cho a+b+c=0 va a^2+b^2+c^2=2 tinh a^4+b^4+c^4
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a+b+c = 0 <=> (a+b+c)^2 = 0
<=> 2(ab+bc+ca) = 0 - (a^2+b^2+c^2) = 0 - 1 = -1
<=> ab+bc+ca = -1/2
<=> (ab+bc+ca)^2 = 1/4
<=> a^2b^2+b^2c^2+c^2a^2 = 1/4 - 2abc.(a+b+c) = 1/4 - 0 = 1/4
Có : a^2+b^2+c^2 = 1
<=> (a^2+b^2+c^2) = 1
<=> A = a^4+b^4+c^4 = 1 - 2.(a^2b^2+b^2c^2+c^2a^2) = 1 - 2.1/4 = 1/2
Vậy A = 1/2
k mk nha
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\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)=2+2\left(ab+bc+ac\right)\)
=> \(0=2+2\left(ab+bc+ac\right)\)=> \(ab+bc+ca=-1\)
=> \(\left(ab+bc+ac\right)^2=1\)
Mà \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+a^2bc+abc^2\right)\)
\(=a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+a^2c^2\)
=> \(a^2b^2+b^2c^2+c^2a^2=1\)
Mặt khác : \(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=> \(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(=4-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=> \(a^4+b^4+c^4=4-2=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
( a + b + c )2 = a2 + b2 + c2 + 2ab + 2 bc+ 2ac = 0
Mà a2 + b 2 + c2 = 1
=> 2ab + 2bc + 2ac = - 1
=> ab + bc + ac = \(\frac{-1}{2}\)
=> ( ab + bc + ac ) 2 = a2b2 + a2c2 + b2c 2 + 2ab2c + 2ac2b + 2a2bc = \(\left(\frac{-1}{2}\right)^2\)=\(\frac{1}{4}\)
=> a2b2 + a2c2 + b2c2 + 2abc ( a + b +c ) = \(\frac{1}{4}\)
mà a + b + c = 0 => 2abc ( a +b +c ) = 0
=> a2b2 + b2c2 + c2a2 = \(\frac{1}{4}\)
Ta có : ( a2 + b2 + c2 )2 = a4 + b4 + c4 + 2 ( a2b2 + b2c2 + c2a2 ) = 1
=> a4 +b4 + c4 + 2. \(\frac{1}{4}\) = 1
=> a4 + b4 + c4 = 1 - \(\frac{1}{2}\)
=> a4 + b4 + c4 = \(\frac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
\(a^2+b^2+c^2=16\Rightarrow\left(a+b+c\right)^2-2ab-2bc-2ac=16\)\(\Leftrightarrow-2\left(ab+bc+ac\right)=16\Rightarrow ab+bc+ac=-8\)\(\Rightarrow\left(ab+bc+ac\right)^2=64\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2=64\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=64\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=64\)
Ta có:
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2b^2c^2-2a^2c^2\)\(=16^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=256-2.64=128\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c=0
=>(a+b+c)2=0
=>a2+b2+c2+2(ab+bc+ca)=0
Do a2+b2+c2=1
=>2(ab+bc+ca)=-1
=>ab+bc+ca=-0,5
=>(ab+bc+ca)2=0,25
=>a2b2+b2c2+c2a2+2abc(a+b+c)=0,25
=>a2b2+b2c2+c2a2=0,25(do a+b+c=0)
Từ a2+b2+c2=1
=>(a2+b2+c2)2=1
=>a4+b4+c4+2(a2b2+b2c2+c2a2)=1
=>a4+b4+c4+2.0,25=1
=>a4+b4+c4+0,5=1
=>a4+b4+c4=0,5
![](https://rs.olm.vn/images/avt/0.png?1311)
a+b+c=0 => (a+b+c)^2=0 <=> a^2+b^2+c^2+2(ab+bc+ca)=0
<=> 2+2(ab+bc+ca)=0 => ab+bc+ca=-1
(ab+bc+ca)^2=(ab)^2+(bc)^2+(ca)^2+2ab^2c+2abc^2+2a^2bc=(ab)^2+(bc)^2+(ca)^2+2abc(a+b+c)
=> (ab)^2+(bc)^2+(ca)^2 = (-1)^2 = 1
(a^2+b^2+c^2)^2 = a^4+b^4+c^4+2[(ab)^2+(bc)^2+(ca)^2] = a^4+b^4+c^4 + 2
<=>4=a^4+b^4+c^4+2 => a^4+b^4+c^4 = 2
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