tc \(0\le a;b;c\le1\)

\(a^3+b^3+c^3+a+b+c=2a^2+2b^2+2c^2=2\)

\(a^3-2a^2+a+b^3-2b^2+b+c^3-2c^2+c=0\)

\(a\left(a-1\right)^2+b\left(b-1\right)^2+c\left(c-1\right)^2=0\)

\(\hept{\begin{cases}a\left(a-1\right)^2=0\\b\left(b-1\right)^2=0\\c\left(c-1\right)^2=0\end{cases}}\)

đến đây lập luận ok

bình phương (1/a+1/b+1/c) rồi áp dụng HĐT tính bình thường

Sử dụng: 

\(A^3+B^3+C^3-3ABC=\left(A+B+C\right)\left(A^2+B^2+C^2-AB-BC-AC\right)\) (1)

Áp dụng vào bài:

\(\left(a-1\right)^3+\left(b-2\right)^3+\left(c-3\right)^3-3\left(a-1\right)\left(b-2\right)\left(c-3\right)\)

\(=\left(a-1+b-2+c-3\right)\)[ \(\left(a-1\right)^2+\left(b-2\right)^2+\left(c-3\right)^2\)

\(+\left(a-1\right)\left(b-2\right)+\left(a-1\right)\left(c-3\right)+\left(b-2\right)\left(c-3\right)\)]

<=> \(0-3\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

( vì \(a-1+b-2+c-3=a+b+c-6=6-6=0\))

<=> \(\left(a-1\right)\left(b-2\right)\left(c-3\right)=0\)

<=>  a = 1 hoặc b = 2 hoặc c = 3.

Không mất tính tổng quát: g/s : a = 1

Khi đó: b + c =5

Ta có:  \(T=\left(b-2\right)^{2n+1}+\left(c-3\right)^{2n+1}\)

\(=\left(b-2+c-3\right).A\)

\(=\left(b+c-5\right).A\)

\(=0.A=0\)

Với \(A=\left(b-2\right)^{2n}-\left(b-2\right)^{2n-1}\left(c-3\right)+\left(b-2\right)^{2n-2}\left(c-3\right)^2-...+\left(c-3\right)^{2n}\)

Tương tự b = 2; c= 3 thì T = 0.

Vậy T = 0.

\(B=\frac{2019}{1}+\frac{2018}{2}+\frac{2017}{3}+......+\frac{1}{2019}\)

\(=\left(\frac{2018}{2}+1\right)+\left(\frac{2017}{3}+1\right)+.....+\left(\frac{1}{2019}+1\right)+1\)

\(=\frac{2020}{2}+\frac{2020}{3}+\frac{2020}{4}+.....+\frac{2020}{2019}+\frac{2020}{2020}\)

\(=2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2020}\right)\)

\(=2020A\)

\(\Rightarrow\frac{A}{B}=\frac{A}{2020A}=\frac{1}{2020}\)

Ta có a + b = 3

=> (a + b)² = 9

=> a² + 2ab + b² = 9

=> a² + b² = 5 (ab = 2)

Khi a² + b² = 5 => a² - 2ab + b² = 1

=> (a - b)² = 1

=> a - b = \(\pm1\)

Đặt A \(\frac{1}{a^3}-\frac{1}{b^3}=\frac{b^3-a^3}{\left(a.b\right)^3}=\frac{\left(b-a\right)\left(b^2+ab+a^2\right)}{\left(ab\right)^3}=-\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(ab\right)^3}\)

Với  a - b = 1 ; ab = 2 ; a² + b² = 5 ta có A = \(-\frac{1.\left(5+2\right)}{2^3}=-\frac{7}{8}\)

Với a - b = - 1 ; ab = 2 ; a² + b² = 5 ta có A = \(-\frac{\left(-1\right).\left(5+2\right)}{2^3}=\frac{7}{8}\)

Ta có: \(\hept{\begin{cases}a+b=3\\ab=2\end{cases}}\Leftrightarrow\hept{\begin{cases}\left(a+b\right)^2=9\\ab=2\end{cases}\Leftrightarrow}\hept{\begin{cases}a^2+2ab+b^2=9\\ab=2\end{cases}}\Leftrightarrow\hept{\begin{cases}a^2+b^2=5\\ab=2\end{cases}}\)

Khi đó: \(\frac{1}{a^3}-\frac{1}{b^3}=\frac{b^3-a^3}{a^3b^3}=\frac{\left(b-a\right)\left(a^2+ab+b^2\right)}{8}=\frac{7\left(b-a\right)}{8}\)

Ta có: \(a+b=3\Rightarrow a=3-b\) thay vào: \(\left(3-b\right)b=2\)

\(\Leftrightarrow b^2-3b+2=0\Leftrightarrow\left(b-1\right)\left(b-2\right)=0\Leftrightarrow\orbr{\begin{cases}b=1\Rightarrow a=2\\b=2\Rightarrow a=1\end{cases}}\)

Nếu \(\hept{\begin{cases}a=2\\b=1\end{cases}\Rightarrow}\frac{1}{a^3}-\frac{1}{b^3}=-\frac{7}{8}\)

Nếu \(\hept{\begin{cases}a=1\\b=2\end{cases}}\Rightarrow\frac{1}{a^3}-\frac{1}{b^3}=\frac{7}{8}\)