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Do a < b < c < d < m < n
=> 2c < c + d
m< n => 2m < m+ n
=> 2c + 2a +2m = 2 ( a + c + m) < a +b + c + d + m + n)
Do đó :
\(\dfrac{\text{(a + c + m)}}{\left(a+b+c+d+m+n\right)}\) < \(\dfrac{1}{2}\)
\(A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\\ \Rightarrow A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\left(1\right)\\ A< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow1< A< B\\ \Rightarrow A\notin Z\)
\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\\ =\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ =\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Vì \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\Rightarrow A.ko.phải.số.nguyên\)
1)Từ đề bài:
`=>a^2+4b+4+b^2+4c+4+c^2+4a+4=0`
`<=>(a+2)^2+(b+2)^2+(c+2)^2=0`
`<=>a=b=c-2`
`ab+bc+ca=abc`
`<=>1/a+1/b+1/c=1`
`<=>(1/a+1/b+1/c)^2=1`
`<=>1/a^2+1/b^2+1/c^2+2/(ab)+2/(bc)+2/(ca)=1`
`<=>1/a^2+1/b^2+1/c^2=1-(2/(ab)+2/(bc)+2/(ca))`
`a+b+c=0`
Chia 2 vế cho `abc`
`=>1/(ab)+1/(bc)+1/(ca)=0`
`=>2/(ab)+2/(bc)+2/(ca)=0`
`=>1/a^2+1/b^2+1/c^2=1-0=1`
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{a+b+c}{b+c+a}=1\)
Khi đó:
\(\dfrac{a}{b}=1\Rightarrow a=b\left(1\right)\)
\(\dfrac{b}{c}=1\Rightarrow b=c\left(2\right)\)
\(\dfrac{c}{a}=1\Rightarrow c=a\left(3\right)\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow a=b=c\)