Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
<=> \(2a^2+2b^2+2c^2=2ab+2bc+2ca< =>\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0< =>\)
a=b=c => 32020 = 3.a2019 <=> 32019 = a2019 => a=b=c=3
A= 12017 + 02018 + (-1)2019 = 0
Ta có : a + b + c = 6
=> ( a + b + c ) ^ 2 = 6 ^ 2 = 36
=> a ^ 2 + b ^ 2 + c ^ 2 + 2 x ( ab + bc + ca ) = 36
=> 12 + 2 x ( ab + bc + ca ) = 36 ( vì a ^ 2 + b ^ 2 + c ^ 2 = 12 )
=> 2 x ( ab + bc + ca ) = 36 - 12
=> 2 x ( ab + bc + ca ) = 24
=> ab + bc + ca = 12
Do đó ab + bc + ca = a ^ 2 + b ^ 2 + c ^ 2
=> a = b = c = 2 ( vì a + b + c = 6 )
Khi đó : P = ( 2 - 3 ) ^ 2020 + ( 2 - 3 ) ^ 2020 + ( 2 - 3 ) ^ 2020
=> P = ( - 1 ) ^ 2020 + ( - 1 ) ^ 2020 + ( - 1 ) ^ 2020
=> P = 1 + 1 + 1 = 3
Vậy P = 3
Cách 2:
Ta có: \(a^2+b^2+c^2=12\)
\(\Rightarrow a^2+b^2+c^2-12=0\)
\(\Rightarrow a^2+b^2+c^2-24+12=0\)
\(\Rightarrow a^2+b^2+c^2-4\left(a+b+c\right)+12=0\)(Vì a+b+c=6)
\(\Rightarrow\left(a^2-4a+4\right)+\left(b^2-4b+4\right)+\left(c^2-4c+4\right)=0\)
\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}}\Rightarrow a=b=c=2\)
Thay a=b=c=2 vào P, ta có:
\(P=\left(2-3\right)^{2020}+\left(2-3\right)^{2020}+\left(2-3\right)^{2020}\)
\(=1+1+1=3\)
P/s: Bài bạn nguyễn tuấn thảo , chỗ để suy ra a=b=c=2 lm tắt quá nhé :))
làm cái đề ra ấy, ngại viết lại đề :P
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)
\(\Rightarrow M=1^{2018}+1^{2019}+1^{2020}=1+1+1=3\)
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
Mình chỉ biết đến đây thôi:
\(\Leftrightarrow\left(b-c\right)\left(a^3-b^3\right)+\left(a-b\right)\left(c^3-b^3\right)=2020^{2019}\)
\(\Leftrightarrow\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(c-b\right)\left(c^2+bc+b^2\right)=2020^{2019}\)
\(\Leftrightarrow\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-c^2-bc-b^2\right)=2020^{2019}\)
\(\Leftrightarrow\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(a+b+c\right)=2020^{2019}\)
Ta có: \(a+b+c=3\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=9\)
\(\Rightarrow2\left(ab+bc+ca\right)=9-\left(a^2+b^2+c^2\right)=6\Rightarrow ab+bc+ca=3\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ca\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\)
Mà a + b + c = 3 nên a = b = c = 1
Suy ra \(P=\left(-1\right)^{2019}+\left(-1\right)^{2020}+\left(-1\right)^{2021}=-1\)