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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)
B1:
a) \(5\left(x^2+y^2\right)-20x^2y^2\)
\(=5\left(x^2-4x^2y^2+y^2\right)\)
b) \(=2\left(x^8-16\right)=2\left(x^4-4\right)\left(x^4+4\right)=2\left(x^2-2\right)\left(x^2+2\right)\left(x^4+4\right)\)
B2:
a) Đặt \(x^2-3x+1=y\)
=> \(y^2-12y+27\)
\(=\left(y^2-12y+36\right)-9\)
\(=\left(y-6\right)^2-3^2\)
\(=\left(y-9\right)\left(y-3\right)\)
\(=\left(x^2-3x-10\right)\left(x^2-3x-4\right)\)
\(=\left(x+1\right)\left(x-4\right)\left(x^2-3x-10\right)\)
b) Đặt \(x^2+7x+11=t\)
Ta có: \(\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
\(Dat:a^2+a+1=b\Rightarrow....=a\left(a+1\right)-12=\left(a+4\right)\left(a-3\right)\)
=
a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\) (1)
Đặt x2 + x +1 = t
Ta có : \(t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12\)
\(=t\left(t-3\right)+4\left(t-3\right)=\left(t-3\right)\left(t+4\right)\)
Thay vào (1), ta được : \(\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)\)
b) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\) (2)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt x2 + 7x + 11 = y
Ta có : \(\left(y-1\right)\left(y+1\right)-24=y^2-1-24=y^2-25=\left(y-5\right)\left(y+5\right)\)
Thay vào (2), ta được : \(\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
1) \(ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)\)
\(=ab\left(a+b\right)-b^2c-bc^2+a^2c-ac^2\)
\(=ab\left(a+b\right)-c\left(b^2-a^2\right)-c^2\left(a+b\right)\)
\(=ab\left(a+b\right)-c\left(a+b\right)\left(a-b\right)-c^2\left(a+b\right)\)
\(=\left(a+b\right)\left(ab-ac+bc-c^2\right)\)
\(=\left(a+b\right)\left[a\left(b-c\right)+c\left(b-c\right)\right]\)
\(=\left(a+b\right)\left(b-c\right)\left(a+c\right)\)
\(Dat:x^2+x=a\Rightarrow....=a^2-2a-15=\left(a-1\right)^2-4^2=\left(a+3\right)\left(a-7\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
\(Dat:x+y=a\Rightarrow....=a^2-a-12=\left(a+3\right)\left(a-4\right)=\left(x+y+3\right)\left(x+y-4\right)\)
a) A= \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Đặt \(x^2+x=a\) .
Khi đó : \(A=a^2-2a-15=a^2-5a+3a-15\)\(=a\left(a-5\right)+3\left(a-5\right)=\left(a+3\right)\left(a-5\right)\)
Mà \(a=x^2+x\) nên \(A=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b) B = \(x^2+2xy+y^2-x-y-12\) \(=\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt x+y = z.
Khi đó : \(B=z^2-z-12=z^2-4z+3z-12=z\left(z-4\right)+3\left(z-4\right)\)\(=\left(z+3\right)\left(z-4\right)\)
Mà z = x+y nên B = (x+y+3)(x+y-4)
phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung(không dùng HĐT)
x3 -3x2y+3xy2 - y3
\(=xy\left(x^2-3x+3y-y^2\right)\)
\(=xy\left[\left(x-y\right)\left(x+y\right)+3\left(x-y\right)\right]\)
\(=xy\left(x-y\right)\left(x+y+3\right)\)
\(Ht\)
nếu sai cho mik xl vì mik chx thành thục cái này
x4 - 3x2 +1
= x4 - 2x2 + 1 - x2
= ( x2 - 1 )2 - x2
= ( x2- 1 +x ) ( x2 - 1 - x )
Có sai đề hông má :??
Tooy sửa đề :\(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\left(1\right)\)
Đặt \(t=x^2+3x+1\)ta được:
\(\left(1\right)=t\left(t+1\right)-6\)
\(=t^2+t-6\)
\(=t^2-2t+3t-6\)
\(=t\left(t-2\right)+3\left(t-2\right)\)
\(=\left(t-2\right)\left(t+3\right)\)
\(=\left(x^2+3x+1-2\right)\left(x^2+3x+1+3\right)\)
\(=\left(x^2+3x-1\right)\left(x^2+3x+4\right)\)