Đặt \(x^2+x+1=t\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)

\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)

= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)

= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)

= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)

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

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

Bạn nên tách bài ra để đăng. Không nên đăng 1 loạt như thế này.

1: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)

=(x^2+x)^2+3(x^2+x)-10

=(x^2+x+5)(x^2+x-2)

=(x^2+x+5)(x+2)(x-1)

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

\(=\left(x^2+5ax\right)^2+10a^2\left(x^2+5ax\right)+25a^2\)

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

3: \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)

\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)

5: \(M=\left(n+1\right)\left(n^2+2n\right)+360\)

=n(n+1)(n+2)+360 chia hết cho 6

6A

7D

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

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

\(=2\left(x^2+x-5\right)\left(x^2+x-6\right)-3\left(x^2+x-6\right)\)

\(=\left(x^2+x-6\right)\left(2x^2+2x-13\right)\)

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)

\(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

Đặt t=\(x^2+x\)

\(\Rightarrow C=2\left(t-5\right)^2-5t+28=2t^2-20t+50-5t+28=2t^2-25t+78=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)\)

Thay t: \(C=2\left(t-\dfrac{13}{2}\right)\left(t-6\right)=2\left(x^2+x-\dfrac{13}{2}\right)\left(x^2+x-6\right)=2\left(x-2\right)\left(x+3\right)\left(x^2+x-\dfrac{13}{2}\right)\)

Ta có: \(C=2\left(x^2+x-5\right)^2-5\left(x^2+x\right)+28\)

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

\(=\left(x-2\right)\left(x+3\right)\left(2x^2+2x-13\right)\)

\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)

Sao đề là phân tích mà lại "= 0" vậy bạn?

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

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

x³+27+(x+3)(x+9)

= (x+3)(x²-3x+9)+(x+3)(x+9)

= (x+3)(x²-3x+9+x+9)

=(x+3)(x²-2x+18)

=(x^2+8x)^2+23(x^2+8x)+135

Cái này ko phân tích được nha bạn

\(\left(x^2+8x+8\right)\left(x^2+8x+15\right)+15\\ \Leftrightarrow\left(x^4+8x^3+15x^2+8x^3+64x^2+120x+8x^2+64x+120\right)+15\\ \Leftrightarrow x^4+16x^3+87x^2+184x+135\)

Gọi `A=(x^2+8x+8)(x^2+8x+15)+15`

Đặt `t=x^2+8x+11,5`

`=>A=(t-3,5)(t+3,5)+15=t^2-3,5^2+15=t^2-2,75=(t-sqrt(2,75))(t+sqrt(2,75))=(x^2+8x+11,5-(sqrt11)/2)(x^2+8x+11,5+(sqrt11)/2)=(x^2+8x+(23-\sqrt11)/2)(x^2+8x+(23+\sqrt11)/2)`