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BC
2
9 tháng 2 2021
Ta có: a+b+c=0
\(\Leftrightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=0-1=-1\)
hay \(ab+bc+ac=-\dfrac{1}{2}\)
\(\Leftrightarrow\left(ab+bc+ac\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(b+c+a\right)=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=\dfrac{1}{4}\)
Ta có: \(M=a^4+b^4+c^4\)
\(\Leftrightarrow M=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2-2a^2b^2-2a^2c^2-2b^2c^2\)
\(\Leftrightarrow M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+a^2c^2+b^2c^2\right)\)
\(\Leftrightarrow M=1^2-2\cdot\dfrac{1}{4}=1-\dfrac{1}{2}=\dfrac{1}{2}\)
Vậy: \(M=\dfrac{1}{2}\)
9 tháng 2 2021
Ta có : \(a+b+c=0\)
\(\Rightarrow\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ac\right)=1\) ( * )
\(\Rightarrow ab+bc+ac=-\dfrac{1}{2}\)
Lại có : \(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\) ( suy ra từ * )
\(\Rightarrow a^4+b^4+c^4=2\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\)
Vậy ...
AQ
8
MT
15 tháng 7 2015
ta có:
(a+b+c)2=a2+b2+c2+2ab+2bc+2ac
<=>(a+b+c)2=a2+b2+c2+2.(ab+bc+ac)
=>02 = 1 +2.(ab+bc+ac)
=>ab+bc+ac = -1/2
(ab+bc+ac)2=a2b2+a2c2+b2c2+ab2c+a2bc+abc2
<=>(ab+bc+ac)2=a2b2+a2c2+b2c2+abc.(a+b+c)
=> (-1/2)2=a2b2+a2c2+b2c2+abc.0
=>a2b2+a2c2+b2c2=1/4
suy ra:
(a2+b2+c2)2=a4+b4+c4+a2b2+a2c2+b2c2
=>12=a4+b4+c4+1/4
=>a4+b4+c4=1-1/4=3/4
K
1
11 tháng 2 2018
ta có:
(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac
<=>(a+b+c)^2=a^2+b^2+c^2+2.(ab+bc+ac)
=>0^2 = 1 +2.(ab+bc+ac)
=>ab+bc+ac = -1/2 (ab+bc+ac)2=a2b 2+a2c 2+b2c 2+ab2c+a2bc+abc2
<=>(ab+bc+ac)2=a2b 2+a2c 2+b2c 2+abc.(a+b+c)
=> (-1/2)2=a2b 2+a2c 2+b2c 2+abc.0 =>a2b 2+a2c 2+b2c 2=1/4
suy ra:
(a2+b2+c2 ) 2=a4+b4+c4+a2b 2+a2c 2+b2c 2
=>12=a4+b4+c4+1/4
=>a4+b4+c4=1-1/4=3/4
:A
VD
1
30 tháng 6 2015
\(a^2+b^2+c^2=1\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=1\Leftrightarrow0-2\left(ab+bc+ca\right)=1\Leftrightarrow ab+bc+ca=-\frac{1}{2}\)
\(M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)
\(=1-2\left(\frac{1}{4}-2abc\left(a+b+c\right)\right)=1-\frac{1}{2}+4abc.0=\frac{1}{2}\)
3 tháng 1 2021
Ta có: a+b+c=0
nên \(\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Leftrightarrow2ab+2ac+2bc=-1\)
\(\Leftrightarrow ab+ac+bc=\dfrac{-1}{2}\)
\(\Leftrightarrow\left(ab+ac+bc\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2a^2bc+2ab^2c+2abc^2=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)
\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2=\dfrac{1}{4}\)
Ta có: \(a^2+b^2+c^2=1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+a^2c^2+b^2c^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\cdot\dfrac{1}{4}=1\)
\(\Leftrightarrow a^4+b^4+c^4=1-\dfrac{1}{2}=\dfrac{1}{2}\)
\(\Leftrightarrow a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{2}{4}+\dfrac{1}{4}=\dfrac{3}{4}\)
Vậy: \(a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}\)
TY
1
30 tháng 6 2015
\(a^2+b^2+c^2=1\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=1\Leftrightarrow0-2\left(ab+bc+ca\right)=1\Leftrightarrow ab+bc+ca=-\frac{1}{2}\)
\(M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)
\(=1-2\left(\frac{1}{4}-2abc\left(a+b+c\right)\right)=1-\frac{1}{2}+4abc.0=\frac{1}{2}\)
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: \(a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+a^2c^2\right)\)
\(\Leftrightarrow a^4+b^4+c^4=1-2.\frac{1}{4}=1-\frac{1}{2}=\frac{1}{2}\)
PA
1 tháng 10 2020
\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)
ND
1
HP
26 tháng 6 2016
Từ \(a+b+c=0=>a+b=-c=>\left(a+b\right)^2=\left(-c\right)^2=>a^2+2ab+b^2=c^2\)
\(=>a^2+2ab+b^2-c^2=0=>a^2+b^2-c^2=-2ab\)
\(=>\left(a^2+b^2-c^2\right)^2=\left(-2ab\right)^2=>a^4+b^4+c^4+2a^2b^2-2b^2c^2-2a^2c^2=4a^2b^2\)
\(=>a^4+b^4+c^4=4a^2b^2-\left(2a^2b^2-2b^2c^2-2a^2c^2\right)=2a^2b^2+2b^2c^2+2a^2c^2\)
\(=>2\left(a^4+b^4+c^4\right)=a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)
\(=>2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2=1^2=1=>a^4+b^4+c^4=\frac{1}{2}\)