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 :)
\(G=\left(x^2-3x-1\right)-12\left(x^2-3x-1\right)+27\)
\(=\left(x^2-3x-1-9\right)\left(x^2-3x-1-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)
Sửa đề
\(F=x^2+2xy+y^2-x-y-12\)
\(F=\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt \(x+y=t\), ta có:
\(F=t^2-t-12\)
\(F=t^2-4t+3t-12\)
\(F=t.\left(t-4\right)+3.\left(t-4\right)\)
\(F=\left(t-4\right).\left(t+3\right)\)
\(F=\left(x+y-4\right).\left(x+y+3\right)\)
\(G=\left(x^2-3x-1\right)^2-12.\left(x^2-3x-1\right)+27\)
Đặt \(x^2-3x-1=t\), ta có:
\(G=t^2-12t+27\)
\(G=t^2-9t-3t+27\)
\(G=t.\left(t-9\right)-3.\left(t-9\right)\)
\(G=\left(t-9\right).\left(t-3\right)\)
\(G=\left(x^2-3x-1-9\right).\left(x^2-3x-1-3\right)\)
\(G=\left(x^2-3x-10\right).\left(x^2-3x-4\right)\)
\(G=\left(x^2-5x+2x-10\right).\left(x^2-4x+x-4\right)\)
\(G=\left[x.\left(x-5\right)+2.\left(x-5\right)\right].\left[x.\left(x-4\right)+\left(x-4\right)\right]\)
\(G=\left(x-5\right).\left(x+2\right).\left(x-4\right).\left(x+1\right)\)
Học tốt
