( ab + bc + ca )^2 = a^2b^2 + b^2c^2 +c^2a^2 + 2abc( a + b + c )

                          =a^2b^2 + b^2c^2 + c^2a^2 + 2abc.0 ( vì a + b + c = 0)

                          =a^2b^2 + b^2c^2 + c^2a^2

Đây nhé! Tích giúp c nha

1: Ta có: \(a^2+b^2+c^2\)

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

\(=5^2-2\cdot174=-323\)

Lời giải:
$a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)$

$=[(a+b+c)-2(ab+bc+ac)]^2-2(a^2b^2+b^2c^2+c^2a^2)$

$=[-2(ab+bc+ac)]^2-2(a^2b^2+b^2c^2+c^2a^2)$

$=4(ab+bc+ac)^2-2[(ab+bc+ac)^2-2abc(a+b+c)]$

$=4(ab+bc+ac)^2-2[(ab+bc+ac)^2]=2(ab+bc+ac)^2$
Ta có đpcm.

Ta có :

\(\left(a+b+c\right)=0\)

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

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

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

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

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

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

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\) (vì \(a+b+c=0\))

\(\left(1\right)+\left(2\right)\Rightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)

\(\Rightarrow dpcm\)

\(\dfrac{1}{a+2}+\dfrac{1}{b+2}+\dfrac{1}{c+2}\ge1\Leftrightarrow\dfrac{2}{a+2}+\dfrac{2}{b+2}+\dfrac{2}{c+2}\ge2\)

\(\Leftrightarrow\dfrac{a}{a+2}+\dfrac{b}{b+2}+\dfrac{c}{c+2}\le1\)

\(\Rightarrow1\ge\dfrac{a^2}{a^2+2a}+\dfrac{b^2}{b^2+2b}+\dfrac{c^2}{c^2+2c}\ge\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2+2\left(a+b+c\right)}\)

\(\Rightarrow a^2+b^2+c^2+2\left(a+b+c\right)\ge a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Rightarrow\) đpcm

a + b + c = 0

=> (a + b + c)² = 0

=> a² + b² + c² + 2ab + 2bc + 2ca = 0

=> a² + b² + c² = -2(ab + 2bc + 2ca)

=> (a² + b² + c²)² = [-2(ab + bc + ca)]²

=> a⁴ + b⁴ + c⁴ + 2a²b² + 2b²c² + 2c²a² = 4(a²b² + b²c² + c²a² + 2ab²c + 2a²bc + 2abc²) 

=> a⁴ + b⁴ + c⁴ = 4a²b² + 4b²c² + 4c²a² + 8a²bc + 8ab²c + 8abc² - 2a²b² - 2b²c² - 2a²c²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 8abc(a + b + c)

=> a⁴ + b⁴ + c⁴= 2a²b² + 2b²c² + c²a²

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 2abc(a + b + c) (Vì a + b + c = 0)

=> a⁴ + b⁴ + c⁴ = 2a²b² + 2b²c² + 2c²a² + 2a²bc + 2ab²c + 2abc²

=>  a⁴ + b⁴ + c⁴ = 2(a²b² + b²c² + c²a² + a²bc + ab²c + abc²) 

=> a⁴ + b⁴ + c⁴ = 2(ab + bc + ca)² (đpcm)

\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+abc^2+a^2bc\right)=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow2\left(ab^2c+abc^2+a^2bc\right)=0\\ \Leftrightarrow abc\left(a+b+c\right)=0\left(đpcm;a+b+c=0\right)\)