\(a^3+b^3+c^3=3abc\)

\(\Leftrightarrow\)\(a^3+b^3+c^3-3abc=0\)

\(\Leftrightarrow\)\(\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc=0\)

\(\Leftrightarrow\)\(\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)=0\)

\(\Leftrightarrow\)\(\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right]=0\)

Do \(a+b+c\ne0\) nên \(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab=0\)

\(\Leftrightarrow\)\(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow\)\(2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\)\(\left(a^2-2ab+b^2\right)+\left(b^2-bc+c^2\right)+\left(c^2-ca+a^2\right)=0\)

\(\Leftrightarrow\)\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}\Leftrightarrow a=b=c}\)

\(\Rightarrow\)\(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{3a^2}{\left(3a\right)^2}=\frac{3a^2}{9a^2}=\frac{1}{3}\)

...

Cảm ơn bạn nha

1

3

Học tốt

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: \(M=\dfrac{1}{2}\)

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 ...

\(x=\dfrac{1}{y}\Rightarrow\dfrac{1}{y}-y=4\\ \Rightarrow y^2+4y-1=0\\ \Leftrightarrow\left[{}\begin{matrix}y=-2-\sqrt{5}\Rightarrow x=2-\sqrt{5}\\y=-2+\sqrt{5}\Rightarrow x=2+\sqrt{5}\end{matrix}\right.\)

Với \(x=2-\sqrt{5};y=-2-\sqrt{5}\)

\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^2=322\)

Với \(x=2+\sqrt{5};y=-2+\sqrt{5}\)

\(A=x^2+y^2=18\\ B=x^3-y^3=76\\ C=x^4+y^4=322\)

A=x^2+y^2

=(x-y)^2+2xy

=4^2+2=18

B=(x-y)^3+3xy(x-y)

=4^3+3*1*4

=64+12=76

C=(x^2+y^2)^2-2x^2y^2

=18^2-2

=322

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: \(a^4+b^4+c^4+\dfrac{1}{4}=\dfrac{3}{4}\)

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

3/4 bạn nhé

3/4 NHA BN

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

\(a,\)\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\)\(\Rightarrow\left(ab+bc+ac\right)^2=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)
Ta có: \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)=196\)\(\Leftrightarrow a^{^{ }4}+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)\(\Leftrightarrow\)\(a^4+b^4+c^4=98\)