1) Phân tích thành nhân tử
x^2 - 25 - 2xy + y^2
2x^3 - 1/4
25x^2 - 64y^2
x^4 + 64
2) Tìm x
x(5-3x) +3x(x-3) = 16
3x(x-10)= x-10
B1:Phân tích đa thức thành nhân tử:
1)x2-7x+10
2)x2+3x-5
3)2x2+3x-5
4)2x2+x-6
5)3x2+4x-4
6)3x2-10x-8
7)15x2-11x+2
8)6x2+5x-6
B2:Phân tích đa thức thành nhân tử:
1)(x2+x+1)(x2+x+2)-12
2)x2+2xy+y2-x-y-12
3)x(x+4)(x+6)(x+10)+128
4)x2-2xy+y2+3x-3y-4
B3:Phân tích đa thức thành nhân tử:
a)x2-xz-9y2+3yz
b)x3-x2-5x+125
c)x4-25x2+20x-4
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
\(x^2-7x+10\)
\(=x^2-2x-5x+10\)
\(=x\left(x-2\right)-5\left(x-2\right)\)
\(=\left(x-2\right)\left(x-5\right)\)
học tốt
1, Làm tính nhân : 3xy(x^2-2xy+5)
Phân tích đa thức thành nhân tử : x^2+2xy-25+y^2\
2.Tìm xy biết a) 4x^2+20x=0
b ) x(x+3)-3x-9=0
Bài 2:
a: =>4x(x+5)=0
=>x=0 hoặc x=-5
b: =>(x+3)(x-3)=0
=>x=-3 hoặc x=3
Bài 2:Phân tích đa thức sau thành nhân tử:
6,(x+2).(x+3).(x+4)
7,x^2-2xy+y^2+3x-3y
8,x^4+4
9,4x.(x+1)^2-5x^2.(x+1)-4.(x+1)
10,(1+2x).(1-2x)-(x+2).(x-2)
11,a^2-2a-46^2-46
7,x2-2xy+y2+3x-3y=(x-y)2+3(x-y)=(x-y)(x-y+3)
8,x4+4=(x4+4x2+4)-4x2=(x2+2)2-(2x)2=(x2-2x+2)(x2+2x+2)
9,4x(x+1)2-5x2(x+1)-4.(x+1)=(x+1)\(\left[4x\left(x+1\right)-5x^2-4\right]\)=(x+1)(4x2+4x-5x2-4)=(x+1)(-x2+4x-4)=-(x+1)(x-2)2
7: \(x^2-2xy+y^2+3x-3y\)
\(=\left(x-y\right)^2+3\cdot\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+3\right)\)
8: \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
9: \(4x\left(x+1\right)^2-5x^2\left(x+1\right)-4\left(x+1\right)\)
\(=\left(x+1\right)\left(4x^2+4x-5x^2-4\right)\)
\(=\left(x+1\right)\left(-x^2+4x-4\right)\)
\(=-\left(x+1\right)\left(x-2\right)^2\)
10: \(\left(1+2x\right)\left(1-2x\right)-\left(x+2\right)\left(x-2\right)\)
\(=1-4x^2-x^2+4\)
\(=-5x^2+5\)
\(=-5\left(x-1\right)\left(x+1\right)\)
phân tích thành nhân tử : a) x^2 + 6x + 9 b) x^3 + 3x^2 + 3x + 1 c) 8x^3 - 1/8 d) 10x - 25 - x^2 e) 1/25x^2 - 64y^2
a) \(x^2\)\(+\)\(6x\)\(+\)\(9\)
\(=\left(x+3\right)^2\)
b) \(x^3\)\(+\)\(3x^2\)\(+\)\(3x\)\(+\)\(1\)
\(=\left(x+1\right)^3\)
c) \(8x^3\)\(-\)\(\frac{1}{8}\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
d) \(10x\)\(-\)\(25\)\(-\)\(x^2\)
\(=\)\(-x^2\)\(+\)\(10\)\(-\)\(25\)
\(=-\left(x^2-10+25\right)\)
\(=-\left(x-5\right)^2\)
e) \(\frac{1}{25}x^2\)\(-\)\(64y^2\)
=\(\left(\frac{1}{25}x-8y\right)\left(\frac{1}{5}x+8y\right)\)
Phân tích thành nhân tử
1) 5x-5y+x(x-y)
2) x^2+4x+3
3) x^2-2xy+y^2-z^2
4) x(x-5)-3x+15
5) y^2-x^2+2x-1
6) 7x^2y+14y+7
7) x^3+x^2-4x-4
8) x^2-2x-15
9) x^2+3y-5
10) 2xy+z+2x+yz
1) \(5x-5y+x\left(x-y\right)\)
\(=5\left(x-y\right)+x\left(x-y\right)\)
\(=\left(x-y\right)\left(x+5\right)\)
2) \(x^2+4x+3\)
\(=\left(x^2+x\right)+\left(3x+3\right)\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
3) \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-z^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
4) \(x\left(x-5\right)-3x+15\)
\(=x\left(x-5\right)-3\left(x-5\right)\)
\(=\left(x-5\right)\left(x-3\right)\)
5) \(y^2-x^2+2x-1\)
\(=y^2-\left(x^2-2x+1\right)\)
\(=y^2-\left(x-1\right)^2\)
\(=\left(x+y-1\right)\left(y-x+1\right)\)
\(1.\left(x-y\right)\left(x+5\right)\)
\(2.\left(x+1\right)\left(x+3\right)\)
\(3.\left(x-y-z\right)\left(x-y+z\right)\)
\(4.\left(x-3\right)\left(x-5\right)\)
\(5.\left(y-x+1\right)\left(y+x+1\right)\)
\(7.\left(x+1\right)\left(x-2\right)^2\)
\(8.\left(x-5\right)\left(x+3\right)\)
\(10.\left(y+1\right)\left(2x+z\right)\)
Phân tích thành nhân tử
1) 5x-5y+x(x-y)
2) x^2+4x+3
3) x^2-2xy+y^2-z^2
4) x(x-5)-3x+15
5) y^2-x^2+2x-1
6) 7x^2y+14y+7
7) x^3+x^2-4x-4
8) x^2-2x-15
9) x^2+3y-5
10) 2xy+z+2x+yz
1)
5x - 5y + x ( x - y ) = (x-y)(5+x)
2)
x2+4x+3=x2+x+3x+3=(x+1)(x+3)
3)x2-2xy+y2-z2=\(\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
4)\(x\left(x-5\right)-3x+15=\left(x-3\right)\left(x-5\right)\)
tt thôi nha dài lắmTruong minh quan cái này dễ mà cũng hỏi
bài 1 phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung
6) 9x^3y^2+3x^2y^2
7) x^3+2x^2+3x
8) 6x^2y +4xy^2+2xy
9) 5x^2.(x-2y)-15x.(x-2y)
10) 3.(x-y)-5x.(y-x)
6) \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)
7) \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)
8) \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)
9) \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=5x\left(x-2y\right)\left(x-3\right)\)
10) \(3\left(x-y\right)-5x\left(y-x\right)=\left(x-y\right)\left(3+5x\right)\)
6) 9x3y2 + 3x2y2 = 3x2y2( 3x + 1 )
7) x3 + 2x2 + 3x = x( x2 + 2x + 3 )
8) 6x2y + 4xy2 + 2xy = 2xy( 3x + 2y + 1 )
9) 5x2( x - 2y ) - 15x( x - 2y ) = 5x( x - 2y )( x - 3 )
10 3( x - y ) - 5x( y - x ) = 3( x - y ) + 5x( x - y ) = ( x - y )( 3 + 5x )
a, \(9x^3y^2+3x^2y^2=3x^2y^2\left(3x+1\right)\)
b, \(x^3+2x^2+3x=x\left(x^2+2x+3\right)\)
c, \(6x^2y+4xy^2+2xy=2xy\left(3x+2y+1\right)\)
d, \(5x^2\left(x-2y\right)-15x\left(x-2y\right)=\left(5x^2-15x\right)\left(x-2y\right)=5x\left(x-3\right)\left(x-2y\right)\)
e, \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)
Phân tích đa thức thành nhân tử:
x\(^3\)-64y\(^3\)+3x\(^2\)+3x+1
= (x3+3x2+3x+1)-(4y)3
=(x+1)3-(4y)3
=(x+1-4y)[(x+1)2+(x+1).4y+16y2 ]
=(x+1-4y)[(x2+2x+1)+(4xy+4y)+16y2]
Phân tích đa thức thành nhân tử( bằng mọi phương pháp đã học)a, x^2 - 2x - 4y^2 - 4y b, x^2-4x^2y^2+y^2+2xy c, x^6-x^4+2x^3+2x^2 d, x^3+3x^2+3x+1-8y^3
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)