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\(\left(a+b+c\right)^2=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=\frac{a^{2020}+1}{a^{2020}+a^{2020}+a^{2020}+3}=\frac{a^{2020}+1}{3\left(a^{2020}+1\right)}=\frac{1}{3}\)
=>a^2+b^2+c^2+3-2a-2b-2c=0
=>(a^2-2a+1)+(b^2-2b+1)+(c62-2c+1)=0
=>(3 hằng dẳng thức của a-1 b-1 c-1)
Suy ra (a-1)^2=0
và (b-1)^2=0
và(c-1)^2=0
thay vào A suy ra A=0
cố gắng trình bày lại nhé bạn!
\(2a^{2020}+2b^{2020}+2c^{2020}-2\left(ab\right)^{1010}-2\left(bc\right)^{1010}-2\left(ca\right)^{1010}=0\)
\(\Leftrightarrow\left(a^{1010}-b^{1010}\right)^2+\left(b^{1010}-c^{1010}\right)^2+\left(c^{1010}-a^{1010}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a^{1010}-b^{1010}=0\\b^{1010}-c^{1010}=0\\c^{1010}-a^{1010}=0\end{matrix}\right.\)
\(\Rightarrow\left|a\right|=\left|b\right|=\left|c\right|\)
Nếu đề không cho a;b;c dương thì không tính được cụ thể giá trị A
Nếu a;b;c dương thì \(a=b=c\Rightarrow A=0\)
Ta có:\(a^{2000}+b^{2000}=a^{2001}+b^{2001}=a^{2002}+b^{2002}\)
\(\Rightarrow a^{2000}+b^{2000}+a^{2002}+b^{2002}=2\left(a^{2001}+b^{2001}\right)\)
\(\Rightarrow a^{2002}-a^{2001}-a^{2001}+a^{2000}+b^{2002}-b^{2001}-b^{2001}+b^{2000}=0\)
\(\Rightarrow a^{2001}\left(a-1\right)-a^{2000}\left(a-1\right)+b^{2001}\left(b-1\right)-b^{2000}\left(b-1\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a^{2001}-a^{2000}\right)+\left(b-1\right)\left(b^{2001}-b^{2000}\right)=0\)
\(\Rightarrow a^{2000}\left(a-1\right)^2+b^{2000}\left(b-1\right)^2=0\)
Dấu"="xảy ra khi \(\orbr{\begin{cases}a^{2000}\left(a-1\right)^2=0\\b^{2000}\left(b-1\right)^2=0\end{cases}}\)Mà \(a,b>0\)
\(\Rightarrow a=b=1\)
Do đó:\(a^{2020}+b^{2020}=1^{2020}+1^{2020}=1+1=2\)
Ta có (a + b + c)(ab + bc + ca) = 2020
<=> \(\left(a+b+c\right)abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2020\)
<=> \(\left(a+b+c\right).2020.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=2020\)
<=> \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{a+b+c}=0\)
<=> \(\frac{a+b}{ab}+\frac{a+b}{c\left(a+b+c\right)}=0\)
<=> \(\left(a+b\right)\left(\frac{1}{ab}+\frac{1}{c\left(a+b+c\right)}\right)=0\)
=> (a + b)(a + c)(b + c) = 0
=> a = -b hoặc a = -c ; b = -c (1)
Khi đó P = (b2c + 2020)(c2a + 2020)(a2b + 2020)
= (b2c + abc)(c2a + abc)(a2b + abc)
= bc(b + a)ac(b + c)ab(a + c)
= (abc)2.0 = 0 (Theo (1))