phan tich thanh nhan tu bang cach tach hang tu
a) x3-x2-x-2
b) x3-6x2-x+30
phan tich thanh nhan tu(bang ki thuat tach hang tu)
1)-8x^2+5x+3
2)-10x^2-17x+6
\(1.-8x^2+5x+3=-8x^2+8x-3x+3=-8x\left(x-1\right)-3\left(x-1\right)=-\left(8x+3\right)\left(x-1\right)\)
\(2.-10x^2-17x+6=-10x^2-20x+3x+6=-10x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(3-10x\right)\)
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;
\(x^8+x^4+1\)
#) 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 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 thanh nhan tu(bang ki nang tach hang tu)
1) 3x^2 +7x-6
2)6x^213x+6
3)8x^2+5x+3
1) 3x^2 +7x-6
=3x2-2x+9x-6
=x.(3x-2)+3.(3x-2)
=(3x-2)(x+3)
2)6x^213x+6
đề khùng
3)8x^2+5x+3
8x2-3x+8x+3
Đề khùng
phan tich da thuc thanh nhan tu bang phuong phap tach
a, x2 - 2x - 3
x2-2x-3 = x2-3x+x-3 = x(x-3) + (x-3) = (x+1)(x-3)
CHọn mình nha :)
\(x^2-2x-3\)
\(=x^2+x-3x-3\)
\(=x\left(x+1\right)-3\left(x+1\right)\)
\(=\left(x+1\right)\left(x-3\right)\)
hk tot
^^
\(x^2-2x-3\)
\(=x^2+x-3x-3\)
\(=x\left(x+1\right)-3\left(x+1\right)\)
\(=\left(x+1\right)\left(x-3\right)\)
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 thanh nhan tu ( bang ki thuat bo sung hang dang thuc):
x^2-4xy-x+3y^2+3y
x^2+4xy+2x+3y^2+6y
6x^2+xy-7x-2y^2+7y-5
6a^2-ab-2b^2+a+4b