phan tich da thuc thanh nhan tu bang cach them bot cung 1 hang tu
:\(x^3+x^2+4\)
phan tich da thuc thanh nhan tu bang cach them bot cung 1 hang tu;
\(x^3-2x-4\)
#) TL :
x3 - 2x - 4
= x3 - 4x + 2x - 4
= x( x2 - 4 ) + 2( x - 2)
= x( x -2 )( x + 2) + 2(x-2)
= (x- 2)( x2 + 2x + 2 )
Chúc bn hok tốt ạ :3
Cách 1: Như bạn kia
Cách 2: Muốn thêm bớt thì thêm bớt:)
\(x^3-2x-4=x^3-2x^2+\left(2x^2-2x-4\right)\)
\(=x^2\left(x-2\right)+2\left(x-2\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
Cách 3: Tách hạng tử:
\(x^3-2x-4=\left(x^3-8\right)-\left(2x-4\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)-2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
Cách 4: Tách hạng tử:
\(x^3-2x-4=\frac{1}{2}x^3-2x+\frac{1}{2}x^3-4\)
\(=\frac{1}{2}x\left(x^2-4\right)+\frac{1}{2}\left(x^3-8\right)\)
Dùng hằng đẳng thức tiếp xem có ra không:D
phan tich da thuc thanh nhan tu bang cach them bot cung 1 hang tu;
\(x^8+x^4+1\)
#) TL :
x8 + x4 + 1
= (x4)2 + 2x4 + 1 - x4
= ( x4 + 1 )2 - x4
= ( x4 - x2 + 1 )(x4 + x2 + 1)
= ( x4 - x2 + 1)( x2 - x + 1)( x2 + x + 1 )
Chúc bn hok tốt ạ :3
phan tich da thuc thanh nhan tu
x^2-x-y^2-y
x^2-2xy+y^2-z^2
bai 32 va 33 sbt
lop 8 bai phan tich da thuc thanh nhan tu bang cach nhom hang tu
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
con bai 32, 33 neu ban tra loi duoc minh h them
tim x bang cach phan tich da thuc thanh nhan tu
x3-3x2+3x-1=0
\(x^3-3x^2+3x-1\) =0
=>\(\left(x-1\right)^3\)=0
=>x-1=0
=>x=1
vậy x =1
\(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
phan tich da thuc sau thanh nhan tu bang cach ha bac-tách để xuất hiện ntc là (x-4)
x^4-4x^3-7x^2+22x+24
\(x^4-4x^3-7x^2+22x+24\)
\(=\left(x^4-4x^3\right)-\left(7x^2-28x\right)-\left(6x-24\right)\)
\(=x^4.\left(x-4\right)-7x.\left(x-4\right)-6.\left(x-4\right)\)
\(=\left(x-4\right).\left(x^4-7x-6\right)\)
Tham khảo nhé~
phan tich da thuc sau thanh nhan tu bang canh nhom hang tu x\(^3\) -2x\(^2\) +2x-13
=x^3-2x^2+2x-4-9
=(x-2)(x^2+2)-9
\(=\left(\sqrt{\left(x-2\right)\left(x^2+2\right)}-3\right)\left(\sqrt{\left(x-2\right)\left(x^2+2\right)}+3\right)\)
phan tich da thuc sau thanh nhan tu (bang cach dat nhan tu chung)
(4x-y)(a+b)(4x-y)(c-1)
\((4x-y)(a+b)(4x-y)(c-1)\)
\(=\left(4x-y\right)\left(4x-y\right)=\left(4x-y\right)^{1+1}=\left(4y-2\right)^2\)
\(=\left(a+b\right)\left(4x-y\right)^2\left(c-1\right)\)
(4x-y)(a+b)(4x-y)(c-1)
= ( 4x - y ) ( 4x - y ) = ( 4x - y ) 1 + 1 = ( 4y - 2 ) 2
= (a + b ) ( 4x - y )2 ( c - 1 )
Bài giải :
(4x-y)(a+b)(4x-y)(c-1)
= ( 4x - y ) ( 4x - y ) = ( 4x - y ) 1 + 1
= ( 4y - 2 ) 2
= (a + b ) ( 4x - y )2 ( c - 1 )
phan tich da thuc thanh nhan tu bang pp dung hang dang thuc:
a) 16x^2 - (x^2 + 4)^2
b) 27x^3 - 54x^2 + 36x - 8
c) (x+y)^3 - (x-y)^3
help me
a) 16x2-(x2+4)2= (4x)2-(x2+4)2
= (4x-x2-4)(4x+x2+4)
\(\text{b) 27x^3-54x^2+36x-8=[(3x)^3-3.(3x)^2.2+3.3x.2^2-2^3}]\)
= (3x-2)3
\(\text{c) (x+y)^3 - (x-y)^3= (x+y-x+y)[(x+y)^2+(x+y)(x-y)+(x-y)^2]}\)
=2y(x2+2xy+y2+x2-y2+x2-2xy+y2)
= 2y(3x2+y2)
phan tich da thuc sau thanh nhan tu: 3(x+5)(x+6)(x+7)-8x(2 cach)