(x^10+y^10)(x^2+y^2)-(x^8+y^8)(x^4+y^4)

=x^12+x^10y^2+y^10x^2+y^12-x^12-x^8y^4-x^4y^8-y^12

=x^10y^2+y^10x^2-x^8y^4-x^4y^8

=x^2y^2(x^8+y^8-x^6y^2-x^2y^6)

=x^2y^2[x^6(x^2-y^2)+y^6(y^2-x^2)]

=x^2y^2[x^6(x-y)(x+y)-y^6(x-y)(x+y)]

=x^2y^2(x^6-y^6)(x-y)(x+y)

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

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

a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)

Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)

\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)

b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)

Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)

c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)

Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)

d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)

Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)

Bài 1:

$5x+10=5(x+2)$

Bài 2:

Tại $x=8$ thì $x^2+4x+4=(x+2)^2=(8+2)^2=10^2=100$

Bài 3:

$x^2-6x+9=x^2-2.3.x+3^2=(x-3)^2$

Bài 4:

Diện tích mảnh đất là:

$(x+5)(x-5)=24$

$\Leftrightarrow x^2-25=24$

$\Leftrightarrow x^2=49$

$\Rightarrow x=7$ (do $x>5$)

Chiều dài mảnh đất là: $x+5=7+5=12$ (m)

Đặt a=x-7;b=x-8. Ta có a+b=2x-15. Thế vào biểu thức rồi làm

Lam ro hon mot chut dk ko de xem ket qua co dung ko

a:Ta có: \(2\left(x-2\right)^3=2-x\)

\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)

\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2+1\right]=0\)

Suy ra: x-2=0

hay x=2

a)  \(x^4+324=\left(x^2-6x+18\right)\left(x^2+6x+18\right)\)

c)  \(x^{13}+x^5+1=\left(x^2+x+1\right)\left(x^{11}-x^{10}+x^8-x^7+x^5-x^4+x^3-x+1\right)\)

d)  \(x^{11}+x+1=\left(x^2+x+1\right)\left(x^9-x^8+x^6-x^5+x^3-x^2+1\right)\)

e)  \(x^8+3x^4+4=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)

\(x^8+3x^4+4\)

\(=x^8+4x^4+4-x^4\)

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

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