Phân tich đa thức thành nhân tử:
a) x3 - 2x - 4
b) x2 + 4x + 3
c) 16x - 5x2 -3
d) 2x2 + 7x +5
e) 2x2 +3x - 5
bài 1 phân tích các đa thức sau thành nhân tử
a) x2 + 4x +3 b) 16x - 5x2 - 3 c) 2x2 + 7x + 5
d) 2x2 + 3x -5 e) x3 - 3x2 + 1 - 3x f ) x2 - 4x - 5
g) (a2 + 1 )2 - 4a2 h) x3 - 3x2 - 4x + 12 i) x4 + x3 + x + 1
k) x4 - x3 - x2 + 1 l ) (2x + 1 )2 - ( x - 1 )
\(a,=\left(x+1\right)\left(x+3\right)\\ b,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ c,=2x^2+2x+5x+5=\left(2x+5\right)\left(x+1\right)\\ d,=2x^2-2x+5x-5=\left(x-1\right)\left(2x+5\right)\\ e,=x^3+x^2-4x^2-4x+x+1=\left(x+1\right)\left(x^2-4x+1\right)\\ f,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
1.Phân tích đa thức sau thành nhân tử
a,x2+4x-3
b,16x-5x2-3
c,2x2+3x-5
d,2x2+3x-5
b) \(16x-5x^2-3=5x\left(3-x\right)-\left(3-x\right)=\left(3-x\right)\left(5x-1\right)\)
c) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
d) \(2x^2+3x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
Phân tích các đa thức sau thành nhân tử:
a,x3+4x-5
b,x3-3x2+4
c,x3+2x2+3x+2
d,x2+2xy+y2+2x-2y-3
e,(x2+3x)2-2(x2+3x)-8
f,(x2+4x+10)2-7(x2+4x+11)+7
a) x3+4x-5 = x3-x2+x2+4x-5=(x3-x2)+(x2-x)+(5x-5)=x2(x-1)+x(x-1)+5(x-1)=(x2+x+5)(x-1)
b) x3-3x2+4=x3-2x2-x2+4=(x3-2x2)-(x2-4)=x2(x-2)-(x-2)(x+2)=(x2-x+2)(x-2)
c) x3+2x2+3x+2=x3+x2+x2+x+2x+2=(x3+x2)+(x2+x)+(2x+2)=x2(x+1)+x(x+1)+2(x+1)=(x2+x+2)(x+1)
d) bạn xem lại đề đúng ko
e) (x2+3x)2-2(x2+3x)-8=x4+6x3+9x2-2x2-6x-8=x4+6x3+7x2-6x-8=x4-x3+7x3-7x2+14x2-14x+8x-8=(x4-x3)+(7x3-7x2)+(14x2-14x)+(8x-8)=x3(x-1)+7x2(x-1)+14x(x-1)+8(x-1)=(x3+7x2+14x+8)(x-1)=(x3+x2+6x2+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)
f) (x2+4x+10)2-7(x2+4x+11)+7=(x2+4x+10)2-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)
a) Ta có: \(x^3+4x-5\)
\(=x^3-x+5x-5\)
\(=x\left(x-1\right)\left(x+1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+5\right)\)
b) Ta có: \(x^3-3x^2+4\)
\(=x^3+x^2-4x^2+4\)
\(=x^2\left(x+1\right)-4\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+4\right)\)
\(=\left(x+1\right)\cdot\left(x-2\right)^2\)
c) Ta có: \(x^3+2x^2+3x+2\)
\(=x^3+x^2+x^2+x+2x+2\)
\(=x^2\left(x+1\right)+x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+x+2\right)\)
d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)
\(=\left(x+y\right)^2+2\left(x+y\right)-3\)
\(=\left(x+y\right)^2+3\left(x+y\right)-\left(x+y\right)-3\)
\(=\left(x+y\right)\left(x+y+3\right)-\left(x+y+3\right)\)
\(=\left(x+y+3\right)\left(x+y-1\right)\)
e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)
\(=\left(x^2+3x\right)\left(x^2+3x-4\right)+2\left(x^2+3x-4\right)\)
\(=\left(x^2+3x-4\right)\left(x^2+3x+2\right)\)
\(=\left(x+4\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)
\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)
\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)
\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)
\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)
Phân tích đa thức thành nhân tử:
a) x3 - 2x2 - 2x - 4
b) xy + 1 - x - y
c) x2 - 4xy + 4y2 - 4y
d) 16 - x2 + 2xy - y2
\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
Phân tích các đa thức sau thành nhân tử:
a) 5x-20xy
b) x2-9
c) x2-2xy+y2-z2
d) 5x.(x-1)-2.(x-1)
e) x2+4x+3
f) x3-x 3x2y+3xy2+y3-y
g) x2-x-y2-y
h) 16x-5x2-3
i) x3-4x
j) 2x2-6x
k) x3- 3x2-4x+12
l) x2-y2-5x+5y
Mn giúp em giải vs em cần gấp để lm bài kiểm tra.Em cảm ơn trc ạ
1. Phân tích thành nhân tử
A) x4 + 2x3 + x2
B) x3 - x + 3x2y + 3xy2 + y3 - y
C) 5x2 - 10xy +5y2 - 20z2
2. Phân tích thành nhân tử
A) x2 + 5x -6
B) 5x2 + 5xy - x - y
C) 7x - 6x2 - 2
3.Phân tích thành nhân tử
A) x2 + 4 + 3
B) 2x2 + 3x -5
C) 16x - 5x2 - 3
4. Tìm x, bt
A) 5x ( x - 1 ) = x -1
B) 2( x + 5 ) -x2 - 5x = 0
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Phân tích đa thức sau thành nhân tử:
a) x2 - 2x - 4y2 - 4y
b) 2x2 + 3x - 5
a: \(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
a)x2-2x-4y2-4y
=x2-2x-4y2-4y+1-1
=(x2-2x+1)-(4y2+4y+1)
=(x-1)2-(2y+1)2
=(x-2y-2)(x+2y)
b)2x2+3x-5
=2x2-2x+5x-5
=2x(x-1)+5(x-1)
=(x-1)(2x+5)
Phân tích đa thức thành nhân tử:
a) xy + y2 – x – y
b) 25 – x2 + 4xy – 4y2
c) 4x3 + 4xy2 + 8x2y – 16x
d) (x2 + x)2 + 4(x2 + x) – 12
e) (x + 1) (x + 2) (x + 3) (x + 4) - 24 g)
h) x2 – 5x + 4
i) x4 – 5x2 + 4
j) x3 – 2x2 + 6x – 5
k) x2 – 4x + 3
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
Bài 1: Phân tích đa thức thành nhân tử:
a) x2y+xy+x+1
b) x2-(a+b)x+ab
c) ax2+ay-bx2-by
d) ax-2x-a2+2a
e) 2x2+4ax+x+2a
f) x3+ax2+x+a
g) x4+2x3-4x-4
a) x2y+xy+x+1= (x2y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)
b) x2-(a+b)x+ab=x2-ax-bx+ab=(x2-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)
c) ax2+ay-bx2-by=(ax2+ay)-(bx2+by)=a(x2+y)-b(x2+y)=(a-b)(x2+y)
d) ax-2x-a2+2a=(ax-2x)-(a2-2a)=x(a-2)-a(a-2)=(a-2)(x-a)
e) 2x2+4ax+x+2a=(2x2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)
f) x3+ax2+x+a=(x3+ax2)+(x+a)=x2(x+a)+(x+a)=(x2+1)(x+a)
g: Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^2-2\right)\left(x^2+2\right)-2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
Phân tích các đa thức sau thành nhân tử:
4x2y2 - ( x2 + y2 - a2)2
x3 - 1 + 5x2 - 5 +3x - 3
( x - y)2 + 4(x-y) + 4
x2 -2x( 3x+1) + (3x+1)2
x4 + 2x2(2x+1) + ( 2x+1)2
\(\left(x-y\right)^2+4\left(x-y\right)+4\)
\(=\left(x-y\right)^2+2.\left(x-y\right).2+2^2\)
\(=\left(x-y+2\right)^2\)
hk tốt
^^