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Ta có: \(2020+c^2=ab+bc+ca+c^2=\left(b+c\right)\left(c+a\right)\)
Tương tự => \(2020+a^2=\left(a+b\right)\left(c+a\right)\)
và \(2020+b^2=\left(a+b\right)\left(b+c\right)\)
=> 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
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
\(ad=bc\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a+b}{c+d}.\)
=> \(\frac{a^{2020}}{c^{2020}}=\frac{b^{2020}}{d^{2020}}=\frac{\left(a+b\right)^{2020}}{\left(b+d\right)^{2020}}\)
Xong lại áp dụng tính chất dãy tỉ số = nhau \(\frac{a^{2020}}{c^{2020}}=\frac{b^{2020}}{d^{2020}}=\frac{a^{2020}-b^{2020}}{c^{2020}-d^{2020}}.\)
Kết hợp lại là ra nhé
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
a)
\(A=\frac{2020^3+1}{2020-2019}=\frac{\left(2020+1\right)\left(2020^2-2020+1\right)}{2020-2020+1}\) \(=2020+1=2021\)
b)
B = \(\frac{2020^3-1}{2020^2+2021}=\frac{\left(2020-1\right)\left(2020^2+2020+1\right)}{2020^2+2020+1}\) \(=2020-1=2019\)
\(a^2+\frac{1}{a^2}\ge2\sqrt{a^2+\frac{1}{a^2}}=2\\ \)(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
\(\left(a+b+c\right)^2=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=3ab+3bc+3ca\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow P=\frac{a^{2020}+1}{a^{2020}+a^{2020}+a^{2020}+3}=\frac{a^{2020}+1}{3\left(a^{2020}+1\right)}=\frac{1}{3}\)