Cho a + b + c = 0 v 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 <=> 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
Bạn kiểm tra lại có sai chỗ nào không nhé
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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
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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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Ta có : \(a+b+c=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-1\)
\(\left(ab+bc+ac\right)^2=1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=1\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=1\)
\(\left(a^2+b^2+c^2\right)^2=4\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\Leftrightarrow a^4+b^4+c^4=4-2\left(a^2+b^2+c^2\right)=4-2=2\)
Ta có:
a+b+c=0 => (a+b+c)2=0 => a2+b2+c2 = -2(ab+bc+ac)
=> a4+b4+c4 + 2(a2b2+b2c2 + a2c2) = 4(a2b2+b2c2 + a2c2)+8(ab2c + abc2 + a2bc)
=> a4+b4+c4 =2(a2b2+b2c2 + a2c2) + 8abc(a+b+c)
=> a4+b4+c4 =2(a2b2+b2c2 + a2c2)
Mặt khác, vì
a2+b2+c2 = -2(ab+bc+ac)=2
=> ab +bc+ac = -1
=>a2b2+b2c2 + a2c2+2(ab2c + abc2 + a2bc) = 1
=> a2b2+b2c2 + a2c2 = 1
=> a4+b4+c4 = 1* 2 =2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](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)
- Ta có : \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=\frac{-\left(a^2+b^2+c^2\right)}{2}=-\frac{4}{2}=-2\)
- Ta có ; \(\left(a^2+b^2+c^2\right)^2=16\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=16\)
\(\Leftrightarrow a^4+b^4+c^4=16-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
Mặt khác : \(\left(ab+bc+ac\right)^2=4\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=4\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=4\)
\(\Rightarrow a^4+b^4+c^4=16-2.4=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có a + b + c = 0
<=> (a + b + c)2 = 0
<=> a2 + b2 + c2 + 2(ab + bc + ca) = 0
<=> ab + bc + ca = \(-\frac{1}{2}\)
=> \(\left(ab+bc+ca\right)^2=\frac{1}{4}\)
<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab^2c+2a^2bc+2abc^2=\frac{1}{4}\)
<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
<=> \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\frac{1}{4}\)
Lại có a2 + b2 + c2 = 1
=> (a2 + b2 + c2)2 = 1
<= > a4 + b4 + c4 + 2[(ab)2 + (bc)2 + (ca)2] = 1
<=> \(a^4+b^4+c^4+2.\frac{1}{4}=1\)
<=> \(a^4+b^4+c^4=\frac{1}{2}\)
Từ a + b + c = 0 => ( a + b + c )2 = 0 <=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
<=> ab + bc + ca = -1/2 => ( ab + bc + ca )2 = 1/4
<=> a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2a2bc = 1/4
<=> a2b2 + b2c2 + c2a2 + 2abc( a + b + c ) = 1/4
<=> a2b2 + b2c2 + c2a2 = 1/4 ( vì a + b + c = 0 )
Từ a2 + b2 + c2 = 1 => ( a2 + b2 + c2 )2 = 1 <=> a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2 = 1
<=> a4 + b4 + c4 + 2( a2b2 + b2c2 + c2a2 ) = 1
<=> a4 + b4 + c4 + 1/2 = 1 <=> a4 + b4 + c4 = 1/2
Vậy A = 1/2