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Từ a+b+c=6 \(\Rightarrow\)a+b=6-c
Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9
\(\Leftrightarrow\)ab=9-c(a+b)
Mà a+b=6-c (cmt)
\(\Rightarrow\)ab=9-c(6-c)
\(\Rightarrow\)ab=9-6c+c2
Ta có: (b-a)2\(\ge\)0 \(\forall\)b, c
\(\Rightarrow\)b2+a2-2ab\(\ge\)0
\(\Rightarrow\)(b+a)2-4ab\(\ge\)0
\(\Rightarrow\)(a+b)2\(\ge\)4ab
Mà a+b=6-c (cmt)
ab= 9-6c+c2 (cmt)
\(\Rightarrow\)(6-c)2\(\ge\)4(9-6c+c2)
\(\Rightarrow\)36+c2-12c\(\ge\)36-24c+4c2
\(\Rightarrow\)36+c2-12c-36+24c-4c2\(\ge\)0
\(\Rightarrow\)-3c2+12c\(\ge\)0
\(\Rightarrow\)3c2-12c\(\le\)0
\(\Rightarrow\)3c(c-4)\(\le\)0
\(\Rightarrow\)c(c-4)\(\le\)0
\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)
*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)
*
Hãy chứng minh \(a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)
Ta có: \(a+b+c=0\)
⇒\(\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
mà \(a^2+b^2+c^2=1\)
nên \(2ab+2ac+2bc=-1\)
\(\Leftrightarrow2\cdot\left(ab+ac+bc\right)=-1\)
\(\Leftrightarrow\left(ab+ac+bc\right)^2=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Ta có: \(a^2+b^2+c^2=1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4+\frac{1}{2}=1\)
hay \(a^4+b^4+c^4=1-\frac{1}{2}=\frac{1}{2}\)(đpcm)
Ta có: a+b+c=0
=> (a+b+c)2 = \(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
mà \(a^2+b^2+c^2=1\) => 1 + 2(ab + bc + ac) = 0
=> 2(ab + bc + ac) = -1 => ab + bc + ac = \(\frac{-1}{2}\)
=> (ab + bc + ac)2 = \(\left(\frac{-1}{2}\right)^2\)
=> a2b2 + b2c2 + a2c2 + 2(ab2c+abc2+a2bc) = \(\frac{1}{4}\)
=> a2b2 + b2c2 + a2c2 + 2abc(a+b+c) = \(\frac{1}{4}\)
mà a+b+c = 0 => a2b2 + b2c2 + a2c2 = \(\frac{1}{4}\)
Do a2 + b2 + c2 =1
=> (a2 + b2 + c2)2 = a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2)=1
=> a4 + b4 + c4 + 2.\(\frac{1}{4}\) = 1
=> a4 + b4 + c4 = 1 - 2.\(\frac{1}{4}\) =\(\frac{1}{2}\)
Bạn không hiểu chỗ nào thì hỏi lại mình nhé
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow1+2\left(ab+bc+ca\right)=0\Rightarrow ab+bc+ca=-\frac{1}{2}\Rightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\Rightarrow a^2b^2+b^2c^2+c^2a^2+2abc.0=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Xét: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Rightarrow a^4+b^4+c^4+2.\frac{1}{4}=1\Rightarrow a^4+b^4+c^4=\frac{1}{2}\)(đpcm)
Bài 2 :
Ta có :
\(\frac{a^2}{b^2+c^2}-\frac{a}{b+c}=\frac{a^2b-ab^2+a^2c-ac^2}{\left(b+c\right)\left(b^2+c^2\right)}=\frac{ab\left(a-b\right)+ac\left(a-c\right)}{\left(b+c\right)\left(b^2+c^2\right)}\)( 1 )
\(\frac{b^2}{c^2+a^2}-\frac{b}{c+a}=\frac{bc\left(b-c\right)+ab\left(b-a\right)}{\left(c+a\right)\left(c^2+a^2\right)}\)( 2 )
\(\frac{c^2}{a^2+b^2}-\frac{c}{a+b}=\frac{ac\left(c-a\right)+bc\left(c-c\right)}{\left(a+b\right)\left(a^2+b^2\right)}\) ( 3 )
Cộng ( 1 ) , ( 2 ) , ( 3 ) ta được :
\(\left(\frac{a^2}{b^2+c^2}+\frac{b^2}{c^2+a^2}+\frac{c^2}{a^2+b^2}\right)-\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\)
\(=ab\left(a-b\right)\left[\frac{1}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+c\right)\left(a^2+c^2\right)}\right]\)
\(+ac\left(a-c\right)\left[\frac{1}{\left(b+c\right)\left(b^2+c^2\right)}-\frac{1}{\left(a+b\right)\left(a^2+b62\right)}\right]\)
\(+bc\left(b-c\right)\left[\frac{1}{\left(a+c\right)\left(a^2+c^2\right)}-\frac{1}{\left(a+b\right)\left(a^2+b^2\right)}\right]\)
Theo đề bài thì \(a,b,c>0\)( các biểu thức trong các dấu ngoặc đều không âm ) \(\Leftrightarrow dpcm\)
Thấy đúng thì tk nka !111
Bài 3:
ta có : \(a^4+b^4\ge2a^2b^2\)
Cộng \(a^4+b^4\) vào 2 vế ta được:
\(2\left(a^4+b^4\right)\ge\left(a^2+b^2\right)^2\)\(\Leftrightarrow a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)^2\)
Ta cũng có : \(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2\)
\(\Leftrightarrow a^4+b^4\ge\frac{1}{8}\left(a+b\right)^4\)
mà theo bài thì \(a+b>1\)\(\Rightarrow dpcm\)
TK MK NKA !!!
Ta có: a+b+c=0
=> \(\left(a+b+c\right)^2=0\)
=> \(a^2+b^2+c^2+2ab+2bc+2ac=0\)
=> 2ab + 2bc + 2ac = -1 (do \(a^2+b^2+c^2=1\) )
=> \(\left(2ab+2bc+2ac\right)^2=\left(-1\right)^2\)
=> \(4a^2b^2+4b^2c^2+4a^2c^2+8ab^2c+8abc^2+8a^2bc=1\)
=>\(4a^2b^2+4b^2c^2+4a^2c^2+8abc\left(a+b+c\right)=1\)
=>\(2\left(2a^2b^2+2b^2c^2+2a^2c^2\right)=1\) (do a+b+c=0)
=>\(2a^2b^2+2b^2c^2+2a^2c^2=\frac{1}{2}\)
Lại có: \(a^2+b^2+c^2=1\)
=> \(\left(a^2+b^2+c^2\right)^2=1\) = 1
=> \(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1\)
=> \(a^4+b^4+c^4+\frac{1}{2}=1\)
=> \(a^4+b^4+c^4=\frac{1}{2}\)
=> ĐPCM
Ta có a+b+c=0=>\(\left(a+b+c\right)^2=0\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)(1)
Vì \(a^2+b^2+c^2=1\)
Thay vào (1) có ab+bc+ca=\(-\frac{1}{2}\)
Ta có\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=1-2\(\left[\left(ab+bc+ca\right)^2-2a^2bc-2ab^2c-2abc^2\right]\)
=1-2\(\left[\frac{1}{4}-2abc\left(a+b+c\right)\right]\)
=1-2\(\left(\frac{1}{4}-0\right)\)
=1-\(\frac{1}{2}\)=\(\frac{1}{2}\)(đpcm