<=> [(x + 2)(x + 5)][(x + 3)(x + 4) - 24 = (x² + 7x + 10) (x² + 7x + 12) - 24 (1)

đặt x² + 7x + 11 = t

=> ( 1 ) <=> (t - 1)(t + 1) - 24 = t² - 1 - 24 = t² - 25 = (t - 5)(t + 5)

=> (x² + 7x + 11 - 5) (x² + 7x + 11 + 5) = (x² + 7x + 6) (x² + 7x + 16) (x + 1) (x + 6) (x² + 7x + 16)

chúc you học tốt!! ^^

ok mk nhé!! 4545454353434636565454676345345346654767567567587676345346334534534565646756

x4 + 4

= (x2)2 + 22

= x4 + 2.x2.2 + 4 – 4x2

(Thêm bớt 2.x2.2 để có HĐT (1))

= (x2 + 2)2 – (2x)2

(Xuất hiện HĐT (3))

= (x2 + 2 – 2x)(x2 + 2 + 2x)

a: \(x^4+2x^3+3x^2+2x+1\)

\(=x^4+x^3+x^2+x^3+x^2+x+x^2+x+1\)

\(=x^2\left(x^2+x+1\right)+x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^2+x+1\right)=\left(x^2+x+1\right)^2\)

b: \(x^4-4x^3+2x^2+4x+1\)

\(=x^4-2x^3-x^2-2x^3+4x^2+2x-x^2+2x+1\)

\(=x^2\left(x^2-2x-1\right)-2x\left(x^2-2x-1\right)-\left(x^2-2x-1\right)\)

\(=\left(x^2-2x-1\right)\left(x^2-2x-1\right)=\left(x^2-2x-1\right)^2\)

c: \(x^4+x^3+2x^2+2x+4\)

\(=x^4-x^3+2x^2+2x^3-2x^2+4x+2x^2-2x+4\)

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

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

d) x4 + 2x3 - 4x – 4 = (x4 – 4) + (2x3 – 4x) = (x2 – 2)(x2 + 2) + 2x(x2 – 2)

= (x2 – 2)(x2 + 2 + 2x) = (x - √2)( x + √2)( x2 + 2 + 2x)

Có sai dấu ko bạn 

Đề : \(a^4+8a^3+14a^2-8a-15\)

\(=a^4-a^2+8a^3-8a+15a^2-15\)

\(=a^2\left(a^2-1\right)+8a\left(a^2-1\right)+15\left(a^2-1\right)\)

\(=\left(a^2-1\right)\left(a^2+8a+15\right)\)

\(=\left(a+1\right)\left(a-1\right)\left(a+3\right)\left(a+5\right)\)

mik chỉ biết (x+2)(x+3)(x+4)(x+5)-24=(x+6)(x+1)(x²+7x+16 bằng cách đặt ẩn phụ thui còn lại ko biết sorry nha

x4 – 2x2

(Có x2 là nhân tử chung)

= x2(x2 – 2)

Sửa đề: x^4+4y^4

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

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

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

Lời giải:

a.

$=(x^2)^2+(\frac{1}{2}y^4)^2+2.x^2.\frac{1}{2}y^4-x^2y^4$

$=(x^2+\frac{1}{2}y^4)^2-(xy^2)^2$
$=(x^2+\frac{1}{2}y^4-xy^2)(x^2+\frac{1}{2}y^4+xy^2)$
b.

$=(\frac{1}{2}x^2)^2+(y^4)^2+2.\frac{1}{2}x^2.y^4-x^2y^4$
$=(\frac{1}{2}x^2+y^4)^2-(xy^2)^2$
$=(\frac{1}{2}x^2+y^4-xy^2)(\frac{1}{2}x^2+y^4+xy^2)$

c.

$=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$

$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$

d.

$=\frac{64x^4+y^4}{64}=\frac{1}{64}(8x^2+y^2-4xy)(8x^2+y^2+4xy)$

c: \(64x^4+y^4\)

\(=64x^4+16x^2y^2+y^4-16x^2y^2\)

\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)

\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)