1.

Cho $a+b+c=0$.

Ta có:

$a+b+c=0\Rightarrow a=-(b+c)$

$\Rightarrow a^2=(b+c)^2=b^2+c^2+2bc$

$\Rightarrow a^2-b^2-c^2=2bc$

Tương tự:

$b^2-c^2-a^2=2ca$

$c^2-a^2-b^2=2ab$

Do đó: $A=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}$

$=\dfrac{a^3+b^3+c^3}{2abc}$

Mà $a+b+c=0$ nên:

$a^3+b^3+c^3=3abc$

Suy ra $A=\dfrac{3abc}{2abc}$

$A=\dfrac32$

2.

Cho $abc=2$.

Ta có:

$A=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}$

Vì $abc=2$ nên:

$ab=\dfrac2c,\quad bc=\dfrac2a,\quad ac=\dfrac2b$

Suy ra:

$A=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{2bc}{2+2bc+2b}$

$=\dfrac{ac}{2+ac+2c}+\dfrac{ab}{2+ab+a}+\dfrac{bc}{1+bc+b}$

Quy đồng và sử dụng $abc=2$:

$A=1$

(a+b+c)³=(a+b)³+3(a+b)²c+3(a+b)c²+c³

=a³+b³+3ab.(a+b)+3(a+b)²c+3(a+b)c²+c³

=a³+b³+c³+3(a+b)(ab+ac+bc+c²)

=a³+b³+c³+3(a+b)[a.(b+c)+c.(b+c)]

=a³+b³+c³+3(a+b)(b+c)(c+a) 

=>dpcm

P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=>2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5²-1)(5²+1)(5⁴+1)(5⁸+1)(5¹⁶+1)

=(5⁴-1)(5⁴+1)(5⁸+1)(5¹⁶+1)

=(5⁸-1)(5⁸+1)(5¹⁶+1)

=(5¹⁶-1)(5¹⁶+1)

=5³²-1

==>P=(5³²-1)/2

khó thế mk ko bít

=\(\frac{b-c-\left(a-c\right)+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

=0

Ta có:

$M=\dfrac{ab+bc+ca-a-b-c}{a^2b-a^2-b+1}$

Vì $abc=1$ nên:

$c=\dfrac1{ab}$

Suy ra: $M=\dfrac{ab+\dfrac1a+\dfrac1b-a-b-\dfrac1{ab}}{a^2b-a^2-b+1}$

$=\dfrac{(a-1)(b-1)(ab-1)}{ab(a^2-1)(b-1)}$

$=\dfrac{(a-1)(b-1)(ab-1)}{ab(a-1)(a+1)(b-1)}$

$=\dfrac{ab-1}{ab(a+1)}$

Mà $ab=\dfrac1c$ nên:

$M=\dfrac{\frac1c-1}{\frac1c(a+1)}$

$=\dfrac{1-c}{a+1}$

Vậy: $M=\dfrac{1-c}{a+1}$

ta có: a + b + c = 0 => a+b = - c => a² + 2ab + b² = c² => a² + b² - c² = - 2ab

tương tự như trên, ta có: b² + c² - a² = -2bc; c² + a² - b² = -2ac

thay vào A, có:

\(A=\frac{1}{-2bc}-\frac{1}{2ca}-\frac{1}{2ab}\)

\(A=-\frac{1}{2}.\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=-\frac{1}{2}.\left(\frac{a+b+c}{abc}\right)=-\frac{1}{2}.\left(\frac{0}{abc}\right)=0\)

KL: A = 0 tại a + b + c = 0