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Ta có:
\(a\left(b+c\right)^2+b\left(c+a\right)^2+c\left(a+b\right)^2=4abc\)
\(\Leftrightarrow\left(ab+ac\right)\left(b+c\right)+b\left(c^2+2ac+a^2\right)+c\left(a^2+2ab+b^2\right)=4abc\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+bc^2+2abc+ba^2+ca^2+2abc+cb^2-4abc=0\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+\left(bc^2+cb^2\right)+\left(ba^2+ca^2\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac\right)+bc\left(b+c\right)+a^2\left(b+c\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left(ab+ac+bc+a^2\right)=0\)
\(\Leftrightarrow\left(b+c\right)\left[b\left(c+a\right)+a\left(a+c\right)\right]=0\)
\(\Leftrightarrow\left(b+c\right)\left(a+b\right)\left(c+a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}b+c=0\\a+b=0\\c+a=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}b=-c\\a=-b\\c=-a\end{matrix}\right.\)
Ta lại có:
\(a^{2013}+b^{2013}+c^{2013}=1\)
Với : \(b=-c\Leftrightarrow a^{2013}-c^{2013}+c^{2013}=1\Leftrightarrow a=1\)
\(\Rightarrow M=\dfrac{1}{a^{2015}}+\dfrac{1}{b^{2015}}+\dfrac{1}{c^{2015}}=\dfrac{1}{1}+\dfrac{-1}{c^{2015}}+\dfrac{1}{c^{2015}}=1\)
Mà do \(a,b,c\) bình đẳng nên với trường hợp nào đều là \(M=1\)
Đặt \(\dfrac{1}{5}+\dfrac{2013}{2014}+\dfrac{2015}{2016}=B;\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}=C\)
\(A=\left(B+1\right)\cdot C-B\cdot\left(C+1\right)\)
\(=BC+C-BC-B\)
=C-B
\(=\dfrac{2013}{2014}+\dfrac{2015}{2016}+\dfrac{1}{10}-\dfrac{1}{5}-\dfrac{2013}{2014}-\dfrac{2015}{2016}=-\dfrac{1}{10}\)
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
ĐKXĐ: \(a,b,c\ne0\)
\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=2013.\dfrac{1}{2013}\)
\(\Leftrightarrow1+1+1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}=1\)
\(\Leftrightarrow\dfrac{a^2c+a^2b+b^2c+ab^2+bc^2+ac^2+2abc}{abc}=0\)
\(\Leftrightarrow a^2c+a^2b+b^2c+ab^2+bc^2+ac^2+2abc=0\)
\(\Leftrightarrow ac\left(a+b\right)+ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)
\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-b\\b=-c\\c=-a\end{matrix}\right.\)
Mà \(a+b+c=2013\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2013\\b=2013\\c=2013\end{matrix}\right.\)(đpcm)
a/ \(A=\dfrac{2012}{\left|x\right|+2013}\)
vì: \(\left|x\right|\ge0\Rightarrow\left|x\right|+2013\ge2013\)
=> \(\dfrac{2012}{\left|x\right|+2013}\le\dfrac{2012}{2013}\)
Dấu ''='' xảy ra khi x = 0
Vậy MAXA = 2012/2013 khi x = 0
b/ \(B=\dfrac{\left|x\right|+2012}{-2013}\)
Vì: \(\left|x\right|\ge0\Rightarrow\left|x\right|+2012\ge2012\)
=> \(\Rightarrow\dfrac{\left|x\right|+2012}{-2013}\le-\dfrac{2012}{2013}\)
Dấu ''='' xảy ra khi x = 0
Vậy.........
Bài 2: Ăn cơm xoq lm cho
Bài 2:
a, Để C nhỏ nhất thì /x/+2012 phải nhỏ nhất
Mà /x/ luôn lớn hơn hoặc bằng 0 => /x/+2012 nhỏ nhất khi /x/ =0
=> x+0, GTNN của C=\(\dfrac{0+2012}{2013}=\dfrac{2012}{2013}\)khi x=0
Bài 2.1
a: \(\dfrac{2012}{\left|x\right|+2013}\le\dfrac{2012}{2013}\)
Dấu '=' xảy ra khi x=0
b: \(\dfrac{\left|x\right|+2012}{-2013}\le-\dfrac{2012}{2013}\)
Dấu '=' xảy ra khi x=0
ta có :\(\dfrac{a}{b}=\dfrac{c}{d}=>\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(=>\dfrac{a^{2013}}{c^{2013}}=\dfrac{b^{2013}}{d^{2013}}=\dfrac{\left(a+b\right)^{2013}}{\left(c+b\right)^{2013}}\left(1\right)\)
Mặt khác:\(\dfrac{a}{c}=\dfrac{b}{d}=>\dfrac{a^{2013}}{c^{2013}}=\dfrac{b^{2013}}{d^{2013}}=\dfrac{2.a^{2013}}{2.c^{2013}}=\dfrac{3.b^{2013}}{3.d^{2013}}=\dfrac{2.a^{2013}-3.b^{2013}}{2.c^{2013}-3.d^{2013}}\left(2\right)\)Từ (1),(2)=>\(\dfrac{\left(a+b\right)^{2013}}{\left(c+d\right)^{2013}}=\dfrac{2.a^{2013}-3.b^{2013}}{2.c^{2013}-3.d^{2013}}\left(đpcm\right)\)
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