2020-\left(a+b+c\right)\)
\(\Rightarrow P=\frac{A}{a+b+c}=\frac{\left(a+b+c\right).2019}{a+b+c}=2019\)
Vậy...
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
Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\Rightarrow x+y+z=0\).
\(A^2=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)\)
\(=4+2.\frac{x+y+z}{xyz}=4+0=4\).
\(\Leftrightarrow A=\pm2\).
Các cao nhân giúp với!!!!!!!!!! Thanks for all
Ta có:\(a+b+c\ne0\)vì nếu \(a+b+c=0\)thế vào giả thiết ta có:
\(\frac{a}{-a}+\frac{b}{-b}+\frac{c}{-c}=1\Leftrightarrow-3=1\)(vô lí)
Khi \(a+b+c\ne0\)ta có:
\(\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right).\left(a+b+c\right)=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{a.\left(b+c\right)}{b+c}+\frac{b.\left(c+a\right)}{c+a}+\frac{b^2}{c+a}+\frac{c.\left(a+b\right)}{a+b}+\frac{c^2}{a+b}=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}+a+b+c=a+b+c\)
\(\Rightarrow\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0\)\(\Rightarrow P=0\)
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