Đặt \(x^2-3x-1=a\), ta có:

\(a^2-12a+27=a^2-9a-3a+27=a\left(a-9\right)-3\left(a-9\right)=\left(a-9\right)\left(a-3\right)\)

\(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

Mà \(x^2-3x-10=x^2-5x+2x-10=x\left(x-5\right)+2\left(x-5\right)=\left(x-5\right)\left(x+1\right)\)

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

\(\Rightarrow\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27=\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+2\right)\)

Đặt x² - 3x - 1 = k

Khi đó, ta có: A = k² - 12k + 27 = k² - 3x - 9x + 27 = k(k - 3) - 9(k - 3) = (k - 9)(k -  3)

=> (x² - 3x - 1 - 9)(x² - 3x - 1 - 3) = (x² - 3x - 10)(x² - 3x - 4)

= (x² - 5x + 2x - 10)(x² - 4x + x - 4)

= [x(x - 5) + 2(x - 5)][x(x - 4) + (x - 4)]

= (x + 2)(x - 5)(x + 1)(x - 4)

ai giúp mình với TT^TT

F=x²+2xy+y²-x-y-12 

= (x + y)^2 - (x + y) - 12 

= (x + y)(x + y - 1) - 12

đặt x + y = t

F = t(t - 1) - 12

= t^2 - t - 12

=  (t - 4)(t + 3)

G=(x²-3x-1)²-12(x²-3x-1)+27

đăth x^2 - 3x - 1 = t

G = t^2 - 12t + 27

= (t - 3)(t - 9)

có t = x^2 - 3x - 1

thay vào 

Câu F ( kiểm tra lại đề )

 Câu G . Đặt x^2 -3x-1=t

 t^2 -12t+27 ( thực hiện pp tách)

\(F=x^2+2xy+y^2-x-y+12\)

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

\(=\left(x+y-\frac{1}{2}\right)^2+\frac{47}{4}>0\) thì làm sao phân tích nhân tử :)

\(G=\left(x^2-3x-1\right)-12\left(x^2-3x-1\right)+27\)

\(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)\)

\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)

\(=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)

Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6=t.\left(t+1\right)-6\)

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

\(=t\left(t-2\right)+3\left(t-2\right)=\left(t-2\right)\left(t+3\right)\)

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

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

\(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)

Đặt \(x^2+3x+1=a\)ta có :

\(a\left(a+1\right)-6\)

\(=a^2+a-6\)

\(=a^2+6a-a-6\)

\(=\left(a^2+6a\right)-\left(a+6\right)\)

\(=a\left(a+6\right)-\left(a+6\right)\)

\(=\left(a+6\right)\left(a-1\right)\)

Thay \(a=x^2+3x+1\)vào A ta có :

\(A=\left(x^2+3x+1+6\right)\left(x^2+3x+1-1\right)\)

\(=\left(x^2+3x+7\right)\left(x^2+3x\right)\)

(x^2+3x+2)(x^2+7x+12)+1

=(x²+x+2x+2)(x²+3x+4x+12)+1

=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]+1

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

=[(x+1)(x+4)][(x+2)(x+3)]+1

=(x²+5x+4)(x²+5x+6)+1

=(x²+5x+4)[(x²+5x+4)+2]+1

=(x²+5x+4)²+2(x²+5x+4)+1

=(x²+5x+4+1)²

=(x²+5x+5)²