phan tich da thuc sau thanh nhan tu bang nhom hang tu x\(^2\) -(a+b).x+ab
ax-2x-a\(^2\) +2a
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 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;
\(x^3-2x-4\)
#) 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
:\(x^3+x^2+4\)
\(x^3+x^2+4\)
\(=x^3-x^2+2x^2+2x-2x+4\)
\(=\left(x^3-x^2+2x\right)+\left(2x^2-2x+4\right)\)
\(=x\left(x^2-x+2\right)+2\left(x^2-x+2\right)\)
\(=\left(x^2-x+2\right)\left(x+2\right)\)
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 sau thanh nhan tu ab(x^2+y^2)-xy(a^2+b^2)
\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)
\(=abx^2+aby^2-a^2xy-b^2xy\)
\(=ax\left(bx-ay\right)+by\left(ay-bx\right)\)
\(=ax\left(bx-ay\right)-by\left(bx-ay\right)\)
\(\left(bx-ay\right)\left(ax-by\right)\)
hãy k nếu bạn thấy đây là câu tl đúng :)
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 thanh nhan tu:
a) (a-b)2012x+2012y-ax-ay
b) (a-b) 144y2-x2+2x-1
a)(a-b)2012x+2012y-ax-ay
=(a-b)[x(2012-a)+y(2012-a0]
=(a-b)(2012-a)(x+y)
b)(a-b)144y2-x2+2x-1
=(a-b)[144y2-(x2-2x+1)]
=(a-b)[(12y)2-(x-1)2]
=(a-b)(12y+x-1)(12y-x+1)
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 cat da thuc sau thanh nhan tu
a) x^4 + 1 - 2x^2
\(x^4+2x^2+1=\left(x^2+1\right)^2\) (Nhớ k cho mình với nhé!)