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ND
1
HP
26 tháng 6 2016
Từ \(a+b+c=0=>a+b=-c=>\left(a+b\right)^2=\left(-c\right)^2=>a^2+2ab+b^2=c^2\)
\(=>a^2+2ab+b^2-c^2=0=>a^2+b^2-c^2=-2ab\)
\(=>\left(a^2+b^2-c^2\right)^2=\left(-2ab\right)^2=>a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2=4a^2b^2\)
\(=>a^4+b^4+c^4=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2\right)=2a^2b^2+2b^2c^2+2a^2c^2\)
\(=>2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
\(=>2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2=1^2=1=>a^4+b^4+c^4=\frac{1}{2}\)
PH
1
XO
17 tháng 10 2020
Ta có a + b + c = 0
=> a + b = -c
=> (a + b)2 = (-c)2
=> a2 + b2 + 2ab = c2
=> a2 + b2 - c2 = -2ab
=> (a2 + b2 - c2)2 = (-2ab)2
=> a4 + b4 + c4 + 2a2b2 - 2a2c2 - 2b2c2 = 4a2b2
=> a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2a2c2
Khi đó a2 + b2 + c2 = 14
<=> (a2 + b2 + c2)2 = 142
=> a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2a2c2 = 196
=> a4 + b4 + c4 + a4 + b4 + c4 = 196 (Vì a4 + b4 + c4 = 2a2b2 + 2b2c2 + 2a2c2)
=> 2(a4 + b4 + c4) = 196
=> a4 + b4 + c4 = 98
DP
2
27 tháng 10 2017
Ta có: \(a+b+c=0
\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow1+2ab+2ac+2bc=0\)
\(\Leftrightarrow ab+ac+bc=-\frac{1}{2}\)
\(\Leftrightarrow\left(ab+ac+bc\right)^2=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}\) Vì ( a+b+c=0)
Mặt khác: \(a^2+b^2+c^2=1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4+2.\frac{1}{4}=1
\)
\(\Leftrightarrow a^4+b^4+c^4=1-\frac{1}{2}=\frac{1}{2}\)
BT
3
MA
23 tháng 9 2016
Có: \(a^2+b^2+c^2=1\Rightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1\)
\(\Rightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
Lại có: \(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow2\left(ab+bc+ac\right)=-1\)
\(\Rightarrow ab+bc+ac=-\frac{1}{2}\)
\(\Rightarrow\left(ab+bc+ac\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2+2a^2bc+2ab^2c+2abc^2=\frac{1}{4}\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}-2abc\left(a+b+c\right)\)
\(\Rightarrow a^2b^2+b^2c^2+a^2c^2=\frac{1}{4}\)
Vậy: \(a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=1-2.\frac{1}{4}=1-\frac{1}{2}=\frac{1}{2}\)
HT
1
XO
22 tháng 12 2020
Ta có a2 + b2 + c2 = 14
=> (a2 + b2 + c2)2 = 196
=> a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2 = 196
=> a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 196
Lại có a + b + c = 0
=> (a + b + c)2 = 0
=> a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
=> 2(ab + bc + ca) = -14
=> ab + bc + ca = -7
=> (ab + bc + ca)2 = 49
=> a2b2 + b2c2 + c2a2 + 2ab2c + 2a2bc + 2abc2 = 49
=> a2b2 + b2c2 + c2a2 + 2abc(a + b + c) = 49
=> a2b2 + b2c2 + c2a2 = 49
Khi đó a4 + b4 + c4 + 2(a2b2 + b2c2 + c2a2) = 196
<=> a4 + b4 + c4 + 2.49 = 196
=> a4 + b4 + c4 + 98 = 196
=> a4 + b4 + c4 = 98
Vậy N = 98
NM
0
HT
0
HH
1
HB
2
HP
27 tháng 6 2016
\(a+b+c=0=>a+b=-c=>\left(a+b\right)^2=\left(-c\right)^2=>a^2+2ab+b^2=c^2\)
\(=>a^2+2ab+b^2-c^2=0=>a^2+b^2-c^2=-2ab\)\(=>\left(a^2+b^2-c^2\right)^2=\left(-2ab\right)^2\)
\(=>a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2=4a^2b^2\)
\(=>a^4+b^4+c^4=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2\right)\)\(=2a^2b^2+2b^2c^2+2a^2c^2\)
\(=>2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=\left(a^2+b^2+c^2\right)^2=1^2\)\(=1\)
\(=>M=a^4+b^4+c^4=\frac{1}{2}\)
26 tháng 7 2017
Ta có: \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(\Rightarrow a^2+2ab+b^2=c^2\)
\(\Rightarrow a^2+2ab+b^2-c^2=0\)
\(\Rightarrow a^2+b^2-c^2=-2ab\)
\(\Rightarrow\left(a^2+b^2-c^2\right)^2=\left(-2ab\right)^2\)
\(\Rightarrow a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2=4a^2b^2\)
\(\Rightarrow a^4+b^4+c^4=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2\right)=2a^2b^2+2b^2c^2+2a^2c^2\)\(\Rightarrow2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=\left(a^2+b^2+c^2\right)^2=1^2\)\(\Rightarrow2\left(a^4+b^4+c^4\right)=1\)
\(\Rightarrow a^4+b^4+c^4=\dfrac{1}{2}\)
Vậy \(a^4+b^4+c^4=\dfrac{1}{2}\)
Ta có: \(a+b+c=0\)
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow1+2\left(ab+bc+ca\right)=0\)
\(\Rightarrow ab+bc+ca=-\frac{1}{2}\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Thay vào ta được:
\(A=a^4+b^4+c^4\)
\(A=\left(a^2+b^2+c^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(A=1-\frac{1}{2}=\frac{1}{2}\)
Từ \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
Vì \(a^2+b^2+c^2=1\)
\(\Rightarrow1+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=-1\)
\(\Leftrightarrow ab+bc+ca=\frac{-1}{2}\)
\(\Rightarrow\left(ab+bc+ca\right)^2=\left(\frac{-1}{2}\right)^2\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2\left(a^2bc+b^2ac+c^2ab\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
Vì \(a+b+c=0\)\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Ta có: \(a^2+b^2+c^2=1\)
\(\Rightarrow\left(a^2+b^2+c^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
Vì \(a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
\(\Rightarrow a^4+b^4+c^4+2.\frac{1}{4}=1\)
\(\Leftrightarrow a^4+b^4+c^4+\frac{1}{2}=1\)
\(\Leftrightarrow a^4+b^4+c^4=\frac{1}{2}\)
hay \(A=a^4+b^4+c^4=\frac{1}{2}\)