Phân tích đa thức thành nhân tử
a)x5+x+1
b) (x2+x)2-2.(x2+x)-15
c) (4x+1).(12x-1).(3x+2).(x+1)-4
d) x5-5x3+4x
e) x3+5x2+3x-9
g) x10+x8-2
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ZH
0
DT
18 tháng 2 2022
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
18 tháng 2 2022
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
12 tháng 7 2023
\(a,=\left(5x^3+10x\right)+\left(x^4-4\right)\\ =5x\left(x^2+2\right)+\left(x^2+2\right)\left(x^2-2\right)\\ =\left(x^2+2\right)\left(x^2+5x-2\right)\\ b,=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\\ =\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\\ =\left(x+y+z\right)\left(x^2+2xy+y-xz-yz+z^2-3xy\right)\\ =\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(c,=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\\ d,=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\\ e,=\left(x^{10}+x^9+x^8\right)-\left(x^9+x^8+x^7\right)+\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^{10}-x^7+x^5-x^4+x^3-x+1\right)\)
12 tháng 7 2023
a: =x^4+2x^2+5x^3+10x-2x^2-4
=(x^2+2)(x^2+5x-2)
b; =(x+y)^3+z^3-3xy(x+y)-3xyz
=(x+y+z)*(x^2+2xy+y^2-xz-yz+z^2)-3xy(x+y+z)
=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
c: =x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1
=(x^2+x+1)(x^6-x^5+x^3-x^2+1)
DN
2
4 tháng 9 2021
Ta có: \(\left(4x+1\right)\left(12x-1\right)\left(3x+2\right)\left(x+1\right)-4\)
\(=\left(12x^2+8x+3x+2\right)\left(12x^2+12x-x-1\right)-4\)
\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)
\(=\left(12x^2+11x\right)^2+\left(12x^2+11x\right)-6\)
\(=\left(12x^2+11x+3\right)\left(12x^2+11x-2\right)\)
CM
5 tháng 12 2017
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2
= x 3 + - 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
P ( x ) − Q ( x ) = − 2 x 3 + x 2 + x − 2 − x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 − x 3 − 3 x − 1 = − 2 x 3 − x 3 + x 2 + ( x − 3 x ) − 2 − 1 = − 3 x 3 + x 2 − 2 x − 3
Chọn đáp án B
21 tháng 9 2021
a) \(7\left(3x-2\right)+y\left(3x-2\right)=\left(3x-2\right)\left(7+y\right)\)
b) \(x\left(y-x\right)-3\left(x-y\right)=x\left(y-x\right)+3\left(y-x\right)=\left(y-x\right)\left(x+3\right)\)
c) \(x^2-6xy+9y^2=\left(x-3y\right)^2\)
13 tháng 11 2021
2: \(=\dfrac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)}{-\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{-\left(x+y\right)\left(x^2+y^2\right)}{x^2+xy+y^2}\)
DH
1
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 12 2021
\(x^2-2xy+x-2y=x\left(x-2y\right)+x-2y=\left(x-2y\right)\left(x+1\right)\)
\(3x^3+6x+3-3y^2=3\left[\left(x^2+2x+1\right)-y^2\right]=3\left[\left(x+1\right)^2-y^2\right]=3\left(x-y+1\right)\left(x+y+1\right)\)
CM
18 tháng 8 2017
Ta có
P ( x ) = 2 x 3 − 3 x + x 5 − 4 x 3 + 4 x − x 5 + x 2 − 2 = x 5 − x 5 + 2 x 3 − 4 x 3 + x 2 + ( 4 x − 3 x ) − 2 = − 2 x 3 + x 2 + x − 2 Và Q ( x ) = x 3 − 2 x 2 + 3 x + 1 + 2 x 2 = x 3 + − 2 x 2 + 2 x 2 + 3 x + 1 = x 3 + 3 x + 1
Khi đó
M ( x ) = P ( x ) + Q ( x ) = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 2 + x − 2 + x 3 + 3 x + 1 = − 2 x 3 + x 3 + x 2 + ( x + 3 x ) − 2 + 1 = − x 3 + x 2 + 4 x − 1
Bậc của M ( x ) = - x 3 + x 2 + 4 x - 1 l à 3
Chọn đáp án C