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

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

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

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

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

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

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

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

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

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

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

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 :

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

Câu 1 nha bạn 

x^4 + x^3 + x^2 + 2014x^2 + 2014x + 2014 + 1 - x^3

=> x^4 + x^3 + x^2 + 2014x^2 + 2014x + 2014 - x^3 - 1

=> x^2 ( x^2 + x + 1 ) + 2014 ( x^2 + x + 1 ) - ( x - 1 )( x^2 + x + 1 ) 

=> ( x^2 + x + 1 )( x^2 + 2014 - x - 1)

giải giúp mình câu 2 với ạ...câu 1 mk đã làm rồi ạ

1)28.122-28.62+30.144

=28(122-62)+30.144

=28.60+30.144

=30.56+30.144

=30(56+144)

=30.200

=6000

2)52.143-52.39-8.26

=52(143-39)-8.26

=52.104-8.26

=8.676-8.26

=8(676-26)

=8.650

=5200

 Bài 2

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

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

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

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

b)để biểu thức A=0 ⇔ x-3=0 hoặc x+2=0

                                ⇔ x=3  hoặc x=-2 

a: ĐKXĐ: x<>-3

a) đkxđ: x+3\(\ne0\Rightarrow x\ne-3\)
b) ta có: M=\(\dfrac{-2}{3}\)
M= \(\dfrac{x+5}{x+3}\)=\(\dfrac{-2}{3}\)
 (x+5)3=(x+3)(-2)
 3x+15=-2x-6
 3x+2x+15=-6
 5x=-6-15
 5x=-21
 x=-21/5
Vậy x= -21/5 khi M có giá trị là -2/3
c) ko bt lm:))