Bài 4: Cho: \(a+b+c=0\)
\(a^2+b^2+c^2=1\)
Tính giá trị biểu thức: \(A=a^4+b^4+c^4\)
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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 ...
NT
3
9 tháng 9 2016
(a2+b2+c2)2=196(a2+b2+c2)2=196
a4+b4+c4+2(a2b2+b2c2+c2a2)=196(1)a4+b4+c4+2(a2b2+b2c2+c2a2)=196(1)
ta lại có a+b+c)^2=0a2+b2+c2=−2(ab+bc+ca)=14a2+b2+c2=−2(ab+bc+ca)=14(ab+bc+ca)2=49(ab+bc+ca)2=49
a2b2+b2c2+c2a2+2abc(a+b+c)=49a2b2+b2c2+c2a2+2abc(a+b+c)=49
a2b2+b2c2+c2a2=49(2)a2b2+b2c2+c2a2=49(2)
Từ (1);(2)a4+b4+c4=196−49.2=98
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 \(a^4+b^4+c^4=\dfrac{1}{2}\)
NM
0
NM
2
MT
6 tháng 7 2015
lại nhầm lần này đúng
(a+b+c)2=a2+b2+c2+2ac+2bc+2ab
=>02=2+2(ac+bc+ab)
=>ac+bc+ab=2:2=-1
=>(-1)2=a2b2+b2c2+a2c2+2a2bc+2b2ac+2c2ab
(-1)2=a2b2+b2c2+a2c2+2abc(a+b+c)
=>1=a2b2+b2c2+a2c2+2abc.0
=>a2b2+b2c2+a2c2=1
(a2+b2+c2)2=a4+b4+c4+2a2b2+2b2c2+2a2c2
(a2+b2+c2)2=a4+b4+c4+2(a2b2+b2c2+a2c2)
22=a4+b4+c4+2.1
4=a4+b4+c4+2
=>a4+b4+c4=2
NM
1
MT
6 tháng 7 2015
(a+b+c)2=a2+b2+c2+2ac+2bc+2ab
=>02=2+2(ac+bc+ab)
=>ac+bc+ab=2:2=-1
=>(-1)2=a2b2+b2c2+a2c2+2a2bc+2b2ac+2c2ab
(-1)2=a2b2+b2c2+a2c2+2abc(a+b+c)
=>1=a2b2+b2c2+a2c2+2abc.0
=>a2b2+b2c2+a2c2=1
(a2+b2+c2)2=a4+b4+c4+2a2b2+2b2c2+2a2c2
(a2+b2+c2)2=a4+b4+c4+2(a2b2+b2c2+a2c2)
22=a4+b4+c4+2.1
4=a4+b4+c4+2
=>a4+b4+c4=2
Ta có: \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=-1\Leftrightarrow ab+bc+ca=\frac{-1}{2}\)
\(\Leftrightarrow\left(ab+bc+ca\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Lại có: \(a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\cdot\frac{1}{4}=1\Leftrightarrow a^4+b^4+c^4=\frac{1}{2}\)