Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Rightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=36\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=36\)

 \(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}=12\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\)

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

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

=> \(\left(\frac{1}{a^2}-\frac{2}{ab}+\frac{1}{b^2}\right)+\left(\frac{1}{b^2}-\frac{2}{bc}+\frac{1}{c^2}\right)+\left(\frac{1}{c^2}-\frac{2}{ac}+\frac{1}{a^2}\right)=0\)

=> \(\left(\frac{1}{a}-\frac{1}{b}\right)^2+\left(\frac{1}{b}-\frac{1}{c}\right)^2+\left(\frac{1}{c}-\frac{1}{a}\right)^2=0\)

=> \(\hept{\begin{cases}\frac{1}{a}-\frac{1}{b}=0\\\frac{1}{b}-\frac{1}{c}=0\\\frac{1}{c}-\frac{1}{a}=0\end{cases}}\Rightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

Khi đó \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=6\Leftrightarrow3\frac{1}{a}=6\Rightarrow\frac{1}{a}=2\Leftrightarrow\frac{1}{a}=\frac{1}{b}=\frac{1}{c}=2\)

Khi đó  Đặt P = \(\left(\frac{1}{a}-3\right)^{2020}+\left(\frac{1}{b}-3\right)^{2020}+\left(\frac{1}{c}-3\right)^{2020}\)

= (2 - 3)²⁰²⁰ + (2 - 3)²⁰²⁰ + (2 - 3)²⁰²⁰

= 1 + 1 + 1 = 3

Vậy P = 3 

Câu hỏi của hanhungquan - Toán lớp 8 - Học toán với OnlineMath tương tự

Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2019}\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{2019}\Leftrightarrow2019\left(ab+bc+ca\right)=abc\)

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

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

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

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

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

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

\(\Rightarrow a+b=0\)hoặc \(b+c=0\)hoặc \(c+a=0\)

Mà \(a+b+c=2019\)

\(\Rightarrow a=2019\)hoặc \(b=2019\)hoặc \(c=2019\)

khó quá nha bn

mk mới chỉ hok lớp 7 thôi

xin lỡi nha

mk tin sẽ có nguoi tra lới cau hoi của bn

hok tot >_<

có vấn đề

bài này có người giải rồi

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:

\(\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\ge\left(a+b+c+d\right).\frac{16}{\left(a+b+c+d\right)}=16\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\ge4\)

Dấu = xảy ra khi \(a=b=c=d=1\)