Cach phan tich da thuc thanh nhan tu
a)X^3+2X^2+2X+1
b)X^3-4X^2+12X-27
c)a^6-a^4+2a^3+2a^2
d)x^4+2x^3+2x^2+2x+1
e)x^5+x^4+x^3+x^2+x+1
phan tich da thuc thanh nhan tu
A=x^6-2x^5-4x^4+6x^3+4x^2-2x-1
a, x^3-x^2-4x^2+8x-4
b, 4x^2-25-(2x-5)2x+7
c, x^3+27+(x+3)(x-9)
d, 2x^2-2y^2+5x-5y
e, x^2-y^2-2y-1
phan tich da thuc thanh nhan tu
phan tich da thuc thanh nhan tu
(ab - 1)2 + (a + b)2
x3 + 2x2 + 2x +1
x3 - 4x2 + 12x - 27
x4 - 2x3 + 2x -1
Ta có : x3 + 2x2 + 2x + 1
= x3 + x2 + (x2 + 2x + 1)
= x2(x + 1) + (x + 1)2
= (x + 1) ( x2 + x + 1)
a)\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2\)
\(=a^2b^2+a^2+b^2+1\)
\(=a^2\left(b^2+1\right)+\left(b^2+1\right)\)
\(\left(a^2+1\right)\left(b^2+1\right)\)
b)\(x^3+2x^2+2x+1=x^3+x^2+x^2+x+x+1\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
\(=\left(x^2+x+1\right)\left(x+1\right)\)
c)\(x^3-4x^2+12x-27=x^3-3x^2-x^2+3x+9x-27\)
\(=x^2\left(x-3\right)-x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(x^2-x+9\right)\left(x-3\right)\)
d)Chịu
phan tich da thuc thanh nhan tu 2x(x + 3) + 2(x + 3)
tìm x
a) 5x(x – 2) – x – 2 = 0
b) 4x(x + 1) = 8( x + 1)
c) x(2x + 1) + 1/2 x 3/3 - = 0
d) x(x – 4) + (x – 4)^2 = 0
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 chung
x^6-2x^3 +1
x^4 +2x^2+1
\(x^6-2x^3+1=\left(x^3-1\right)^2\)
\(x^4+2x^2+1=\left(x^2+1\right)^2\)
a) x6 - 2x3 + 1
= (x3)2 - 2x3 + 1
= ( x3 - 1)2
b) x4 + 2x2 + 1
= ( x2)2 + 2x2 + 1
= ( x2 + 1)2
Đặt A= x^4+2x^3+4x^2+2x+1
=> A/x^2 = x^2+2x+4+2/x+1/x^2 = (x^2+1/x^2)+2(x+1/x)+4
đặt y= x+1/x => A/x^2= y^2-2+2y+4 = y^2 +2y+2= (y+1)^2 +1 >0
=>PT y^2+2y+2=0 vô nghiệm => A không thể phân tích thành nhân tử
bạn xem lại đề xem có phải sai đề ko? Hoặc cũng có thể mình nhầm hic.. hic *^*
phan tich cac da thuc sau thanh nhan tu
x^2-x-12
x^2+8x+15
x^3-x^2+x+3
x^8+3x^4+4
x^6-x^4-2x^3+2x^2
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) \(x^2+8x+15\)
\(=x^2+3x+5x+15\)
\(=x\left(x+3\right)+5\left(x+3\right)\)
\(=\left(x+3\right)\left(x+5\right)\)
câu 1 |: phan tich da thuc thanh nhan tử x^3 +3x^2-3x-1
\câu 2 làm tính chia
a,( x^4 -2x^3 +2x-1 ) : (x^2-1)
\b, (x^6 -2x^5+2x^4+6x^3-4x^2) : (6x^2)
\cau3 rút gọn phân thức \(\frac{3x^2+6x^2+12}{x^3-8}\)
\mọi người làm gấp với a! lam dc cau nào nhờ giai hộ
câu 1:
x3-1+3x2-3x =(x-1)(x^2+x+1)+3x(x-1)=(x-1)(x^2+x+1+3x)=(x-1)(x^2+4x=1)
Câu 2 :
a) \(\left(x^4-2x^3+2x-1\right):\left(x^2-1\right)\)
\(=\left(x^4-x^2-2x^3+2x+x^2-1\right):\left(x^2-1\right)\)
\(=\left[x^2\left(x^2-1\right)-2x\left(x^2-1\right)+\left(x^2-1\right)\right]:\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-2x+1\right):\left(x^2-1\right)\)
\(=x^2-2x+1\)
b) \(\left(x^6-2x^5+2x^4+6x^3-4x^2\right):6x^2\)
\(=\frac{1}{6}x^4-\frac{1}{3}x^3+\frac{1}{3}x^2+x-\frac{2}{3}\)
Câu 3 :
Sửa đề :
\(\frac{3x^2+6x+12}{x^3-8}=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x-2}\)
X^4+4x^3+5x^2+2x+1
Phan tich da thuc thanh nhan tu