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Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
(x-1).(2x+1) + 3.(x-1).(x+2).(2x+1)
= (x-1).(2x+1).[1+3.(x+2)]
chúc bn học tốt
(x-1).(2x+1) + 3.(x-1).(x+2).(2x+1)
= (x-1).(2x+1).[1+3.(x+2)]
#
\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2-5x+6\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2-5x+6\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)\(A=\left(x-1\right)\left(x^2-5x+6+x-2\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x^2-4x+4\right)-\left(x-1\right)\)
\(A=\left(x-1\right)\left(x-2\right)^2-\left(x-1\right)\)
\(A=\left(x-1\right)\left[\left(x-2\right)^2-1\right]\)
\(A=\left(x-3\right)\left(x-1\right)^2\)
link tham khảo
https://olm.vn/hoi-dap/detail/9212510579.html
hok tót
= (x4 + 2x2 + 1) + (2x4 + x2 + 2) - (x2 + x+1)2
= [(x2 + 1)2 - (x2 + x+1)2 ] + (2x4 + x2 + 2)
= (x2 + 1 + x2 + x + 1). (x2 + 1 - x2 - x- 1) + (2x4 + x2 + 2)
= (2x2 + x + 2) (-x) + (2x4 + x2 + 2) = -2x3 - x2 - 2x + 2x4 + x2 + 2 = -2x3 + 2x4 - 2x + 2
= -2x3. (1 - x) + 2.(1 - x) = (1- x). (-2x3 + 2) = 2.(1 - x)(1- x3) = 2. (1- x). (1- x) .(1 + x + x2) = 2.(1-x)2. (1 + x + x2)
\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left[x\left(x+3\right)\right]\left[\left(x+1\right)\left(x+2\right)\right]+1\)
\(=\left(x^2+3x\right)\left(x^2+3x+2\right)+1\)
Đặt \(x^2+3x=t\) ta có :
\(\left(x^2+3x\right)\left(x^2+3x+2\right)+1=t\left(t+2\right)+1=t^2+2t+1=\left(t+1\right)^2\)
\(=\left(x^2+3x+1\right)^2\)
Vậy \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)+1\) phân tích thành nhân tử là \(\left(x^2+3x+1\right)^2\)
Ta phân tích như sau e nhé :)
\(3x^4+3x^2+3-\left(x^4+x^2+2x+1+2x^2\left(x+1\right)\right)\)
\(=3x^4+3x^2+3-\left(x^4+x^2+2x+1+2x^3+2x^2\right)\)
\(=2x^4-2x^3-2x+2=2\left[x^3\left(x-1\right)-\left(x-1\right)\right]=2\left(x-1\right)^2\left(x^2+x+1\right)\)
(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)2 +2(x2 -5x+4)+1
=(x2 -5x+4+1)2
=
\(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)\left(x-2\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left[\left(x-2\right)\left(x-3\right)+x-2-1\right]\)
\(=\left(x-1\right)\left(x^2-5x+6+x-3\right)\)
\(=\left(x-1\right)\left(x^2-4x-3\right)\)
\(=\left(x-1\right)\left(x^2-x-3x+3\right)\)
\(=\left(x-1\right)\left[x\left(x-1\right)-3\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x-3\right)\)
\(=\left(x-1\right)^2\left(x-3\right)\)