\(1-\dfrac{1}{1+a}\ge\dfrac{2017}{b+2017}+\dfrac{2018}{c+2018}\ge2\sqrt{\dfrac{2017.2018}{\left(b+2017\right)\left(c+2018\right)}}\)

\(1-\dfrac{2017}{b+2017}\ge\dfrac{1}{1+a}+\dfrac{2018}{b+2018}\ge2\sqrt{\dfrac{2018}{\left(1+a\right)\left(b+2018\right)}}\)

\(1-\dfrac{2018}{c+2018}\ge\dfrac{1}{1+a}+\dfrac{2017}{b+2017}\ge2\sqrt{\dfrac{2017}{\left(1+a\right)\left(b+2017\right)}}\)

Nhân vế:

\(\dfrac{abc}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\ge\dfrac{8.2017.2018}{\left(a+1\right)\left(b+2017\right)\left(c+2018\right)}\)

\(\Rightarrow abc\ge8.2017.2018\)

pip install pygame

Bạn vào đây tham khảo sau đó áp dụng vào bài của bạn nhé: Câu hỏi của Võ Khánh Lê - Toán lớp 0 | Học trực tuyến

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{2018}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\left(a+b+c=2018\right)\)

\(\Leftrightarrow\dfrac{a+b}{ab}+\dfrac{a+b+c-c}{c\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right]\left(a+b\right)=0\)

\(\Leftrightarrow\dfrac{ac+bc+c^2+ab}{abc\left(a+b+c\right)}\times\left(a+b\right)=0\)

\(\Leftrightarrow\dfrac{\left(a+c\right)\left(b+c\right)\left(a+b\right)}{abc\left(a+b+c\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=-c\\b=-c\\a=-b\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}b=2018\\a=2018\\c=2018\end{matrix}\right.\)

\(\Rightarrow P=\dfrac{1}{2018^{2017}}\)

Ta có:

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

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

Tới đây thì đơn giản rồi nhé.

máu biếng tới tận não:

\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)

\(\left[\left(a+b\right)^3+c^2\right]-ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a-b=b-c=c-a\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)

Mà a,b,c >0

=> a = b = c

=> S = 3

\(\)

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

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a+c+a-c}{b+d+b-d}=\dfrac{2a}{2b}=\dfrac{a}{b}\left(1\right)\)

\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a+c-a+c}{b+d-b+d}=\dfrac{2c}{2d}=\dfrac{c}{d}\left(1\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) ta có:

\(\dfrac{a}{b}=\dfrac{c}{d}\)

Đặt:

\(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Thay vào tính