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Ta có \(E=\frac{a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=a+b+c=1\)
Ở đây chú ý rằng \(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)=-\left(a+b+c\right)\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Đặt \(\left(\frac{a-b}{c};\frac{b-c}{a};\frac{c-a}{b}\right)\rightarrow\left(x;y;z\right)\)
Khi đó:
\(S=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\)
Ta có:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-cb+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{\left(b-a\right)\left(b+a\right)-c\left(a-b\right)}{ab}\cdot\frac{c}{a-b}=\frac{\left(b-a\right)\left(b+a-c\right)}{ab}\cdot\frac{c}{a-b}=\frac{c\left(b+a-c\right)}{ab}\)
\(=\frac{2c^2}{ab}=\frac{2c^3}{abc}\)
Một cách tương tự khi đó:\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{2\left(a^3+b^3+c^3\right)}{abc}=\frac{2\cdot3abc}{abc}=6\)
Khi đó:\(S=3+6=9\) Bạn để ý rằng \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
sao \(\frac{c\left(b+a-c\right)}{ab}\) lại bằng \(\frac{2c^2}{ab}\)
\(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)+c^2-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b\right)-3\left(a+b\right).c\left(a+b+c\right)-3abc=0\)
\(\Leftrightarrow\left(a+b+c\right)^3-3ab\left(a+b+c\right)-3\left(a+b\right).c\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b+c\right)^2-3ab-3ab-3bc\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]=0\)
Ta có:
\(a;b;c>0\)
\(\Rightarrow a+b+c>0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
\(A=2020\left(1-\dfrac{a}{b}\right)\left(1-\dfrac{b}{c}\right)\left(1-\dfrac{c}{a}\right)-2021\left(\dfrac{a}{b}-\dfrac{b}{c}+\dfrac{c}{a}\right)^3\)
\(\Rightarrow A=2020.\left(1-1\right)\left(1-1\right)\left(1-1\right)-2021\left(1-1+1\right)^3\)
\(\Rightarrow A=-2021\).
Gọi A= \(\frac{a-b}{c}\)+ \(\frac{b-c}{a}\)+ \(\frac{c-a}{b}\), ta có:
A*\(\frac{c}{a-b}\)= 1+\(\frac{c}{a-b}\)(\(\frac{b-c}{a}\)+\(\frac{c-a}{b}\))
= 1+ \(\frac{c}{a-b}\)* \(\frac{b^2-bc+ac-a^2}{ab}\)= 1 +\(\frac{c}{a-b}\)*\(\frac{\left(a-b\right)\left(c-a-b\right)}{ab}\)= 1+\(\frac{2c^2}{ab}\)= 1-+\(\frac{2c^3}{abc}\)
Tương tụ A* \(\frac{a}{b-c}\)= 1+\(\frac{2a^3}{abc}\)
A*\(\frac{b}{c-a}\)= 1+ \(\frac{2b^3}{abc}\)
Vậy S = 3 +\(\frac{2\left(a^3+b^3+c^3\right)}{abc}\)= 9
ở phần a3 + b3 + c3 thì tổng đấy sẽ bằng 3abc , đoạn đấy mk làm tắt nhé, bạn tự thay vào hehe
Ta có: \(a+b+c=0\Rightarrow a^2=\left(b+c\right)^2\Rightarrow a^2-2bc=b^2+c^2\)
\(\Rightarrow a^2-b^2-c^2=a^2-a^2+2bc=2bc\)
Tương tự: \(b^2-c^2-a^2=2ca;c^2-a^2-b^2=2ab\)
\(A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}\)
Lại có: \(a+b+c=0\Rightarrow-a=b+c\)
\(\Rightarrow-a^3=b^3+c^3+3bc\left(b+c\right)\)
\(\Rightarrow a^3+b^3+c^3=-3bc\left(b+c\right)=3abc\left(b+c=-a\right)\)
=> \(A=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
đặt S=(a+1)(b+1)(c+1)
ta có:
\(\Leftrightarrow\frac{a+c}{a+1}-1=b-1\Leftrightarrow\frac{c-1}{a+1}=b-1\)
\(\frac{c+b}{c+1}=a\Leftrightarrow\frac{b-1}{c+1}=a-1\)
\(\frac{b+a}{b+1}=c\Leftrightarrow\frac{a-1}{b+1}=c-1\)
\(\Rightarrow\frac{\left(a-1\right)\left(b-1\right)\left(c-1\right)}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
=>(a+1)(b+1)(c+1)=1