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

\(a^2+b^2+c^2=1\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=1\Leftrightarrow0-2\left(ab+bc+ca\right)=1\Leftrightarrow ab+bc+ca=-\frac{1}{2}\)

\(M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)

\(=1-2\left(\frac{1}{4}-2abc\left(a+b+c\right)\right)=1-\frac{1}{2}+4abc.0=\frac{1}{2}\)

a + b +c =0 => ( a +b + c)^2 =0 => a^2 +b^2 +c^2 + 2ab +2bc + 2ac = 0

=> 1 + 2(ab + bc +ac) = 0 => 2(ab +bc +ac) = -1 ==> ab + bc +ac = -1/2

( ab + bc+ac)^2 = 1/4 => a^2.b^2 + b^2.c^2 + c^2.a^2 + 2ab^2.c +2ab.c^2 + 2 a^2.b.c = 1/4 

=> a^2 . b^2 + b^2 . c^2 + c^2 . a^2 + 2abc ( a+ b+ c) = 1/4

=> a^2 . b^2  + b^2 . c^2 + c^2 . a^2  + 2abc . 0 = 1/4

=> 2( a^2 . b^2 +  + b^2 . c^2 + c^2 . a^2 ) = 2.1/4 = 1/2 

=> 2a^2 . b^2 +  2 b^2 . c^2 + 2c^2 . a^2 = 1/2  

( a^2 + b^2 + c^2 )^2 = 1

=> a^4 + b^4 + c^4 + 2a^2.b^2 + 2b^2.c^2 + 2 c^2 . a^2 = 1

=> a^4 + b^ 4 + c^4 + 1/2 = 1 

=> a^4 + b^4 + c^4 = 1/2

(a+b+c)² = 0

<=> a² + b² + c² + 2ab + 2bc + 2ac = 0

<=> 2ab + 2bc + 2ac = -1

<=> ab + bc + ac = -1/2

<=> a²b² + b²c² + c²a² + 2ab²c + 2abc² + 2a²bc = 1/4

<=> a²b² + b²c² + c²a² + 2abc(a+b+c) = 1/4

<=> a²b² + b²c² + c²a² = 1/4

(a² + b² + c²)² = 1

<=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2a²c² = 1

<=> a⁴ + b⁴ + c⁴ + 2.1/4 = 1

<=> a⁴ + b⁴ + c⁴ = 1 - 1/2 = 1/2.

Vậy M = 1/2

\(a^2+b^2+c^2=1\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)=1\Leftrightarrow0-2\left(ab+bc+ca\right)=1\Leftrightarrow ab+bc+ca=-\frac{1}{2}\)

\(M=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+a^2c^2\right)=1^2-2\left[\left(ab+bc+ca\right)^2-2\left(ab^2c+abc^2+a^2bc\right)\right]\)

\(=1-2\left(\frac{1}{4}-2abc\left(a+b+c\right)\right)=1-\frac{1}{2}+4abc.0=\frac{1}{2}\)

\(\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)abc=\frac{3}{4}8\Rightarrow\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}=\frac{3.8}{4}\Leftrightarrow\)\(\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=6\)

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