1)

$2x=a+b+c$

$\Rightarrow x-a=\dfrac{b+c-a}{2},\quad x-b=\dfrac{c+a-b}{2},\quad x-c=\dfrac{a+b-c}{2}$

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)$

$=\dfrac{(b+c-a)(c+a-b)+(c+a-b)(a+b-c)+(a+b-c)(b+c-a)}{4}$

$=\dfrac{2ab+2bc-a^2-b^2-c^2+2bc+2ca-a^2-b^2-c^2+2ca+2ab-a^2-b^2-c^2}{4}$

$=\dfrac{4ab+4bc+4ca-3(a^2+b^2+c^2)}{4}$

$=\dfrac{(a+b+c)^2-2(a^2+b^2+c^2)}{4}$

$=\dfrac{(2x)^2-2(a^2+b^2+c^2)}{4}$

$=ab+bc+ca-x^2$

$\therefore\ (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=ab+ac+bc-x^2.$

2)

$ab+bc+ca=abc,\quad a+b+c=1$

$(a-1)(b-1)(c-1)$$=abc-ab-bc-ca+a+b+c-1$$=abc-ab-bc-ca+1-1$$=abc-(ab+bc+ca)$$=abc-abc$$=0$

$\therefore\ (a-1)(b-1)(c-1)=0.$

Bài 1: 

Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\) 

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\)

\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{256}+1\right)+1\)

\(............................\)

\(A=\left[\left(2^{256}\right)^2-1\right]+1=2^{512}\)

Dap số A=0

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

1) =0

3)=(3⁷⁹⁸⁸-1)/2

3) Q=(3+1)(3^2+1)(3^4+1)....(3^3994+1)

=(3-1)(3+1)(3^2+1)(3^4+1)...(3^3994+1)

=(3^2-1)(3^2+1)(3^4+1)...(3^3994+1)

=(3^4-1)(3^4+1)...(3^3994+1)

=.........

=(3^3994-1)(3^3994+1)

=3^7988-1

1. \(A=x^6-x^4+x^3-x-x^4+x^2=x^3\left(x^3-x\right)+\left(x^3-x\right)-x\left(x^3-x\right)=\left(x^3-x\right)^2+\left(x^3-x\right)\)

Thay \(x^3-x=6\) vào A, ta được:

\(A=36+6=42\)

KL : A=42

2.

a) đa thức đã cho \(=ab^2+ac^2+abc+bc^2+ba^2+abc+ca^2+cb^2+abc\)

\(=\left(ab^2+ba^2+abc\right)+\left(ac^2+ca^2+abc\right)+\left(bc^2+cb^2+abc\right)\)

\(=ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)\)

\(=\left(ab+bc+ac\right)\left(a+b+c\right)\)

b) đa thức đã cho \(=\left(a^2c+abc\right)+\left(abc+b^2c\right)+\left(ac^2+bc^2\right)+\left(a^2b+ab^2\right)\)

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

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

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

\(=\left[a\left(c+b\right)+c\left(b+c\right)\right]\left(a+b\right)\)

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