1.phân tích đa thức thành nhân tử
a. 8x^3-32x
b.y^3+64+(y+4)(y-16)
bài 2
a.tìm x biết: 4x^3-9x=0
b . tính giá tri biêu thức
A=x^3-9x^2+27x-27 tại x=203
Phân tích đa thức thành nhân tử
a, 1-4x^2
b, 8-27x^3
c, 27+27x+9x^2+x^3
d, 8x^3-12x^2y+6xy^2-y^3
e, x^2+4x-5
Bài làm:
a, 1-4x2
=1-(2x)2
=(1-2x).(1+2x)
b, 8-27x3
=23-(3x)3
=(2-3x).(4+6x+9x2)
Các câu còn lại bạn dùng hằng đẳng thức là phân tích được ra thôi
1 - 4x^2
= 1^2 - ( 2x )^2
= ( 1 - 2x ) ( 1 + 2x )
8 - 27x^ 3
= 2^3 - ( 3x )^3
= ( 2 - 3x ) [ 2^2 + 2 * 3x + ( 3x )^2 ]
= ( 2 - 3x ) ( 4 + 6x + 9x^2 )
= ( 2 - 3x ) ( 9x^2 + 6x + 4 )
27 + 27x + 9x^2 + x^3
= x^3 + 9x^2 + 27x + 27
= x^3 + 3x^2 + 6x^2 + 18x + 9x + 27
= x^2 ( x + 3 ) + 6x ( x + 3 ) + 9 ( x + 3 )
= ( x + 3 ) ( x^2 + 6x + 9 )
= ( x + 3 ) ( x + 3 )^2
= ( x + 3 )^3
x^2 + 4x - 5
= x^2 - x + 5x - 5
= x ( x - 1 ) + 5 ( x - 1 )
= ( x + 1 ) ( x - 5 )
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
9 Phân tích đa thức sau thành nhân tử:
a) 9xy^2-18x^2y ; b) 6x^2-2y ; c)7x(x-y)-14y(y-x)
d)7-x^2 ; e) 16+8x+x^2 ; f)1-27x^3
g) x^3-9x^2+27x-27 ; h) (x+2y)^2-16y^2 ; i) x^3-64y^3
Phân tích các đa thức sau đây thành nhân tử
a, 36x^2 - ( 3x -2 ) ^2
b, 16(4x+5)^5 - 25 (2x+2)^2
c, ( x - y + 4 )^2
d, (x+1)^4 - (x-1)^4
e, 16x^2 - 24xy + 9y^2
f, -x^4/4 + 2x^2y^3 - 4y^6
g , 64x^3 +1
h, x^3y^6z^9 - 125
k, 27x^6 - 8x^3
I , x^6 - y^6
m, 27x^3 - 54x^2y + 36xy^2 - 8y^3
n, y^9 - 9x^2y^6 + 27x^4y^3 - 27x^6
làm ơn giải chi tiết giúp mik vs ạ , cảm ơn
a: =(6x)^2-(3x-2)^2
=(6x-3x+2)(6x+3x-2)
=(9x-2)(3x+2)
d: \(=\left[\left(x+1\right)^2-\left(x-1\right)^2\right]\left[\left(x+1\right)^2+\left(x-1\right)^2\right]\)
\(=4x\cdot\left[x^2+2x+1+x^2-2x+1\right]\)
=8x(x^2+1)
e: =(4x)^2-2*4x*3y+(3y)^2
=(4x-3y)^2
f: \(=-\left(\dfrac{1}{4}x^4-2\cdot\dfrac{1}{2}x^2\cdot2y^3+4y^6\right)\)
\(=-\left(\dfrac{1}{2}x^2-2y^3\right)^2\)
g: =(4x)^3+1^3
=(4x+1)(16x^2-4x+1)
k: =x^3(27x^3-8)
=x^3(3x-2)(9x^2+6x+4)
l: =(x^3-y^3)(x^3+y^3)
=(x-y)(x+y)(x^2-xy+y^2)(x^2+xy+y^2)
Phân tích các đa thức sau thành nhân tử
a.1-2y+y^2
b.(x+1)^2 - 25
c.1-4x^2
d.8-27x^3
e.27+27x+9x^2+6xy
f.8x^3-12x^2y+6xy^2-y^3
g.x^3+8y^3
\(\left(x-1\right)^2-25\)
\(=x^2-2x+1-25\)
\(=x^2-2x-24\)
\(=x^2-6x+4x-24\)
\(=x.\left(x-6\right)+4.\left(x-6\right)\)
\(=\left(x+4\right).\left(x-6\right)\)
a, \(1-2y+y^2=\left(y+1\right)^2=\left(y+1\right)\left(y+1\right)\)
b, \(\left(x+1\right)^2-25=\left(x+1\right)^2-5^2=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
c, \(1-4x^2=1^2-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
d, \(8-27x^3=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a)=(1-y)2
b)=(x+1)2-52
=(x+1+5)(x+1-5)
=(x+6)(x-4)
c)=12-(2x)2
=(1+2x)(1-2x)
d)=23-(3x)3
=(2-3x)(4+6x+9x2)
e)=33+3.9.x+3.3.x2+x3
=(3+x)3
`1-27x^3`
`x-3^3 +27`
`27x^3 +27x^2 +9x+1`
`(x^6)/27 - (x^4 y)/3 +x^2 y-y^3`
Phân tích thành nhân tử
\(1-27x^3\)
\(=1-\left(3x\right)^3\)
\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)
\(---\)
\(x-3^3+27\)
\(=x-27+27=x\)
\(---\)
\(27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(---\)
\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)
\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)
\(=\left(\dfrac{x^2}{3}-y\right)^3\)
#Ayumu
1-27x\(^3\)
=(1-3x)(1+3x+9x\(^2\)
phân tích các đa thức sau thành nhân tử
a, 4x^4 + 4x^3 - x^2 - x
b, x^6 - x^4 - 9x^3 + 9x^2
c, x^4 - 4x^3 + 8x^2 - 16x + 16
a) \(4x^4+4x^3-x^2-x=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=\left(4x^3-x\right)\left(x+1\right)=x\left(4x^2-1\right)\left(x+1\right)\)
\(=x\left\{\left(2x\right)^2-1\right\}\left(x+1\right)=x\left(2x-1\right)\left(2x+1\right) \left(x+1\right)\)
c) \(x^4-4x^3+8x^2-16x+16=x^4+8x^2+16-\left(4x^3+16x\right)\)
\(=\left(x^2+4\right)^2-4x\left(x^2+4\right)=\left(x^2-4x+4\right)\left(x^2+4\right)=\left(x-2\right)^2\left(x^2+4\right)\)
b) \(x^6-x^4-9x^3+9x^2=x^4\left(x^2-1\right)-\left(9x^3-9x^2\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)-9x^2\left(x-1\right)\)
\(=\left(x^5+x^4-9x^2\right)\left(x-1\right)=\left(x-1\right)x^2\left(x^3+x^2-9\right)\)
Phân tích đa thức thành nhân tử: 1, x^3+2x^2-6x-27 2, 9x^2+6x-4y^2-4y 3, 12x^3+4x^2-27x-9
1. \(x^3+2x^2-6x-27=\left(x-3\right)\left(x^2+5x+9\right)\)
2. \(9x^2+6x-4y^2-4y=\left(9x^2-4y^2\right)+\left(6x-4y\right)\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)=\left(3x-2y\right)\left(3x+2y+2\right)\)
3. \(12x^3+4x^2-27x-9=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x^2-\dfrac{9}{4}\right)=\left(x+\dfrac{1}{3}\right)\left(x+\dfrac{3}{2}\right)\left(x-\dfrac{3}{2}\right)\)
1) Ta có: \(x^3+2x^2-6x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)+2x\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
2: Ta có: \(9x^2+6x-4y^2-4y\)
\(=\left(3x-2y\right)\left(3x+2y\right)+2\left(3x-2y\right)\)
\(=\left(3x-2y\right)\left(3x+2y+2\right)\)
Phân tích đa thức thành nhân tử
a.\(x^4+4\)
b.\(x^3-9x^2+6x+16\)
c.(x+2)(x+3)(x+4)(x+5)-24
a.
\(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b.
\(x^3-9x^2+6x+16=\left(x^3-7x^2-8x\right)-\left(2x^2-14x-16\right)\)
\(=x\left(x^2-7x-8\right)-2\left(x^2-7x-8\right)\)
\(=\left(x-2\right)\left(x^2-7x-8\right)=\left(x-2\right)\left(x^2+x-8x-8\right)\)
\(=\left(x-2\right)\left[x\left(x+1\right)-8\left(x+1\right)\right]=\left(x-2\right)\left(x+1\right)\left(x-8\right)\)
c.
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10+2\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10\right)^2-4\left(x^2+7x+10\right)+6\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+10-4\right)+6\left(x^2+7x+10-4\right)\)
\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)