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https://olm.vn/hoi-dap/detail/907307687798.html
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Ta có : \(a+b+c=3\Rightarrow\left(a+b+c\right)^2=9\)
\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)
\(\Rightarrow a^2+b^2+c^2=9-2\left(ab+bc+ca\right)=9-2\times6=3\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\)
\(\Rightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow a=b=c\)
Mà \(a+b+c=3\Rightarrow a=b=c=1\)
\(\Rightarrow A=\left(1-1\right)^{2019}+\left(1^2-1\right)^{2020}+\left(1^3-1\right)^{2021}\)
\(=0^{2019}+0^{2020}+0^{2021}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
oh no bài thứ nhất là dạng chứng minh cs đúng ko ,
ko thể nào là dạng tìm a,b,c đc-.-
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(a+b+c=1\Leftrightarrow-b+b+c=1\Rightarrow c=1\)
=> \(P=-\frac{1}{b^{2021}}+\frac{1}{b^{2021}}+\frac{1}{c^{2021}}=\frac{1}{1^{2021}}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=\frac{2a+3b+3c-1}{2015+a}+\frac{3a+2b+3c}{2016+b}+\frac{3a+3b+2c+1}{2017+c}\)
\(=\frac{6047-a}{2015+a}+\frac{6048-b}{2016+b}+\frac{6049-c}{2017+c}\)
\(=\frac{8062}{2015+a}+\frac{8064}{2016+b}+\frac{8066}{2017+c}-3\)
\(\ge\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{2015+2016+2017+a+b+c}-3=\frac{\left(\sqrt{8062}+\sqrt{8064}+\sqrt{8066}\right)^2}{8064}-3\)
Dấu = xảy ra khi ....
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
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:
\(\left(3a-2b+c\right)^2=9a^2+4b^2+c^2+2\left(3ac-6ab-2bc\right)\)
\(\Rightarrow b^2=9a^2+4b^2+c^2\)
(vì \(3a-3b+c=0\Leftrightarrow3a-2b+c=-b\), \(6ab+2bc-3ac=0\))
\(\Leftrightarrow9a^2+3b^2+c^2=0\)
\(\Leftrightarrow a=b=c=0\).
Khi đó: \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)
Ta có:
(3a−2b+c)2=9a2+4b2+c2+2(3ac−6ab−2bc)
⇒b2=9a2+4b2+c2
(vì 3a−3b+c=0⇔3a−2b+c=−b, 6ab+2bc−3ac=0)
⇔9a2+3b2+c2=0
⇔a=b=c=0.
Khi đó: P=(−1)2019+(−1)2020+(−1)2021=−1