1) Phân tích thành nhân tử
a) 2x^2 ( x-1 ) - 1( 1-x ). 4x
b) 9x^2y^2 + 15 x^2y^3- 21x^3y^2
c) ( 3x-1)^2 - 9( x+2)^2
d) 9 - x^2 - 2xy - y^2
e) 16x^2y - 4xy^2 - 4x^3 + x^2y
f) x^3 + x^2 - 2x - 8
Nhờ mọi người giúp mk với nhé
Phân tích mỗi đa thức sau thành nhân tử
a)x^3-2x^2y+xy^2+xy
b)x^3+4x^2y+4xy^2-9x
c)x^3-y^3+x-y
d)4x^2-4xy+2x-y+y^2
e)9x^2-3x+2y-4y^2
f)3x^2-6xy+3y^2-5x+5y
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Phân tích đa thức thành nhân tử
a) x2(x-2y)+5x(x-2y)
b)x2(x-1)+2x(x-1)+x-1
c)(x2+1)2-4x2
d)3(x+3)-x2+9
e)x4- 1/9x2
f)2x2-2xy-16x+16y
g) x3+4x2y-16x+4xy2
a) \(=\left(x-2y\right)\left(x^2+5x\right)\)
b) \(=\left(x-1\right)\left(x^2+2x+1\right)=\left(x-1\right)\left(x+1\right)^2\)
c) \(=\left(x^2+1-2x\right)\left(x^2+1+2x\right)\)
\(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)^2\)
d) \(=3\left(x+3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x+3\right)\left(3-x+3\right)\)
\(=\left(x+3\right)\left(6-x\right)\)
e) \(=\left(x^2-\frac{1}{3}x\right)\left(x^2+\frac{1}{3}x\right)\)
f) \(=2x\left(x-y\right)-16\left(x-y\right)\)
\(=2\left(x-y\right)\left(x-8\right)\)
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, \(3\left(x-y\right)-5x\left(y-x\right)=3\left(x-y\right)+5x\left(x-y\right)=\left(3+5x\right)\left(x-y\right)\)
B1: quy đồng mẫu số các phân thức:
a. 5/ 6x^2y ; 7/ 12xy^2 ; 11/ 18xy
b. 4x+2/ 15x^3y ; 5y - 3/ 9x^2y ; x+1/5xy^3
c. 3/2x ; 3x-3/2x-1 ; 3x-2/2x- 4x^2
d. x^3 + 2x / x^3+1 ; 2x/ x^2 - x +1 ; 1/ x+1
e. y/ 2x^2 - xy ; 4x/ y^2 - 2xy
f. 1/x+2 ; 3/ x^2 - 4 ; x-14/ ( x^2 + 4x + 4 ) (x-2)
g. 1/x+2 ; 1/ (x+2)(4x+7) ;
h. 1/x+3 ; 1/ (x+3)(x+2) ; 1/ (x+2)(4x+7)
B2: dùng quy tắc đổi dấu để tìm mẫu thức chung :
a.4/ x+2 ; 2/x-2 ; 5x-6/4-x^2
b. 1-3x/2x ; 3x-2/2x-1 ; 3x-2/2x-4x^2
c. 1/ x^2 + 6x + 9 ; 1/ 6x-x^2-9 ; x/ x^2 -9
d. x^2 + 2/ x^3 - 1 ; 2/ x^2 + x +1 ; 1/ 1-x
e. x/ - 2y ; x/ x+2y ; 4xy/ 4y^2 - x^2
Ai làm xong trước mình tick 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 1 : Phân tích đa thức thành nhân tử
a) 5x^2y-20xy^2
b) 1-8x+16x^2-y^2
c) 4x-4-x^2
d) x^3-2x^2+x-xy^2
e)27-3x^2
f) 2x^2+4x+2-2y^2
Bài 2: tìm x, biết
a) x^2(x-2023)-2023+x=0
b) -x(x-4)+(2x^3-4x^2-9x):x=0
c) x^2+2x-3x-6=0
d) 3x(x-10)-2x+20=0
Bài 1
a) 5x²y - 20xy²
= 5xy(x - 4y)
b) 1 - 8x + 16x² - y²
= (1 - 8x + 16x²) - y²
= (1 - 4x)² - y²
= (1 - 4x - y)(1 - 4x + y)
c) 4x - 4 - x²
= -(x² - 4x + 4)
= -(x - 2)²
d) x³ - 2x² + x - xy²
= x(x² - 2x + 1 - y²)
= x[(x² - 2x+ 1) - y²]
= x[(x - 1)² - y²]
= x(x - 1 - y)(x - 1 + y)
= x(x - y - 1)(x + y - 1)
e) 27 - 3x²
= 3(9 - x²)
= 3(3 - x)(3 + x)
f) 2x² + 4x + 2 - 2y²
= 2(x² + 2x + 1 - y²)
= 2[(x² + 2x + 1) - y²]
= 2[(x + 1)² - y²]
= 2(x + 1 - y)(x + 1 + y)
= 2(x - y + 1)(x + y + 1)
Bài 2:
a: \(x^2\left(x-2023\right)+x-2023=0\)
=>\(\left(x-2023\right)\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x-2023=0
=>x=2023
b:
ĐKXĐ: x<>0
\(-x\left(x-4\right)+\left(2x^3-4x^2-9x\right):x=0\)
=>\(-x\left(x-4\right)+2x^2-4x-9=0\)
=>\(-x^2+4x+2x^2-4x-9=0\)
=>\(x^2-9=0\)
=>(x-3)(x+3)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: \(x^2+2x-3x-6=0\)
=>\(\left(x^2+2x\right)-\left(3x+6\right)=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>(x+2)(x-3)=0
=>\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d: 3x(x-10)-2x+20=0
=>\(3x\left(x-10\right)-\left(2x-20\right)=0\)
=>\(3x\left(x-10\right)-2\left(x-10\right)=0\)
=>\(\left(x-10\right)\left(3x-2\right)=0\)
=>\(\left[{}\begin{matrix}x-10=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=10\end{matrix}\right.\)
Câu 1:
a: \(5x^2y-20xy^2\)
\(=5xy\cdot x-5xy\cdot4y\)
\(=5xy\left(x-4y\right)\)
b: \(1-8x+16x^2-y^2\)
\(=\left(16x^2-8x+1\right)-y^2\)
\(=\left(4x-1\right)^2-y^2\)
\(=\left(4x-1-y\right)\left(4x-1+y\right)\)
c: \(4x-4-x^2\)
\(=-\left(x^2-4x+4\right)\)
\(=-\left(x-2\right)^2\)
d: \(x^3-2x^2+x-xy^2\)
\(=x\left(x^2-2x+1-y^2\right)\)
\(=x\left[\left(x^2-2x+1\right)-y^2\right]\)
\(=x\left[\left(x-1\right)^2-y^2\right]\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
e: \(27-3x^2\)
\(=3\left(9-x^2\right)\)
\(=3\left(3-x\right)\left(3+x\right)\)
f: \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left[\left(x^2+2x+1\right)-y^2\right]\)
\(=2\left[\left(x+1\right)^2-y^2\right]\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
Bài 2
a) x²(x - 2023) - 2023 + x = 0
x²(x - 2023) - (x - 2023) = 0
(x - 2023)(x² - 1) = 0
x - 2023 = 0 hoặc x² - 1 = 0
*) x - 2023 = 0
x = 2023
*) x² - 1 = 0
x² = 1
x = 1 hoặc x = -1
Vậy x = -1; x = 1; x = 2023
b) -x(x - 4) + (2x³ - 4x² - 9x) : x = 0
-x² + 4x + 2x² - 4x - 9 = 0
x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
Vậy x = 3; x = -3
c) x² + 2x - 3x - 6 = 0
(x² + 2x) - (3x + 6) = 0
x(x + 2) - 3(x + 2) = 0
(x + 2)(x - 3) = 0
x + 2 = 0 hoặc x - 3 = 0
*) x + 2 = 0
x = -2
*) x - 3 = 0
x = 3
Vậy x = -2; x = 3
d) 3x(x - 10) - 2x + 20 = 0
3x(x - 10) - (2x - 20) = 0
3x(x - 10) - 2(x - 10) = 0
(x - 10)(3x - 2) = 0
x - 10 = 0 hoặc 3x - 2 = 0
*) x - 10 = 0
x = 10
*) 3x - 2 = 0
3x = 2
x = 2/3
Vậy x = 2/3; x = 10
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)
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
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
Bài1: phân tích đa thức thành nhân tử
1) 21x^2y - 12xy^2
2) x^3 + x^2 - 2x
3) 3x. (x - 1) + 7x^2. (x - 1)
4) 3x. (x-a) + 4a. (a-x)
5) 1/2x. (x-2) + 4a. (a-x)
6) 21. (x-y)^2 - 7.(y-x)
7) x^2yz + xy^2z^2 + x^2yz^2
8) 9x^2y^2 + 15x^2y - 21xy^2
9) x^2y^2 - 1
10) x^4y^4 - z^4
11) (x+1)^2 - 24
12) (x+1)^2 - (y+6)^2
13) x^6 + 1
14) -4y^2 + 4y - 1
15) (2a + 3)^2 - (2a + 1)^2
Bài2: tìm x, biết:
a) x^4 - 16x =0
b) x. (x-3) - x +3 =0
c) 4x^2 - 1/4 =0
d) x^3 - 3x^2 + 3x - 1=0
e) 8x^3 - 36x^2 + 54x - 27=0
f) x^2 + 4x = -4
g) x^2 = 6x - 9
Bài 2;
\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)
\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)
\(x.\left(x-3\right)-x+3=0\)
\(x.\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
\(x^3-3x^2+3x-1=0\)
\(\left(x-1\right)^3=0\)( hằng đẳng thức số 5 )
\(\Rightarrow x=1\)
Vậy \(x=1\)