Bài 1. Phân tích đa thức sau thành nhân tử
\(x^4-8x\)
K
Khách
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24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
NL
1
KV
30 tháng 11 2023
a) x⁴ - y⁴
= (x²)² - (y²)²
= (x² - y²)(x² + y²)
= (x - y)(x + y)(x² + y²)
b) 1 - 8x³y⁶
= 1³ - (2xy²)³
= (1 - 2xy²)(1 + 2xy² + 4x²y⁴)
NK
2
XO
22 tháng 8 2021
x4 - 4x3 - 8x2 + 8x
= x(x3 - 4x2 - 8x + 8)
= x[x3 + 8 - 4x(x + 2)]
= x[(x + 2)(x2 - 2x + 4) - 4x(x + 2)]
= x(x + 2)(x2 - 6x + 4)
= x(x + 2)(x2 - 6x + 9 - 5)
= \(x\left(x+2\right)\left[\left(x-3\right)^2-5\right]=x\left(x+2\right)\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)\)
22 tháng 8 2021
\(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left(x^3-6x^2+2x^2+4x-12x+8\right)\)
\(=x\left[\left(x^3-6x^2+4x\right)+\left(2x^2-12x+8\right)\right]\)
\(=x\left[x\left(x^2-6x+4\right)+2\left(x^2-6x+4\right)\right]\)
\(=x\left(x^2-6x+4\right)\left(x+2\right)\)
\(=x\left[\left(x-3\right)^2-\left(\sqrt{5}\right)^2\right]\left(x+2\right)\)
\(=x\left(x-3-\sqrt{5}\right)\left(x-3+\sqrt{5}\right)\left(x+2\right)\)
T
1
4 tháng 11 2021
\(=\left(x^2+x+4\right)^2+3x\left(x^2+x+4\right)+5x\left(x^2+x+4\right)+15x^2\\ =\left(x^2+x+4\right)\left(x^2+x+4+3x\right)+5x\left(x^2+x+4+3x\right)\\ =\left(x^2+x+4+3x\right)\left(x^2+x+4+5x\right)\\ =\left(x^2+4x+4\right)\left(x^2+6x+4\right)\\ =\left(x+2\right)^2\left(x^2+6x+4\right)\)
QA
29 tháng 7 2021
Trả lời:
a, x4 + 3x3 + x2 + 3x
= ( x4 + 3x3 ) + ( x2 + 3x )
= x3 ( x + 3 ) + x ( x + 3 )
= ( x3 + x ) ( x + 3 )
= x ( x2 + 1 ) ( x + 3 )
b, Sửa đề: x4 - x2 + 8x - 8
= ( x4 - x2 ) + ( 8x - 8 )
= x2 ( x2 - 1 ) + 8 ( x - 1 )
= x2 ( x - 1 ) ( x + 1 ) + 8 ( x - 1 )
= ( x - 1 ) [ x2 ( x + 1 ) + 8 ]
= ( x - 1 ) ( x3 + x2 + 8 )
VA
1
TN
20 tháng 2 2018
\(x^4+8x^3+28x^2+48x-13\)
\(=x^4+4x^3+13x^2+4x^3+16x^2+52x-x^2-4x-13\)
\(=x^2\left(x^2+4x+13\right)+4x\left(x^2+4x+13\right)-\left(x^2+4x+13\right)\)
\(=\left(x^2+4x-1\right)\left(x^2+4x+13\right)\)
DT
2
7 tháng 8 2019
= \(x^4-2x^3-6x^3+12x^2-x^2+2x+6x-12\)
= \(x^3\left(x-2\right)-6x^2\left(x-2\right)-x\left(x-2\right)+6\left(x-2\right)\)
= \(\left(x-2\right)\left(x^3-6x^2-x+6\right)\)
= \(\left(x-2\right)\left(x^2\left(x-6\right)-\left(x-6\right)\right)\)
= \(\left(x-2\right)\left(x-6\right)\left(x-1\right)\left(x+1\right)\)
7 tháng 8 2019
x4 - 8x3 + 11x2 + 8x - 12
= (x3 - 7x2 + 4x + 12)(x - 1)
= (x3 - 8x + 12)(x + 1)(x - 1)
= (x - 6)(x - 2)(x + 1)(x - 1)
KV
9 tháng 12 2023
Bài 3
a) x² + 10x + 25
= x² + 2.x.5 + 5²
= (x + 5)²
b) 8x - 16 - x²
= -(x² - 8x + 16)
= -(x² - 2.x.4 + 4²)
= -(x - 4)²
c) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
d) (x + y)² - 9x²
= (x + y)² - (3x)²
= (x + y - 3x)(x + y + 3x)
= (y - 2x)(4x + y)
e) (x + 5)² - (2x - 1)²
= (x + 5 - 2x + 1)(x + 5 + 2x - 1)
= (6 - x)(3x + 4)
KV
9 tháng 12 2023
Bài 4
a) x² - 9 = 0
x² = 9
x = 3 hoặc x = -3
b) (x - 4)² - 36 = 0
(x - 4 - 6)(x - 4 + 6) = 0
(x - 10)(x + 2) = 0
x - 10 = 0 hoặc x + 2 = 0
*) x - 10 = 0
x = 10
*) x + 2 = 0
x = -2
Vậy x = -2; x = 10
c) x² - 10x = -25
x² - 10x + 25 = 0
(x - 5)² = 0
x - 5 = 0
x = 5
d) x² + 5x + 6 = 0
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 = -3; x = -2
28 tháng 10 2021
Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
\(x^4-8x=x\left(x^3-8\right)=x\left(x-2\right)\left(x^2+2x+4\right)\)
x4 - 8x = x.(x3 - 8) = x. (x3 - 23) = x.(x - 2).(x2 + 2x +4)