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Ta có: \(2020+c^2=ab+bc+ca+c^2=\left(b+c\right)\left(c+a\right)\)

Tương tự => \(2020+a^2=\left(a+b\right)\left(c+a\right)\)

\(2020+b^2=\left(a+b\right)\left(b+c\right)\)

=> PT = \(\frac{a-b}{\left(b+c\right)\left(c+a\right)}+\frac{b-c}{\left(a+b\right)\left(c+a\right)}+\frac{c-a}{\left(a+b\right)\left(b+c\right)}\)

= \(\frac{\left(a-b\right)\left(a+b\right)+\left(b-c\right)\left(b+c\right)+\left(c-a\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) = \(\frac{a^2-b^2+b^2-c^2+c^2-a^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) = 0

2 tháng 5 2020

Cmr biểu thức đó bằng 0

NV
18 tháng 8 2020

\(\left(a+b+c\right)^2=3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)

\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow a=b=c\)

\(\Rightarrow P=\frac{a^{2020}+1}{a^{2020}+a^{2020}+a^{2020}+3}=\frac{a^{2020}+1}{3\left(a^{2020}+1\right)}=\frac{1}{3}\)

26 tháng 10 2019

\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2+\frac{1}{a^2}}=2\\ \)(do Bđt cosi)=> \(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge6\\ \)

Dấu "=" xảy ra <=> a=b=c=1

=>B=3

26 tháng 10 2019

Bất đẳng thức cosi mình chưa học

25 tháng 10 2019

\(a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge2\sqrt{\frac{a^2}{a^2}}+2\sqrt{\frac{b^2}{b^2}}+2\sqrt{\frac{c^2}{c^2}}=6\)

Dấu = xảy ra khi a^4=b^4=c^4=1 <=> \(a=\pm1;b=\pm1;c\pm1\)

-> B = 3

22 tháng 9 2019

\(a+b+c=2020\Rightarrow\frac{1}{a+b+c}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(a+b+c\right)-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a\left(ab+ac\right)+abc-abc=0\)

\(\Leftrightarrow\left(ab+bc+ac\right)\left(b+c\right)+a^2\left(b+c\right)=0\)

\(\Leftrightarrow\left(ab+bc+ac+a^2\right)\left(b+c\right)=0\)

\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

Nếu a + b = 0 thì c = 2020

Nếu b + c = 0 thì a = 2020

Nếu a + c = 0 thì b = 2020

22 tháng 9 2019

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2020}\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Rightarrow\frac{bc+ac+ab}{abc}=\frac{1}{a+b+c}\)

\(\Rightarrow\left(a+b+c\right)\left(ab+ac+bc\right)=abc\)

\(\Rightarrow a^2b+a^2c+abc+ab^2+abc+b^2c+abc+ac^2+bc^2=abc\)

\(\Rightarrow...\)

\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

\(TH1:a=-b\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a}-\frac{1}{a}+\frac{1}{c}=\frac{1}{c}\)

Mà \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2020}\Rightarrow\frac{1}{c}=\frac{1}{2020}\Leftrightarrow c=2020\)

Các trường hợp kia tương tự

16 tháng 2 2020

thay 2020 = abc vào biểu thức A ta được :

\(A=\frac{2020a}{ab+2020a+2020}+\frac{b}{bc+b+2020}+\frac{c}{ac+c+1}\)

\(\Rightarrow A=\frac{abc.a}{ab+abc.a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(\Rightarrow A=\frac{abc.a}{ab\left(1+ac+c\right)}+\frac{b}{b\left(c+1+ac\right)}+\frac{c}{ac+c+1}\)

\(\Rightarrow A=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)

\(\Rightarrow A=\frac{ac+1+c}{ac+c+1}=1\)

VẬy A=1

21 tháng 4 2021

\(a^2+b^2+c^2-ab-bc-ac=0\)

\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\) (1)

Mà: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\) 

Nên PT (1) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)

=> a = b = c

\(P=\left(a-b\right)^{2020}+\left(b-c\right)^{2021}+\left(c-a\right)^{2022}\)

\(=\left(a-a\right)^{2020}+\left(b-b\right)^{2021}+\left(c-c\right)^{2022}\)

= 0

 

25 tháng 8 2020

Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)

 \(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

\(\Rightarrow\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}=\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}\)

=> \(\frac{2}{a^2}+\frac{2}{b^2}+\frac{2}{c^2}-\frac{2}{ab}-\frac{2}{bc}-\frac{2}{ca}=0\)

=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)

=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)

=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)

Khi đó  Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)

= (2 - 3)2020 + (2 - 3)2020 + (2 - 3)2020

= 1 + 1 + 1 = 3

Vậy P = 3