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Ta có a2+b2+c2=2
=> a4+b4+c4+2a2b2+2b2c2+2c2a2=4 (bình phương 2 vế)
=> 2a4+2b4+2c4+4a2b2+4b2c2+4c2a2=8 (1)
Ta lại có a+b+c=0
=> a2+b2+c2+2ab+2bc+2ca=0
=> 2 +2ab+2bc+2ca=0
=> 2ab+2bc+2ca=-2
=> 4a2b2+4b2c2+4c2a2+4ab2c+4ac2b+4ca2b=4
=> ---------------------------- +0 =4
=>----------------------------- =4 (2)
Từ (1) và (2) => 2a4+2b4+2c4=4
=> a4+b4+c4=2
Câu b tương tự nhé,, ra 1/2
1,cho a+b - c = 0
a2 + b2 + c2 = 10
tính a4 +b4 +c4
2, cho a- b- c =0
a2 + b2 + c2 = 16
tính a4 + b4+ c4
a+b+c=0=>a+b=-c
=>(a+b)2=(-c)2
=>a2+2ab+b2=(-c)2=c2
=>a2+2ab+b2-c2=0
=>a2+b2-c2=-2ab
=>(a2+b2-c2)2=(-2ab)2
=>a4+b4+c4+2a2b2-2b2c2-2a2c2=4a2b2
=>a4+b4+c4=4a2b2-2a2b2+2b2c2+2a2c2=2a2b2+2b2c2+2a2c2
=>2(a4+b4+c4)=a4+b4+c4+2(a2b2+b2c2+a2c2)=(a2+b2+c2)2=4
=>a4+b4+c4=4:2=2
Vậy.....
Có: \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
Thay: \(a^2+b^2+c^2=2\), có:
\(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow2+2\left(ab+bc+ac\right)=0\)
\(\Rightarrow ab+bc+ac=-1\)
Xét: \(a^2+b^2+c^2=2\)
\(\Rightarrow\left(a^2+b^2+c^2\right)^2=4\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\) (1)
Mặt khác: \(\left(ab+bc+ac\right)^2=1\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2=1\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=1\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1-2abc\left(a+b+c\right)\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=1\) (2)
Từ (1) và (2) suy ra: \(a^4+b^4+c^4=4-1=3\)
Vậy:...
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\)
\(\Rightarrow a=b=c\left(đpcm\right)\)
Ta có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(3\left(a^2+b^2+c^2\right)=3a^2+3b^2+3c^2\)
mà \(\left(a+b+c\right)^2=3\left(a^2+b^2+c^2\right)\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=3a^2+3b^2+3c^2\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ac+a^2=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(c-a\right)^2\ge0\forall a,c\end{matrix}\right.\)
Mà \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(c-a\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=a\end{matrix}\right.\Leftrightarrow}a=b=c\Rightarrowđpcm}\)
a) Ta có: `a + b + c = 0`
`=> (a+b+c)^2 = 0`
`=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 0`
`=> 2 + 2 (ab+bc+ca) = 0`
`=> 1 + ab + bc + ca = 0`
`=> ab+bc+ca = -1`
`=> (ab+bc+ca)^2 = 1`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2ab . bc + 2bc . ca + 2. ca. ab = 1`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc(b+c+a) = 1`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc . 0 = 1`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 0 = 1`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 = 1`
Mà `(a^2 + b^2 + c^2)^2 = 4`
`=> a^4 + b^4 + c^4 + 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 = 4`
`=> a^4 + b^4 + c^4 + 2 (a^2 b^2 + b^2 c^2 + c^2 a^2) = 4`
`=> a^4 + b^4 + c^4 + 2 . 1 = 4`
`=> a^4 + b^4 + c^4 = 2`
Vậy `a^4 + b^4 + c^4 = 2`
--------------------------
b) Ta có: `a + b + c = 0`
`=> (a+b+c)^2 = 0`
`=> a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 0`
`=> 1 + 2 (ab+bc+ca) = 0`
`=> 0,5 + ab + bc + ca = 0`
`=> ab+bc+ca = -0,5`
`=> (ab+bc+ca)^2 = 0,25`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2ab . bc + 2bc . ca + 2. ca. ab = 0,25`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc(b+c+a) = 0,25`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 2abc . 0 = 0,25`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 + 0 = 0,25`
`=> a^2 b^2 + b^2 c^2 + c^2 a^2 = 0,25`
Mà `(a^2 + b^2 + c^2)^2 = 1`
`=> a^4 + b^4 + c^4 + 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 = 1`
`=> a^4 + b^4 + c^4 + 2 (a^2 b^2 + b^2 c^2 + c^2 a^2) = 1`
`=> a^4 + b^4 + c^4 + 2 . 0,25 = 1`
`=> a^4 + b^4 + c^4 = 0,5`
Vậy `a^4 + b^4 + c^4 = 0,5`