cho a,b,c thỏa a/2012=b/2013=c/2014
tính A= 4.(a-b).(b-c)-(c-a)^2
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\(a^{2012}+b^{2012}+c^{2012}\ge3\sqrt[3]{\left(abc\right)^{2012}}=3\)
\(\Rightarrow\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\le\dfrac{1}{3}\)
\(\Rightarrow-\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge-\dfrac{1}{3}\)
Lại có:
\(a^{2013}+a^{2013}+...+a^{2013}\left(\text{2012 số hạng}\right)+1\ge2013\sqrt[2013]{\left(a^{2013}\right)^{2012}}=2013.a^{2012}\)
\(\Rightarrow2012.a^{2013}+1\ge2013.a^{2012}\)
Tương tự: \(2012.b^{2013}+1\ge2013.b^{2012}\) ; \(2012.c^{2013}+1\ge2013.c^{2012}\)
Cộng vế với vế:
\(\Rightarrow a^{2013}+b^{2013}+c^{2013}\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012}\)
\(\Rightarrow A\ge\dfrac{2013\left(a^{2012}+b^{2012}+c^{2012}\right)-3}{2012\left(a^{2012}+b^{2012}+c^{2012}\right)}=\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{a^{2012}+b^{2012}+c^{2012}}\ge\dfrac{2013}{2012}-\dfrac{3}{2012}.\dfrac{1}{3}=1\)
\(A_{min}=1\) khi \(a=b=c=1\)
Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :
\(\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}=\frac{a-b}{2012-2013}=\frac{b-c}{2013-2014}=\frac{c-a}{2014-2012}\)
\(\Rightarrow\frac{a-b}{-1}=\frac{b-c}{-1}=\frac{c-a}{2}\)
\(\Rightarrow\left(\frac{a-b}{-1}\right)\left(\frac{b-c}{-1}\right)=\left(\frac{c-a}{2}\right)^2\)
hay \(\left(a-b\right)\left(b-c\right)=\frac{\left(c-a\right)^2}{4}\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(c-a\right)^2\)
Đặt \(\frac{a}{2012}=\frac{b}{2013}=\frac{c}{2014}=k\Rightarrow\hept{\begin{cases}a=2012k\\b=2013k\\c=2014k\end{cases}}\)
A = 4( a - b )( b - c ) - ( c - a )2
= 4( 2012k - 2013k )( 2013k - 2014k ) - ( 2014k - 2012k )2
= 4.( -k ).( -k ) - ( 2k )2
= 4k2 - 4k2 = 0
Ta có \(\frac{a}{2012}\)= \(\frac{b}{2013}\)= \(\frac{c}{2014}\)= b - a = c - b = \(\frac{c\:-a}{2}\)
Từ đó ta có A= 4(a-b)(b-c)-(c-a)2 = 4(-\(\frac{c\:-a}{2}\))(-\(\frac{c\:-a}{2}\)) - (c - a)2 = ) (c - a)2 - (c - a)2 = 0
Đặt: \(\dfrac{a}{2012}=\dfrac{b}{2013}=\dfrac{c}{2014}=k\)
\(\rightarrow a=2012k,b=2013k,c=2014k\)
Vế trái: \(4.\left(2012k-2013k\right)\left(2013k-2014k\right)=4.\left(-1k\right).\left(-1k\right)=4k^2\)
Vế phải: \(\left(2014k-2012k\right)^2=\left(2k\right)^2=4k^2\)
\(\rightarrow\) Vế trái = vế phải = \(4k^2\)
Ta có: \(a^2+b^2+c^2=1\)
\(\Rightarrow\left\{{}\begin{matrix}\left|a\right|\le1\\\left|b\right|\le1\\\left|c\right|\le1\end{matrix}\right.\)
Ta lại có:
\(a^3+b^3+c^3=a^2+b^2+c^2\)
\(\Leftrightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)=0\)
Vì \(\left\{{}\begin{matrix}1-a\ge0\\1-b\ge0\\1-c\ge0\end{matrix}\right.\)
\(\Rightarrow a^2\left(1-a\right)+b^2\left(1-b\right)+c^2\left(1-c\right)\ge0\)
Dấu = xảy ra khi: \(\left(a,b,c\right)=\left(1,0,0;0,1,0;0,0,1\right)\)
\(\Rightarrow S=1\)
chán ghê hk ai giúp hết