c)\(\Leftrightarrow x^7+x^6-x^6-x^5+2x^5+2x^4-x^4-x^3+2x^3+2x^2-x^2-x+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^6-x^5+2x^4-x^3+2x^2-x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^6-x^5+2x^4-x^3+2x^2-x+1=0\end{matrix}\right.\)

d)\(x^{10}+x^8+x^6+x^4+x^2+1=0\)

\(\Leftrightarrow\left(x^2+x+1\right)x^6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+x+1=0\left(vl\right)\end{matrix}\right.\)

gợi ý nha (mik lm còn j là hok nx )   (x+a)(x+b)(x+c)=x2+(a+b+c)x2+(ab+bc+ac)x+abc

            Muốn chứng minh được ta phải chứng minh vế trái    

(x2+bx+ax+ab)(x+c)=x3+ax2+bx2+cx2+abx+bcx+acx+abc

     x3+ax2+bx2+cx2+abx+bcx+acx+abc=x3+ax2+bx2+cx2+abx+bcx+acx+abc(1)

Vì hai biểu thức trên (1) giông nhau

               Do đó (x+a)(x+b)(x+c)=x2+(a+b+c)x2+(ab+bc+ac)x+abc

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$

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

giải mấy caj này mệt khủng

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

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

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

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

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

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

a) \(\orbr{\begin{cases}x-5=0\\x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=0\end{cases}}\)

b) \(\Leftrightarrow x^2-2x+1=0\)

<=> (x - 1)² = 0

<=> x -1 = 0 

<=> x = 1

sssssssssssAZ