\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4.\)

\(=\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4.\)

\(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4.\)

\(=\left(x+5ax+4a^2+a^2\right)^2.\)

\(=\left(x+5ax+5a^2\right)^2.\)

\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

\(=\)\(\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\)

\(=\)\(\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)

\(=\)\(\left[\left(x^2+5ax+5a^2\right)-a^2\right].\left[\left(x^2+5ax+5a^2\right)-a^2\right]+a^4\)

\(=\)\(\left(x^2+5ax+5a^2\right)^2-a^4+a^4\)

\(=\)\(\left(x^2+5ax+5a^2\right)^2\)

Chúc bạn học tốt ~ 

(x + a)(x + 2a)(x + 3a)(x + 4a) + a⁴

= (x + a)(x + 4a)(x + 2a)(x + 3a) + a⁴

= (x² + 4ax + ax + 4a²)(x² + 3ax + 2ax + 6a²) + a⁴

= (x² + 5ax + 4a²)(x² + 5ax + 6a²) + a⁴

Đặt x² + 5ax + 4a² = t

= t(t + 2a²) + a⁴

= (t + a²)²

= (x² + 5ax + 4a² + a²)²

= (x² + 5ax + 5a²)²

đơn giản

Cách đặt khác ez hơn :))

\(A=\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)

\(A=\left[\left(x+a\right)\left(x+4a\right)\right]\left[\left(x+2a\right)\left(x+3a\right)\right]+a^4\)

\(A=\left(4a^2+5ax+x^2\right)\left(6a^2+5ax+x^2\right)+a^4\)

Đặt \(p=5a^2+5ax+x^2\)

\(\Rightarrow A=\left(p-a^2\right)\left(p+a^2\right)+a^4\)

\(\Rightarrow A=p^2-a^4+a^4\)

\(\Rightarrow A=p^2\)

Thay \(p=5a^2+5ax+x^2\)vào A ta có :

\(A=\left(5a^2+5ax+x^2\right)^2\)

1. \(4x^2-17xy+13y^2=4x^2-4xy-13xy+13y^2=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

2. \(2x\left(x-5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

3. \(A=\left(2a-3b\right)^2+2\left(2a-3b\right)\left(3a-2b\right)+\left(2b-3a\right)^2\)

\(\Leftrightarrow\left(2a-3b\right)^2-2\left(2a-3b\right)\left(2b-3a\right)+\left(2b-3a\right)^2=\left(2a-3b-2b+3a\right)^2=\left(5a-5b\right)^2\)

\(=25\left(a-b\right)^2=25\cdot100=2500\)

=\(\left(x+a-3\right)\left(x^2-2ax-2x+4a-12\right)\)

c) Ta có: \(6x^4-11x^2+3\)

\(=6x^4-2x^2-9x^2+3\)

\(=\left(6x^4-2x^2\right)-\left(9x^2-3\right)\)

\(=2x^2\left(3x^2-1\right)-3\left(3x^2-1\right)\)

\(=\left(3x^2-1\right)\left(2x^2-3\right)\)

d) Ta có: \(\left(x^2+x\right)+3\left(x^2+x\right)+2\)

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

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

a: \(a\left(x-y\right)-b\left(y-x\right)+c\left(x-y\right)\)

\(=a\left(x-y\right)+b\left(x-y\right)+c\left(x-y\right)\)

\(=\left(x-y\right)\left(a+b+c\right)\)

b: \(a^m-a^{m+2}\)

\(=a^m-a^m\cdot a^2\)

\(=a^m\left(1-a^2\right)\)

\(=a^m\left(1-a\right)\left(1+a\right)\)

@Thục Trinh giải đi

1.

\(3x^2-16x+5\\ =3x^2-x-15x+5\\ =x\left(3x-1\right)-5\left(3x-1\right)\\ =\left(x-5\right)\left(3x-1\right)\)

2.

\(3x^3-14x^2+4x+3\\ =\left(3x^3+x^2\right)-\left(15x^2+5x\right)+\left(9x+3\right)\\ =x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\\ =\left(x^2-5x+3\right)\left(3x+1\right)\)

3. \(x^8+x^7+1\\ =\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\\ =x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x^3-1\right)+x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x\left(x^3+1\right)\left(x+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)[x^2\left(x^3+1\right)\left(x-1\right)+x\left(x^3+1\right)\left(x-1\right)+1]\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+x^5-x^4+x^2-x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)4.

\(64x^4+y^4\\ =\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\\ =\left(8x^2+y^2\right)^2-16x^2y^2\\ =\left(8x^2+y^2-4xy\right)\left(8x^2+y+4xy\right)\)

5.

\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\\ =\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\\ =\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+4a^2+2a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)+2a^2\left(x^2+5ax+4a^2\right)+a^4\\ =\left(x^2+5ax+5a^2\right)^2\)

6. Đề bài này bị sai! Bởi vì vốn dĩ cái đề đã được phân thích thành nhân tử sẵn rồi!