mình mới học lớp 6 thôi

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

\(\Leftrightarrow14+2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=-7\)

Suy ra : \(\left(ab+bc+ac\right)^2=49\Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+a^2c^2=49\)

\(a^2+b^2+c^2=14\Leftrightarrow\left(a^2+b^2+c^2\right)^2=196\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=256\)  \(\Leftrightarrow a^4+b^4+c^4=98\)

Vậy ... 

\(a+b+c=0\)

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

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

\(\Leftrightarrow2ab+2bc+2ca=-14\)

\(\Leftrightarrow ab+bc+ca=-7\)

\(\Rightarrow\left(ab+bc+ca\right)^2=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=49\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=49\).

\(a^2+b^2+c^2=14\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=14^2=196\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=196\)

\(\Leftrightarrow a^4+b^4+c^4+2.49=196\)

\(\Leftrightarrow a^4+b^4+c^4=98\)

1.

Ta có: \(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)\left(a^2+b^2\right)\ge ab\left(a^2+b^2\right)\)

\(\Rightarrow VT\le\frac{a}{a+bc\left(b^2+c^2\right)}+\frac{b}{b+ca\left(c^2+a^2\right)}+\frac{c}{c+ab\left(a^2+b^2\right)}\)

\(\Rightarrow VT\le\frac{a^2}{a^2+abc\left(b^2+c^2\right)}+\frac{b^2}{b^2+abc\left(a^2+c^2\right)}+\frac{c^2}{c^2+abc\left(a^2+b^2\right)}\)

\(\Rightarrow VT\le\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(a+b+c=0\)

⇔\(\left(a+b+c\right)^2=0\)

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

⇔\(2018+2\left(ab+bc+ca\right)=0\)

⇔\(ab+bc+ca=-1009\)

⇔\(\left(ab+bc+ca\right)^2=\left(-1009\right)^2=1009^2\)

⇔\(a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+abc^2+a^2bc\right)=1009^2\)

⇔\(a^2b^2+b^2c^2+c^2a^2+2abc\left(b+c+a\right)=1009^2\)

⇔\(a^2b^2+b^2c^2+c^2a^2=1009^2\)

\(a^2+b^2+c^2=2018\)

⇔\(\left(a^2+b^2+c^2\right)^2=2018^2\)

⇔\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=2018^2\)

⇔\(a^4+b^4+c^4+2\cdot1009^2=2018^2\)

⇔\(a^4+b^4+c^4=2018^2-2\cdot1009^2=2036162\)

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 ...

Có: \(a+b+c=0\)

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

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

\(\Leftrightarrow2\left(ab+bc+ca\right)=-1\) (do \(a^2+b^2+c^2=1\) )

\(\Leftrightarrow ab+bc+ca=-\dfrac{1}{2}\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2ab.bc+2bc.ca+2ca.ab=\dfrac{1}{4}\)

\(\Leftrightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=\dfrac{1}{4}\)

\(\Leftrightarrow \left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\) (do \(a+b+c=0\))

Lại có: \(M=a^4+b^4+c^4\)

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

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

\(=1-2.\dfrac{1}{4}\)(do \(\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=\dfrac{1}{4}\))

\(=1-\dfrac{1}{2}=\dfrac{1}{2}\)

Vậy \(M=\dfrac{1}{2}\)

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

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