bài 51
\(x^3-2x^2+x\)
\(2x^2+4x+2-2y^2\)
\(2xy-x^2-y^2+16\)
a)x^3-2x62+x
b)2x^2+4x+2-2y^2
c)2xy-x^2-4^2+16
d)x^3+2x^2y+xy^2-yx
e)2x-2y-x^2+2xy-y^2
Bài 1: Phân tích đa thức thành nhân tử
1. 5x-10-xy+2y
2.2x^2+2y^2-4xy-xz+yz
3.5x^2y-10xy^2
4.3x^2-6xy+3y^2-12z^2
5.x^2+4xy-16+4y^2
6.7x-6x^2-2
7.(2x+y)^2+x(2x+y)
8.x(x-y)+5x-5y
9.x^2-y^2+2x+1
10.x^3-9x
11.xy-2y+x-2
12.x^3-3x^2-4x+12
13.3x-x^2-2xy+3y-y^2
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
Bài 1: Rút gọn biểu thức
A, ( x – 3 )^2 – ( x + 2 )^2
B, ( 4x^2 + 2xy + y^2 )( 2x – y ) – ( 2x + y )( 4x^2 – 2xy + y^2 )
C, ( 2x + 1 )^2 + 2( 4x^2 – 1 ) + ( 2x – 1 )^2
D, ( x – 3 )( x + 3 ) – ( x – 3 )
Bài 2: Phân tích đa thức thành nhân tử
A, a^2 – ab + a – b
B, m^4 – n^6
C, x^2 + 6x + 8
D, 2x^2 + 4x + 2 – 2y^2
Bài 3: Tìm x
A, x^2 – 16 = 0
B, x^4 – 2x^3 + 10x^2 – 20x = 0
C, 15 – 2x – x^2 = 0
D, ( x^2 – 1/2x ) : 2x – ( 3x – 1 ) : ( 3x – 1 ) = 0
Giúp em với ạ !!!
A) \(\left(x-3\right)^2-\left(x+2\right)^2\)
\(=\left(x-3-x-2\right)\left(x-3+x+2\right)\)
\(=-5.\left(2x-1\right)\)
B) \(\left(4x^2+2xy+y^2\right)\left(2x-y\right)-\left(2x+y\right)\left(4x^2-2xy+y^2\right)\)
\(=\left(2x\right)^3-y^3-\left[\left(2x\right)^3+y^3\right]\)
\(=8x^3-y^3-8x^3-y^3\)
\(=-2y^3\)
C) \(x^2+6x+8\)
\(=x^2+6x+9-1\)
\(=\left(x+3\right)^2-1\)
\(=\left(x+3-1\right)\left(x+3+1\right)\)
\(=\left(x+2\right)\left(x+4\right)\)
bài 3 A) \(x^2-16=0\)
\(\left(x-4\right)\left(x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
vậy \(\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
B) \(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^3+10x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x\left(x^2+10\right)=0\\x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
vậy \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
phân tích thành nhân tử
`3x^2 -3xy-5x+5y`
`2x^3 y-2xy^3 -4xy^2 -2xy`
`x^2 -1+2x-y^2`
`x^2 +4x-2xy-4y+4y^2`
`x^3 -2x^2 +x`
`2x^2 +4x+2-2y^2`
a) \(3x^2-3xy-5x+5y\)
\(=\left(3x^2-3xy\right)-\left(5x-5y\right)\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
b) \(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left[x^2-\left(y+1\right)^2\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
c) \(x^2+1+2x-y^2\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
d) \(x^2+4x-2xy-4y+y^2\)
\(=\left(x^2-2xy+y^2\right)+\left(4x-4y\right)\)
\(=\left(x-y\right)^2+4\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y+4\right)\)
e) \(x^3-2x^2+x\)
\(=x\left(x^2-2x+1\right)\)
\(=x\left(x-1\right)^2\)
f) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)+y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x-y+1\right)\left(x+y+1\right)\)
a: =3x(x-y)-5(x-y)
=(x-y)(3x-5)
b: \(=2xy\left(x^2-y^2-2y-1\right)\)
\(=2xy\left[x^2-\left(y^2+2y+1\right)\right]\)
\(=2xy\left(x-y-1\right)\left(x+y+1\right)\)
d:
Sửa đề: x^2+4x-2xy-4y+y^2
=x^2-2xy+y^2+4x-4y
=(x-y)^2+4(x-y)
=(x-y)(x-y+4)
e: =x(x^2-2x+1)
=x(x-1)^2
f: =2(x^2+2x+1-y^2)
=2[(x+1)^2-y^2]
=2(x+1+y)(x+1-y)
phân tích các đa thức sau thành nhân tử bằng nhiều phương pháp
1) x^2-1+2yx+y^2
2) x^4-x^3-x+1
3) 7x^2-63y^2
4) 4x^2+1-4x-y^2
5) x^2(y-x)+x-y
6) x^2+1+2x-y^2
7) x^3-2x^2+x
8) 2x^2+4x+2-2y^2
9) 2xy-x^2-y^2+16
10) 2x-2y-x^2+2xy-y^2
1) x^2-1+2xy+y^2 = (x^2+2xy+y^2)-1 = (x+y)^2 - 1^2 = (x+y-1)*(x+y+1)
2) x^4-x^3-x+1 = (x^4-x)-(x^3-1) = x*(x^3-1)-(x^3-1) = (x^3-1)*(x-1)
3) 7x^2-63y^2 = 7*(x^2-9y^2) = 7*[x^2-(3y)^2] = 7*(x-3y)*(x+3y)
còn lại bn tự tính ik nha
Bài 1: Tính giá trị:
A= x^2+4y^2-2x+10+4xy-4y tại x+2y=5
B= (x^2+4xy+4y^2)-2(x+2y)(y-1)+y^2-2y+1 tại x+y=5
C= x^2-y^2-4x tại x+y=2
D= x^2+y^2+2xy-4x-4y-3 tại x+y=4
E= 2x^6+3x^3y^3+y^6+y^3 tại x^3+y^3=1
Bài 2: Chứng minh rằng
a) -9x^2+12x-5<0
b) 4/9x^2-4x+9/2>0
Bài 3: Tìm giá trị lớn nhất:
A= 4-2x^2
B=(1-x)(2+x)(3+x)(6+x)
C=-2x^2-y^2-2xy+4x+2y+5
D=-9x^2+24x-18
E=-x^4+2x^3-3x^2+4x-1
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
a) x^2+4x+4-y^2
b) x^2-16-4xy+4y^2
c) x^3+2x^2y +xy^2
d) 5x+5y-x^2-2xy-y^2
e) x^5-x^4+x^3-x^2
a) \(x^2+4x+4-y^2\)
\(=\left(x^2+2.x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(a,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ b=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\\ c,=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\\ d,=5\left(x+y\right)-\left(x+y\right)^2=\left(5-x-y\right)\left(x+y\right)\\ e,=x^4\left(x-1\right)+x^2\left(x-1\right)\\ =x^2\left(x^2+1\right)\left(x-1\right)\)
a: \(x^2+4x+4-y^2=\left(x+2-y\right)\left(x+2+y\right)\)
b: \(x^2-4xy+4y^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\)
c: \(x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\)
a. 3x^2-3y^2-x-y
b. 2x^2+4xy-16+2y^2
c. -x^2-x+2
d. 3x^2-7x+4
e.-2x^2+3x-1
f. x^2+2xy+y^2-2x-2y
g.x^3-2x^2+1
h.4x^2-3x-1
k. 2x^2+5x+3
l. x^2-2x-y^2+1
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)
a) \(3x^2-3y^2-x-y\)
\(\Leftrightarrow3\left(x^2-y^2\right)-x-y\)
\(\Leftrightarrow3\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(\Leftrightarrow3\left(x-y\right)\)
d) \(3x^2-7x+4\)
\(\Leftrightarrow3x^2-7x+7-3\)
\(\Leftrightarrow\left(3x^2-3\right)-\left(7x-7\right)\)
\(\Leftrightarrow3\left(x^2-1\right)-7\left(x-1\right)\)
\(\Leftrightarrow3\left(x-1\right)\left(x+1\right)-7\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(3\left(x+1\right)-7\right)\)
\(\Leftrightarrow\left(x+1\right)\left(3x-6\right)\)
e) \(-2x^2+3x-1\)
\(\Leftrightarrow\left(-2x^2-1^2\right)+3x\)
\(\Leftrightarrow\left(-2x-1\right)\left(-2x+1\right)+3x\)
f) \(x^2+2xy+y^2-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
k) \(2x^2+5x+3\)
\(\Leftrightarrow2x^2+2x+3x+3\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)\)
l) \(x^2-2x-y^2+1\)
\(\Leftrightarrow\left(x^2-2x+1\right)-y^2\)
\(\Leftrightarrow\left(x-1\right)^2-y^2\)
\(\Leftrightarrow\left(x-1-y\right)\left(x-1+y\right)\)