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\(a^3+b^3+c^3=3abc\Leftrightarrow a^3+b^3+c^3-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-ac-bc+c^2-3ab\right]=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(a;b;c>0\Rightarrow a+b+c>0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
\(P=0\)
Có \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge\left(\sqrt{a}.\frac{1}{\sqrt{a}}+\sqrt{b}.\frac{1}{\sqrt{b}}+\sqrt{c}.\frac{1}{\sqrt{c}}\right)^2\)(BĐT Bunhiacopxki)
\(=\left(1+1+1\right)^2=9\)
Vậy \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)>1\)
Vậy bài toán ko giải đc; Nếu mk làm sai thì thứ lỗi vì mk năm nay mới lên lớp 8
1/x + 1/y = 1/2018
<=> 1/x = 1/2018 - 1/y = (y - 2018)/(2018y)
<=> x = 2018y/(y - 2018)
=> x + y = 2018y/(y - 2018) + y = y^2/(y - 2018)
=> x - 2018 = 2018y/(y - 2018) - 2018 = 2018^2/(y - 2018)
=> P = 1
\(\frac{a^4}{2018}+\frac{b^4}{2019}=\frac{1}{4037}\)
\(\Leftrightarrow\frac{2019a^4+2018b^4}{2018\cdot2019}=\frac{a^2+b^2}{2018+2019}\)
\(\Leftrightarrow\left(2018+2019\right)\left(2019a^4+2018b^4\right)=2018\cdot2019\left(a^2+b^2\right)\)
\(\Leftrightarrow2019^2\cdot a^4+2018^2\cdot b^4+2018\cdot2019\cdot a^4+2018\cdot2019b^4=2018\cdot2019\cdot a^2+2018\cdot2019\cdot b^2\)
\(\Leftrightarrow2019^2\cdot a^4+2018^2\cdot b^4=2018\cdot2019\cdot a^2\left(1-a^2\right)+2018\cdot2019\cdot b^2\left(1-b^2\right)\)
\(\Leftrightarrow\left(2019a^2\right)^2+\left(2018b^2\right)^2=2\cdot2018\cdot2019\cdot a^2\cdot b^2\)
\(\Leftrightarrow\left(2019a^2-2018b^2\right)=0\)
\(\Leftrightarrow2019a^2=2018b^2\Leftrightarrow\frac{a^2}{2018}=\frac{b^2}{2019}=\frac{a^2+b^2}{2018+2019}=\frac{1}{4037}\)
\(\Rightarrow\frac{a^{2018}}{2018^{10009}}=\frac{b^{2018}}{2019^{1009}}=\frac{1}{4037^{1009}}\)
\(\Rightarrow P=\frac{2}{4037^{1009}}\)