phan tich da thuc thanh nhan tu
x^5+x+1
phan tich da thuc thanh nhan tu
x(x-4)+5x-20
\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)
phan tich da thuc sau thanh nhan tu: x5+x+1
\(x5+x-1 = x5-x4+x3+x4-x3+x2-x2+x-1 = x3(x2-x+1)+x2(x2-x+1)-(x2-x+1) = (x2-x+1)(x3+x2-1) \)
hc tốt nha !!!!!!!!!
phan tich da thuc thanh nhan tu x^5+x+1
\(x^5+x+1=x^5-x^2+x^2+x+1=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
Phan tich da thuc thanh nhan tu
x^5+x+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^7+x^5+1
Phan tich da da thuc thanh nhan phan tu
(x^2+x+1)(x^2+x+2)-12
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
phan tich da thuc thanh nhan tu :x^5+2x^4+3x^3+2x^2+2x+1
x^5+2x^4+2x^3+2x^2+2x+1
=(x^5+x^4)+(x^4+x^3)+(x^3+x^2)+(x^2+x)+(x+1)
=x^4(x+1)+x^3(x+1)+x^2(x+1)+x(x+1)+(x+1)
=(x+1)(x^4+x^3+x^2+x+1)
Phan tich da thuc thanh nhan tu: (x+y+z)^5 - x^5 - y^5 - z^5