Ta có \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{a-b}{c-d}\)

=> \(\left(\frac{a}{c}\right)^{2021}=\left(\frac{b}{d}\right)^{2021}=\left(\frac{a-b}{c-d}\right)^{2021}\)

=> \(\frac{a^{2021}}{c^{2021}}=\frac{b^{2021}}{d^{2021}}=\left(\frac{a-b}{c-d}\right)^{2021}=\frac{a^{2021}+b^{2021}}{c^{2021}+d^{2021}}\)

=>\(\left(\frac{a-b}{c-d}\right)^{2021}=\frac{a^{2021}+b^{2021}}{c^{2021}+d^{2021}}\)(đpcm)

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)

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

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

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

=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0

=> Hoặc a=-b hoặc b=-c hoặc c=-a

Ko mất tổng quát, g/s a=-b

a) Ta có: vì a=-b thay vào ta được:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)

=> đpcm

b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)

=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)

Ta có a + b + c = 6

=> (a + b + c)² = 36

=> a² + b² + c² + 2ab + 2bc + 2ca = 36

=> 12 + 2ab + 2bc + 2ca = 36

=> 2ab + 2bc + 2ca = 24

=> ab + bc + ca = 12 

Khi đó a² + b² + c² = ab + bc + ca (= 12)

<=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca 

<=>  2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0 

<=> (a² - 2ab + b²) + (b² - 2bc + c²) + (c² - 2ca + a²) = 0

<=> (a - b)² + (b - c)² + (c - a)² = 0

<=> \(\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}}\Leftrightarrow a=b=c\)

=> a = b = c = 2 

Khi đó A = (2 - 3)²⁰²¹ + (2 - 3)²⁰²¹ + (2 - 3)²⁰²¹

= -1 + (-1) + (-1) 

= -3

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

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

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

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

camon camon 

ơ đang chờ mấy bạn top bxh vô trả lời mà hỏng thấy đou

hộ mình với:(

= mìnk ko biết

sorry

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow a^2b+ab^2+c^2a+ca^2+b^2c+bc^2+2abc=0\)

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

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

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

=> Hoặc a+b=0 hoặc b+c=0 hoặc c+a=0

=> Hoặc a=-b hoặc b=-c hoặc c=-a

Ko mất tổng quát, g/s a=-b

a) Ta có: vì a=-b thay vào ta được:

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{1}{b^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{1}{c^3}\)

\(\frac{1}{a^3+b^3+c^3}=\frac{1}{-b^3+b^3+c^3}=\frac{1}{c^3}\)

=> đpcm

b) Ta có: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)

=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)

B

B