phan tich da thuc thanh nhan tu
a, x^7+x^2+1
b, x^7+x^5+1
phan tich da thuc sau thanh nhan tu (x-1)(x-3)(x-5)(x-7)-20
\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left[\left(x-1\right)\left(x-7\right)\right].\left[\left(x-3\right)\left(x-5\right)\right]-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x+11=t\) \(\Rightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=\left(t-4\right)\left(t+4\right)-20=t^2-16-20=t^2-36=\left(t-6\right)\left(t+6\right)\)\(\Rightarrow\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20=\left(x^2-8x+11-6\right)\left(x^2-8x+11+6\right)=\left(x^2-8x+17\right)\left(x^2-8x+5\right)\)
Phan tich da thuc thanh nhan tu x^2*[(x^2+1/x^2)+6*(x-1/x)+7]
phan tich da thuc sau thanh nhan tu: 3(x+5)(x+6)(x+7)-8x(2 cach)
Phan tich da thuc thanh nhan tu
x^7+x^5+1
( x+2)(x+3)(x-7)(x-8)-144 phan tich da thuc thanh nhan tu
bạn có biết viết dấu ko nếu ko biết mik bảo cho s là sắc f là huyền x là ngã r là hỏi j là nặng
phan tich da thuc thanh nhan tu
a) x^7+x^2+1
b) x^8+x+7
c) x^8+3x^4+1
d) x^10+x^5+1
\(x^7+x^2+1\)
\(=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
phan tich da thuc thanh nhan tu (x^2-x+2)^2- 2x^2 +2x -7
phan tich da thuc thanh nhan tu: 7*x-6*x^2-2
7x - 6x2 -2 = -6x2 +3x + 4x -2= -3x( 2x -1) + 2 ( 2x-1) = (2x-1)(2-3x)
x7+x5+x4+x3+x2+1 phan tich da thuc thanh nhan tu
\(x^7+x^5+x^4+x^3+x^2+1\)
\(=x^7+x^6-x^6-x^5+2x^5+2x^4-x^4-x^3+2x^3+2x^2-x^2-x+x+1\)
\(=\left(x^7+x^6\right)-\left(x^6+x^5\right)+\left(2x^5+2x^4\right)-\left(x^4+x^3\right)+\left(2x^3+2x^2\right)-\left(x^2+x\right)+\left(x+1\right)\)
\(=x^6.\left(x+1\right)-x^5.\left(x+1\right)+2x^4\left(x+1\right)-x^3\left(x+1\right)+2x^2\left(x+1\right)-x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^6-x^5+2x^4-x^3+2x^2-x+1\right)\)