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a) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2\)
= \(\left(x^2+4x+8\right)^2+x\left(x^2+4x+8\right)+2x\left(x^2+4x+8\right)+2x^2\)
= \(\left(x^2+4x+8\right)\left(x^2+4x+8+x\right)+2x\left(x^2+4x+8+x\right)\)
= \(\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)
= \(\left(x^2+2x+4x+8\right)\left(x^2+5x+8\right)\)
= \(\left(x+2\right)\left(x+4\right)\left(x^2+5x+8\right)\)
Bạn nên tách bài ra để đăng. Không nên đăng 1 loạt như thế này.
1: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
=(x^2+x)^2+3(x^2+x)-10
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
2: \(=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\)
\(=\left(x^2+5ax\right)^2+10a^2\left(x^2+5ax\right)+25a^2\)
\(=\left(x^2+5ax+5a^2\right)^2\)
3: \(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
5: \(M=\left(n+1\right)\left(n^2+2n\right)+360\)
=n(n+1)(n+2)+360 chia hết cho 6
6A
7D
ta có: \(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x+4\right)^2.\left(x^2-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(\left(x+4\right)^2-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+4-1\right)\left(x+4+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
Cho mình nhé hihi!!!
x2(x+4)2-(x+4)2-(x2-1)
=(x+4)2 (x2-1)-(x2-1)
=(x2-1)(x2+8x+16-1)
=(x-1)(x+1)(x2+8x+15)
A/ \(16x-5x^2-3=\left(15x-3\right)-\left(5x^2-x\right)=3\left(5x-1\right)-x\left(5x-1\right)=\left(5x-1\right)\left(3-x\right)\)
B/ \(x^3-3x^2+1-3x=\left(x^3-4x^2+x\right)+\left(x^2-4x+1\right)=x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
C/ \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
D/ \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
Sai đề rồi đa thức này không có nghiêm làm sao phân tích được
Đặt: \(x^2-6x+1=a;x^2+1=b\)
Khi đó đa thức này có dạng:
\(2a^2+5ab+2b^2=2a^2+4ab+ab+2b^2\)
\(=2a\left(a+2b\right)+b\left(a+2b\right)=\left(a+2b\right)\left(2a+b\right)\)
Thay lại a và b thì được:
\(\left(a+2b\right)\left(2a+b\right)=\left(x^2-6x+1+2x^2+2\right)\left(2x^2-12x+2+x^2+1\right)\)
\(=\left(3x^2-6x+3\right)\left(3x^2-12x+3\right)\)
\(=9\left(x-1\right)^2\left(x^2-4x+1\right)\)
Vậy ...
Đặt \(x^2+x+1=t\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=t\left(t+1\right)-12=t^2+t-12=\left(t^2+t+\dfrac{1}{4}\right)-\dfrac{49}{4}=\left(t+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(t+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(t+\dfrac{1}{2}+\dfrac{7}{2}\right)=\left(t-3\right)\left(t+4\right)=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
= \(\left(x^2+x+1\right)\left[\left(x^2+x+1\right)+1\right]-12\)
= \(\left(x^2+x+1\right)^2\left(x^2+x+1\right)-12\)
= \(\left(x^2+x+1\right)\left(x^2+x+1\right)-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-4.3\)
= \(\left(x^2+x+1\right)\left(x^2+x-2\right)+4\left(x^2+x-2\right)\)
= \(\left(x^2+x+5\right)\left(x^2+x-2\right)\)