pip install pygame

Lời giải:

Với $a,b,c>0$ ta có:

$M> \frac{a}{a+b+c}+\frac{b}{b+c+a}+\frac{c}{c+a+b}=\frac{a+b+c}{a+b+c}{a+b+c}=1(*)$

Mặt khác:
Xét hiệu: $\frac{a}{a+b}-\frac{a+c}{a+b+c}=\frac{-bc}{(a+b)(a+b+c)}<0$ với mọi $a,b,c>0$

$\Rightarrow \frac{a}{a+b}< \frac{a+c}{a+b+c}$

Tương tự ta cũng có: $\frac{b}{b+c}< \frac{b+a}{a+b+c}; \frac{c}{c+a}< \frac{c+b}{a+b+c}$

Cộng lại ta được: $M< \frac{a+c+b+a+c+b}{a+b+c}=\frac{2(a+b+c)}{a+b+c}=2(**)$

Từ $(*); (**)\Rightarrow 1< M< 2$ nên $M$ không là số nguyên.

Lời giải:
\(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}=0\)

\(\Rightarrow \frac{a}{b-c}=-(\frac{b}{c-a}+\frac{c}{a-b})=-\frac{ba-b^2+c^2-ca}{(c-a)(a-b)}\)

\(\Rightarrow \frac{a}{(b-c)^2}=-\frac{ba-b^2+c^2-ca}{(a-b)(b-c)(c-a)}\)

Tương tự:

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

\(\frac{c}{(a-b)^2}=-\frac{ac-a^2+b^2-bc}{(a-b)(b-c)(c-a)}\)

Do đó:

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

Nếu $a,b,c$ đều âm, khi đó \(\frac{a}{(b-c)^2}< 0; \frac{b}{(c-a)^2}< 0; \frac{c}{(a-b)^2}< 0\)

\(\Rightarrow \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}< 0\) (mâu thuẫn)

Nếu $a,b,c$ đều dương, khi đó \(\frac{a}{(b-c)^2}> 0; \frac{b}{(c-a)^2}> 0; \frac{c}{(a-b)^2}> 0\)

\(\Rightarrow \frac{a}{(b-c)^2}+\frac{b}{(c-a)^2}+\frac{c}{(a-b)^2}>0\) (mâu thuẫn)

Trường hợp có từ 2 số trở lên bằng $0$ thì hoàn toàn vô lý.

Do đó, trong 3 số $a,b,c$ phải có một số âm và một số dương.

\(A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\\ \Rightarrow A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\left(1\right)\\ A< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow1< A< B\\ \Rightarrow A\notin Z\)

\(\dfrac{a}{2021-c}+\dfrac{b}{2021-a}+\dfrac{c}{2021-b}\\ =\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ =\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\)

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)

\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}+\dfrac{c+a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

Vì \(1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\Rightarrow A.ko.phải.số.nguyên\)

Có: \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{c}\Leftrightarrow 2ab-2bc-2ca=0\)

\(\Rightarrow A=\sqrt{a^2+b^2+c^2+2ab-2bc-2ca}=\sqrt{(a+b-c)^2}=|a+b-c|\)

⇒ A là số hữu tỉ

Lời giải:

Ta có:

\(M=\frac{a}{a+b+c}+\frac{b}{a+b+d}+\frac{c}{b+c+d}+\frac{d}{a+d+c}\)

\(> \frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}\)

\(\Leftrightarrow M>\frac{a+b+c+d}{a+b+c+d}=1(1)\)

Mặt khác:

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

\(\Leftrightarrow M=4-\underbrace{\left(\frac{b+c}{a+b+c}+\frac{a+d}{a+b+d}+\frac{b+d}{b+c+d}+\frac{a+c}{a+d+c}\right)}_{N}\)

Có: \(N>\frac{b+c}{a+b+c+d}+\frac{a+d}{a+b+c+d}+\frac{b+d}{a+b+c+d}+\frac{a+c}{a+b+c+d}\)

\(\Leftrightarrow N>\frac{2(a+b+c+d)}{a+b+c+d}=2\)

\(\Rightarrow M=4-N< 4-2\Leftrightarrow M< 2(2)\)

Từ \((1);(2)\Rightarrow 1< M< 2\Rightarrow M\not\in \mathbb{N}\)