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

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

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

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

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

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

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

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

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

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

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

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 :

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

\(a.\left(x^2+2x+x+2\right)\left(x^2+5x+6x+30\right)-5\)

\(=\left(x+1\right)\left(x+2\right)\left(x+5\right)\left(x+6\right)-5=\left(x^2+7x+6\right)\left(x^2+7x+10\right)\)

Đặt \(x^2+7x+8=a\Rightarrow\text{Biểu thức }=\left(a-2\right)\left(a+2\right)-5=a^2-9=\left(a-3\right)\left(a+3\right)\)

nên : \(BT=\left(x^2+7x+5\right)\left(x^2+7x+11\right)\)

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

Đặt \(x^2+5ax+5a^2=y\Rightarrow BT=\left(y-a^2\right)\left(y+a^2\right)+a^4=y^2=\left(x^2+5ax+5a^2\right)^2\)

= (-2a ^2 +4a)(1-x)

= -2 (a^2 -2a)(1-x)

= -2a(a-2)(1-x)

= 2a(a-2)(x-1)

\(-2a^2\left(x-1\right)+4a\left(1-x\right)\)

\(=-a\cdot2a\left(x-1\right)-2\cdot2a\left(x-1\right)\)

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

Ta có:

(x+a)(x+2a)(x+3a)(x+4a) + a⁴

=(x+a)(x+4a)(x+3a)(x+2a) +a⁴

=(x²+5ax+4a²)(x²+5ax+6a²) + a⁴

Đặt x²+5ax+5a²=y

=>(x²+5ax+4a²)(x²+5ax+6a²) + a⁴=(y-a²)(y+a²)+a⁴

=y²-a⁴+a⁴

=y²

=(x²+5ax+5a²)²

k mik nha

Hai câu đầu tham khảo

Câu hỏi của Bangtan Sonyeondan - Toán lớp 8 - Học toán với OnlineMath

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

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

\(=\left(x^2+5ax+4a^2\right)\left(a^2+5ax+6a^2\right)+a^4\)(1)

Đặt \(x^2+5ax+4a^2=t\)

\(\Rightarrow\left(1\right)=t\left(t+2a^2\right)+a^4\)

\(=t^2+2a^2t+a^4=\left(t+a^2\right)^2\)(2)

Mà \(x^2+5ax+4a^2=t\)

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

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)