x^4 + 4

= x^4 + 4x^2 + 4 - 4x^2

= (x^2 + 2)^2 - 4x^2

= (x^2 + 2 - 2x)(x^2 + 2 + 2x)

Ta có \(x^4+4=\left(x^4+4x^2+4\right)-\left(2x\right)^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

a^3+b^3+c^3-3abc

=a^3 + b^3 + 3ba^2 + 3ab^2 + b^3 - 3ab(a+b) +c^3 _ 3abc

=(a+b)^3+c^3 -3ab(a+b+c)

=(a+b+c)(a^2+2ab+b^2-bc-ca+c^2)-3ab(a+b+c)

=(a+b+c)(a^2+b^2+c^2-ab-ab-bc)

\(x^{200}\)+ \(x^{100}\)+\(1\)

<=>\(x^{100}\)(\(x^{100}\)+\(1\)) +\(1\)

<=> (\(x^{100}\)+\(1\))(\(x^{100}\)+\(1\))

<=> \(\left(x^{100}+1\right)^2\)

Chúc bạn học tốt ~<>

Ta có :

\(x^{200}+x^{100}+1\)

\(\Rightarrow x^{100}.\left(x^{100}+1^1\right)+1\)

\(\Rightarrow\left(x^{100}+1\right).\left(x^{100}+1\right)\)( bạn nhân phân phối là ra nhé )

\(\Leftrightarrow\left(x^{100}+1\right)^2\)

Vậy nhân tử của đa thức \(x^{200}+x^{100}+1\)là \((x^{100}+1)^2\)

x²⁰⁰+x¹⁰⁰+1 = x²⁰⁰+x¹⁰⁰+x¹⁰⁰+1-x¹⁰⁰=(x¹⁰⁰+1)²-x¹⁰⁰

⁼(x¹⁰⁰+1)²-(x⁵⁰)² =(x¹⁰⁰+1-x⁵⁰)(x¹⁰⁰+1+x⁵⁰)

x²⁰⁰+x1⁰⁰

=m^3-3m^2-3m^2+9n+2m-6

=m^2(m-3)-3m(m3)+2(m-3)

=(m-3)(m^2-3m+2)=(m-3)(m^2-m-2m+2)

=(m-3)[m(m-1)-2(m-1)]

=(m-3)(m-2)(m-1)

\(m^3-6m^2+11m-6\)

\(m^3-6m^2+11m-6\)

\(=\left(m-1\right)\left(m-3\right)\left(m-2\right)\)

x^4 + y^4=(x^2)^2+(y^2)^2
=(x^2+y^2)^2-2x^2y^2
=(x^2+y^2)^2-(√2xy)^2
=(x^2+y^2-√2 xy)(x^2+y^2+√2 xy)

\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}-\frac{3}{abc}=\left(\frac{1}{a}+\frac{1}{b}\right)^3+\left(\frac{1}{c}\right)^3-3.\frac{1}{a}.\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)-\frac{3}{abc}\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2-\left(\frac{1}{a}+\frac{1}{b}\right).\frac{1}{c}+\frac{1}{c^2}\right]-3.\frac{1}{a}.\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{2}{ab}-\frac{1}{ac}-\frac{1}{bc}+\frac{1}{c^2}\right)-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{ac}-\frac{1}{bc}\right)\)

\(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)

\(1^2-2^2+3^2-....-100^2=\left(1^2-2^2\right)+...+\left(99^2-100^2\right)=\)

\(-1\left(1+2\right)+\left(-1\right)\left(3+4\right)+...+\left(-1\right)\left(99+100\right)=\frac{-100.101}{2}=-5050\)

đặt a=x^2-5x

(x^2-5x)^2+10(x^2-5x+24)

=a^2+10(a+24)

=a^2+10a+24

=a^2+6a+4a+24

=a(a+6)+4(a+6)

=(a+6)(a+4)

=(x^2-5x+6)(x^2-5x+4)

đây nha bạn  

(x² - x + 1)² - 5x(x² - x + 1) + 4x²

Đặt x² - x + 1 = a

<=> a² - 5xa + 4x² = x² - 4xa - xa + 4x² 

 = a(a - 4x) - x(a - 4x) = (a - x)(a - 4x)

= (x² - x + 1 - x)(x² - x + 1 - 4x)

= (x² - 2x + 1)(x² - 5x + 1) = (x - 1)²(x² - 5x + 1)

Đặt x² - x + 1 = y

đthức <=> y² - 5xy + 4x²

= y² - xy - 4xy + 4x²

= y( y - x ) - 4x( y - x )

= ( y - x )( y - 4x )

= ( x² - x + 1 - x )( x² - x + 1 - 4x )

= ( x² - 2x + 1 )( x² - 5x + 1 ) 

= ( x - 1 )²( x² - 5x + 1 )