Cho \(a+b+c=0\)và \(a^2+b^2+c^2=14\).Tính giá trị của \(a^4+b^4+c^4\)
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TC
1
15 tháng 8 2017
a2 + b2 + c2=14
hay(a + b + c)2 = 14
a4 + b4 + c4 =(a2 + b2 + c2).(a2 + b2 + c2)=(a+b+c)2 . (a+b+c)2 =14.14=196
k mk nha bạn kb nữa
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: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)
NM
0
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
LH
0
NQ
0
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
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 ...
NQ
1
TM
20 tháng 11 2016
\(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-7\Rightarrow\left(ab+bc+ac\right)^2=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc0=49\)
\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+0=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Xét \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)^2=196\)
\(\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=196\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)
\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)\(\Leftrightarrow a^4+b^4+c^4+98=196\)
\(\Leftrightarrow a^4+b^4+c^4=98\)
T
1
8 tháng 8 2018
\(\left(a+b+c\right)=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Rightarrow2ab+2bc+2ac=-2\)
\(\Rightarrow ab+bc+ac=-1\Rightarrow\left(ab+bc+ac\right)^2=1\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2+2abc\left(a+b+c\right)=4\)
\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+0=4\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=4\)
Có \(\left(a^2+b^2+c^2\right)^2=4\Leftrightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\)
\(\Rightarrow a^4+b^4+c^4+2.4=4\)
Bn làm phần kết quả nhé
Cách 1:
\(+\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(+0=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=14+2\left(ab+bc+ca\right)\)
\(\Rightarrow ab+bc+ca=-7\)
\(+\left(-7\right)^2=\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab.bc+bc.ca+ca.ab\right)\)
\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2+2abc.0\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=49\)
Từ các điều trên suy ra:
\(14^2=a^4+b^4+c^4+2.49\)
\(\Rightarrow a^4+b^4+c^4=14^2-2.49=98\)
Cách 2:
\(+a+b+c=0\Rightarrow a+b=-c\)
\(+14=a^2+b^2+c^2=a^2+b^2+\left(-a-b\right)^2=a^2+b^2+a^2+b^2+2ab=2\left(a^2+b^2+ab\right)\)
\(\Rightarrow a^2+b^2+ab=7\)
\(+a^4+b^4+c^4=a^4+b^4+\left[-\left(a+b\right)\right]^4=\left(a^2+b^2\right)^2-2a^2b^2+\left(a^2+b^2+2ab\right)^2\)
\(=\left(a^2+b^2\right)^2-2a^2b^2+\left(a^2+b^2\right)^2+4\left(a^2+b^2\right).ab+4a^2b^2\)
\(=2\left(a^2+b^2\right)^2+4\left(a^2+b^2\right).ab+2a^2b^2\)
\(=2\left(a^2+b^2+ab\right)^2\)
\(=2.7^2=98\)