Phan tich thanh tu nhan tu
a)x^2-3x+2
b)8x^3-27+(x+1)×(3-2x)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Câu a :
\(\dfrac{3x^2-12x+12}{x^4-8x}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
Câu b :
\(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
c)\(x^3-x^2+x+3=x^2+x-2x^2-2x+3x+3\)
\(=x\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+3\right)\)
d)\(x^8+3x^4+4=\left(x^8+4x^4+4\right)-x^4=\left(x^4+2\right)^2-\left(x^2\right)^2\)
\(=\left(x^4-x^2+2\right)\left(x^4+x^2+2\right)\)
e)\(x^6-x^4-2x^3+2x^2=x^4\left(x^2-1\right)-2x^2\left(x-1\right)=x^4\left(x-1\right)\left(x+1\right)-2x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^3+x^2\right)-2x^2\left(x-1\right)=x^2\left(x-1\right)\left(x^3+x^2-2\right)\)
\(=x^2\left(x-1\right)\left[\left(x^3-1\right)+\left(x^2-1\right)\right]=x^2\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(x+1\right)\right]\)
\(=x^2\left(x-1\right)\left(x-1\right)\left(x^2+2x+2\right)=x^2\left(x-1\right)^2\left(x^2+2x+2\right)\)
a)\(x^2-x-12\)
\(=x^2+4x-3x-12\)
\(=x\left(x+4\right)-3\left(x+4\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
b) \(x^2+8x+15\)
\(=x^2+3x+5x+15\)
\(=x\left(x+3\right)+5\left(x+3\right)\)
\(=\left(x+3\right)\left(x+5\right)\)
A, m2-7m+12=m2-3m+12-4m=m(m-3)-4(m-3)=(m-4)(m-3)
B, 2x4-x3+27-54x=x3(2-x)-27(2-x)=(x3-27)(2-x)=(x-3)(x2+3x+9)(2-x)
a) m^2 -7m +12 = m^2 -3m -4m +12
=m(m -3)-4 (m- 3)
=(m-4)(m-3)
b) 2x^4-x^3 -54x+ 27
=(2m^4-x^3)- (54x - 27)
=x^3(2x-1)-27(2x-1)
=(x^3-27)(2x-1)
\(3x^4-48\)
\(=\left(3x^4-6x^3\right)+\left(6x^3-12x^2\right)+\left(12x^2-24x\right)+\left(24x-48\right)\)
\(=3x^3\left(x-2\right)+6x^2\left(x-2\right)+12x\left(x-2\right)+24\left(x-2\right)\)
\(=\left(x-2\right)\left[\left(3x^3+6x^2\right)+\left(12x+24\right)\right]\)
\(=\left(x-2\right)\left[3x^2\left(x+2\right)+12\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(3x^2+12\right)\)
\(x^4-8x\)
\(=x\left(x^3-8\right)\)
\(=x\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(4x-8\right)\right]\)
\(=x\left[x^2\left(x-2\right)+2x\left(x-2\right)+4\left(x-2\right)\right]\)
\(=x\left(x-2\right)\left(x^2+2x+4\right)\)
2x(x-1)-3x+3
=2x(x-1)-(3x-3)
=2x(x-1)-3(x-1)
=(x-1)(2x-3)
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
Câu a :
\(x^2-3x+2\)
\(=x^2-x-2x+2\)
\(=\left(x^2-x\right)-\left(2x-2\right)\)
\(=x\left(x-1\right)-2\left(x-1\right)\)
\(=\left(x-1\right)\left(x-2\right)\)