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7x2 - 28 = 7(x2 - 4) = 7(x - 2)(x + 2)
2x3 + 3x2 - 18x - 27 = x2(2x + 3) - 9(2x + 3) = (2x + 3)(x2 - 9) = (2x + 3)(x - 3)(x + 3)
Đặt x^2 + x = t
=> D = 2 ( t - 5 )^2 - 5t + 28
=> D = 2 ( t^2 - 10t + 25 ) - 5t + 28
=> D =2t^2 - 20t + 25 - 5t + 28
=> D = 2t^2 - 25t + 53
ĐẾn đây tự phân tích
1)
=3(x-y)+(x-y)(x+y)
=(x-y)(3x+3y)
2)
=x^2+2x+x+2
=x(x+2)+x+2
=(x+1)(x+2).
\(3x-3y+x^2-y^2\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(3+x+y\right)\)
\(x^2+3x+2\)
\(=x^2+2x+x+2\)
\(=x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+1\right)\left(x+2\right)\)
Trả lời:
x4 - 3x3 + 3x2 - x
= x ( x3 - 3x2 + 3x - 1 )
= x ( x - 1 )3
Ta có :
\(x^4-3x^3+3x^2-x\)
\(=x\left(x^3-3x^2+3x-1\right)\)
\(=x\left(x-1\right)^3\)
Vậy ..........
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 \(x^2-3x-4=x^2+x-4x-4=x\left(x+1\right)-4\left(x+1\right)=\left(x+1\right)\left(x-4\right)\)
\(\Rightarrow\left(x^2-3x-1\right)^2-12\left(x^2-3x-1\right)+27=\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+2\right)\)
\(A=\left(x^2+3x+1\right)\left(x^2+3x-3\right)-5\)
Đặt \(t=x^2+3x+1\) thì A thành
\(t\left(t-4\right)-5=t^2-4t-5\)
\(t^2-5t+t-5=t\left(t-5\right)+\left(t-5\right)\)
\(=\left(t-5\right)\left(t+1\right)=\left(x^2+3x+1-5\right)\left(x^2+3x+1+1\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
= (x3-1)+3x(x-1) = (x-1)(x2+x+1)+3x(x-1)
=(x-1)(x2+x+1+3x)
=(x-1)(x2+4x+1)
\(=x^2+4x-7x-28=\left(x+4\right)\left(x-7\right)\)
=(x+4)(x-7)