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
Phân tích các đa thức sau thành nhân tử
a)4x4+4x3-x2-x
b)x6-x4-9x3+9x2
c)x4-4x3+8x2-16x+16
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) x^6-x^4-9x^3+9x^2
b) x^4-4x^3+8x^2-16x+16
c) (xy+4)^2-4(x+y)^2
d) (a+b+c)^2+(a-b+c)^2-4b^2
Nhờ mng mình đag cần gấp
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 đa thức thành nhân tử
a)8x^3+27
b) 4x^2-4x+1-y^2
c) x^4-2x^3+x^2-2x
d) x^2-4y^2+2x+4y
a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
b) \(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
Phân tích đa thức thành nhân tử
a) x^8+ x^7+ 1
b) 8x^4- 4x^3+ 2x^2+ 9x- 45
c) 16x^4- 16x^3+ 7x^2+ 9x- 9
d) 24x^4- 4x^3- 158x^2+ 25x+ 50
e) 2x^3- x^2- 15x- 22x+ 8
phân tích đa thức thành nhân tử
4x^4+4x^3-x^2-x
x^6-x^4-9x^3+9x^2
x^4-4x^3+8x^2-16x+16
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
\(x^6-x^4-9x^3+9x^2\)
\(=x^2\left(x^4-x^2\right)-x^2\left(9x-9\right)\)
\(=x^2\left(x^4-x^2-9x+9\right)\)
\(=x^2\left(x^4+x^3-9x-x^3-x^2+9\right)\)
\(=x^2\left[x\left(x^3+x^2-9\right)-\left(x^3+x^2-9\right)\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
\(x^4-4x^3+8x^2-16x+16\)
\(=x^4-4x^3+4x^2+4x^2-16x+16\)
\(=x^2\left(x^2-4x+4\right)+4\left(x^2-4x+4\right)\)
\(=\left(x^2-4x+4\right)\left(x^2+4\right)\)
\(=\left(x-2\right)^2\left(x^2+4\right)\)
Phân tích các đa tử sau thành nhân tử
a. 15x^2 – 5x^3
b. 8x^3 +4x^2y – y^3 – 2xy^2
c. x^8 + 64y^4
a: \(15x^2-5x^3=5x^2\left(3-x\right)\)
b: \(8x^3-y^3+4x^2y-2xy^2\)
\(=\left(2x-y\right)\left(4x^2+2xy+y^2\right)+2xy\left(2x-y\right)\)
\(=\left(2x-y\right)\left(4x^2+4xy+y^2\right)\)
\(=\left(2x-y\right)\left(2x+y\right)^2\)
c: Ta có: \(x^8+64y^4\)
\(=x^8+16x^4y^2+64y^4-16x^4y^2\)
\(=\left(x^4+8y^2\right)^2-\left(4x^2y\right)^2\)
\(=\left(x^2-4x^2y+8y^2\right)\left(x^2+4x^2y+8y^2\right)\)
Phân tích đa thức thành nhân tử
a.\(16x^3+0,25yz^3\)
b.\(x^4-4x^3+4x^2\)
c.\(x^3+x^2y-xy^2-y^3\)
d.\(x^3+x^2+x+1\)
e.\(x^4-x^2+2x-1\)
f.\(2x^2-18\)
g.\(x^2+8x+7\)
h.\(x^4y^4+4\)
i.\(x^4+4y^4\)
k.\(x^2-2x-15\)
a: \(16x^3+0,25yz^3\)
\(=0,25\cdot x^3\cdot64+0,25\cdot yz^3\)
\(=0,25\left(64x^3+yz^3\right)\)
b: \(x^4-4x^3+4x^2\)
\(=x^2\cdot x^2-x^2\cdot4x+x^2\cdot4\)
\(=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)
c: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)^2\)
d: \(x^3+x^2+x+1\)
\(=x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+1\right)\)
e: \(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
f: \(2x^2-18\)
\(=2\cdot x^2-2\cdot9\)
\(=2\left(x^2-9\right)=2\left(x-3\right)\left(x+3\right)\)
g: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\cdot\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
h: \(x^4y^4+4\)
\(=x^4y^4+4x^2y^2+4-4x^2y^2\)
\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2y^2+2-2xy\right)\left(x^2y^2+2+2xy\right)\)
i: \(x^4+4y^4\)
\(=x^4+4x^2y^2+4y^4-4x^2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)
k: \(x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)