tim x bang cach phan tich da thuc thanh nhan tu
x3-3x2+3x-1=0
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
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 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 cach them bot cung 1 hang tu
:\(x^3+x^2+4\)
\(x^3+x^2+4\)
\(=x^3-x^2+2x^2+2x-2x+4\)
\(=\left(x^3-x^2+2x\right)+\left(2x^2-2x+4\right)\)
\(=x\left(x^2-x+2\right)+2\left(x^2-x+2\right)\)
\(=\left(x^2-x+2\right)\left(x+2\right)\)
x3 + x2 + 4
= x3+ x2 + 4 + 43 - 43
= (x + 4)3 - 43
= [(x+ 4 - 4)] [(x+4)2+ (x+4).4 + 42]
#) TL :
x3 + x2 + 4
= x3 - x2 + 2x2 + 2x - 2x + 4
= x( x2 - x + 2 ) + 2( x2 - x + 2 )
= ( x + 2 )( x2- x + 2 )
Chúc bn hok tốt ạ ;3
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 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 thanh nhan tu 1-3x-x^3+3x^2
\(1-3x-x^3+3x^2\)\(=\left(1-x^3\right)+\left(3x^2-3x\right)\)
\(=\left(1-x\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(3x-x^2-x-1\right)=\left(x-1\right)\left(2x-x^2-1\right)\)
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 :
2x(x-1)-3x+3
2x(x-1)-3x+3
=2x(x-1)-(3x-3)
=2x(x-1)-3(x-1)
=(x-1)(2x-3)