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\(a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=9-2\cdot4=1\)
Do đó \(K=1+2021=2022\)
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\) ; \(\forall a;b;c\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Rightarrow ab+bc+ca\le1\)
\(\Rightarrow P_{max}=1\) khi \(a=b=c\)
Lại có:
\(\left(a+b+c\right)^2\ge0\) ; \(\forall a;b;c\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow ab+bc+ca\ge-\dfrac{a^2+b^2+c^2}{2}=-\dfrac{1}{2}\)
\(P_{min}=-\dfrac{1}{2}\) khi \(a+b+c=0\)
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 \(a+b+c=3\Rightarrow a=b=c=1\)
\(\Rightarrow A=\left(1-1\right)^{2019}+\left(1^2-1\right)^{2020}+\left(1^3-1\right)^{2021}\)
\(=0^{2019}+0^{2020}+0^{2021}=0\)
\(a+b\ge a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\)
\(\Rightarrow2\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le1\)
Xét \(Q=\dfrac{a}{a+1}+\dfrac{b}{b+1}=\dfrac{a\left(b+1\right)+b\left(a+1\right)}{\left(a+1\right)\left(b+1\right)}=\dfrac{a+b+2ab}{\left(a+1\right)\left(b+1\right)}\)
\(Q=\dfrac{a+b+ab+ab}{\left(a+1\right)\left(b+1\right)}\le\dfrac{a+b+ab+1}{\left(a+1\right)\left(b+1\right)}=\dfrac{\left(a+1\right)\left(b+1\right)}{\left(a+1\right)\left(b+1\right)}=1\)
\(\Rightarrow P\le2020+1^{2021}=2021\)
Dấu "=" xảy ra khi \(a=b=1\)
Ta có
D = a ( b 2 + c 2 ) – b ( c 2 + a 2 ) + c ( a 2 + b 2 ) – 2 a b c = a b 2 + a c 2 – b c 2 – b a 2 + c a 2 + c b 2 – 2 a b c = ( a b 2 – a 2 b ) + ( a c 2 – b c 2 ) + ( a 2 c – 2 a b c + b 2 c ) = a b ( b – a ) + c 2 ( a – b ) + c ( a 2 – 2 a b + b 2 ) = - a b ( a – b ) + c 2 ( a – b ) + c ( a – b ) 2 = ( a – b ) ( - a b + c 2 + c ( a – b ) ) = ( a – b ) ( - a b + c 2 + a c – b c ) = ( a – b ) [ ( - a b + a c ) + ( c 2 – b c ) ]
= (a – b)[a(c – b) + c(c – b)]
= (a – b)(a + c)(c – b)
Với a = 99; b = -9; c = 1, ta có
D = (99 - (-9))(99 + 1) (1 - (-9)) = 108.100.10 = 108000
Đáp án cần chọn là: B
mới ăn miếng cơm cà ngon nhức nách luôn ai thèm cơm cà không điểm danh nào
Ta có
\(a+b+c=6\)
\(\Leftrightarrow\left(a+b+c\right)^2=36\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ca=36\)
Mà \(a^2+b^2+c^2=ab+bc+ca\)
Khi đó ta có
\(3\left(ab+bc+ca\right)=36\)
\(\Leftrightarrow ab+bc+ca=12\)
\(\Leftrightarrow\hept{\begin{cases}2ab+2bc+2ca=24\\2a^2+2b^2+2c^2=24\end{cases}}\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}}\Leftrightarrow a=b=c=\frac{6}{3}=2\) ( 1 )
Thay (1) vào C ta có
\(C=\left(1-2\right)^{2021}+\left(2-1\right)^{2021}+\left(2-2\right)^{2021}\)
\(=-1+1+0=0\)
Vậy ......................