Phân tích
a,A=100x^2+20x+1
b, B=5x^2+13x+6
c,C=8x^2-14x+3
d,D=x^4+4y^4
e,E=x^4+x^2+1
K
Khách
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nhờ mn là giúp mình với ạ , minh đang cần gấp :(
17 tháng 10 2021
\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)
LT
phân tích đa thức thành nhân tử
a) 4x^2+8xy-3x-6y
b)x^4y-3x^3y^2+3x^2y^3+xy^4
c)x^3-5x^2-14x
d)x^4+4y^4
1
PA
13 tháng 10 2016
\(4x^2+8xy-3x-6y=4x\left(x+2y\right)-3\left(x+2y\right)=\left(4x-3\right)\left(x+2y\right)\)
\(x^4y-3x^3y^2+3x^2y^3-xy^4=xy\left(x^3-3x^2y+3xy^2-y^3\right)=xy\left(x-y\right)^3\)
\(x^3-5x^2-14x=x\left(x^2-5x-14\right)=x\left(x^2-7x+2x-14\right)=x\left[x\left(x-7\right)+2\left(x-7\right)\right]=x\left(x-7\right)\left(x+2\right)\)
\(x^4+4y^4=\left(x^2\right)^2+2\times x^2\times2y^2+\left(2y^2\right)^2-4x^2y^2=\left(x^2+2y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)
10 tháng 8 2023
a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)
=3x+4
b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)
\(=\dfrac{x\left(5x-2\right)+3\left(5x-2\right)}{5x-2}=x+3\)
c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)
d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)
=7x+1
e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)
\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)
f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)
g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)
Y
17 tháng 8 2018
a) \(5x^2-20x+20y-5y^2\)
\(=5\left(x^2-y^2\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y\right)-20\left(x-y\right)\)
\(=5\left(x-y\right)\left(x+y-4\right)\)
b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
\(=\left(x-y+5-1\right)^2\)
\(=\left(x-y+4\right)^2\)
c) \(7x-7y+5x-5y\)
\(=7\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(7+5\right)\)
\(=12\left(x-y\right)\)
d) \(16x^2-8x+1\)
\(=\left(4x\right)^2-2.4x+1\)
\(=\left(4x-1\right)^2\)
e) \(100x^2-\left(x^2+25\right)^2\)
\(=\left(10x\right)^2-\left(x^2+25\right)^2\)
\(=\left(10x-x^2+25\right)\left(10x+x^2+25\right)\)
\(=\left(10x-x^2+25\right)\left(x+5\right)^2\)
DD
8 tháng 9 2016
\(\text{Tìm x:}\)
\(a.x\left(x-1\right)-3x+3x=0\)
\(x\left(x-1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
\(b.3x\left(x-2\right)+10-5x=0\)
\(3x^2-6x+10-5x=0\)
\(3x^2-11x+10=0\)
\(3x^2-11x=-10\)(bn xem lại đề nhé)
\(c.x^3-5x^2+x-5=0\)
\(x^3-5x^2+x=5\)
\(d.x^4-2x^3+10x^2-20x=0\)
CW
8 tháng 9 2016
bài 1:phân tích thành phân tử
a> x^2-6x-y^2+9
= (x-3)^2 -y^2
= (x-3 -y) (x-3+y)
b>x^2-xy-8x+8y
= x(x-y) - 8(x-y)
= (x-8) (x-y)
c>25-4x^2-4xy-y^2
= 5^2 - (2x + y)^2
= (5 - 2x -y) (5 +2x+y)
d>xy-xz-y+z
= x(y-z) - (y-z)
= (x-1) (y-z)
e>x^2-xz-yz+2xy+y^2
= (x+y)^2 - z(x+y)
= (x+y-z) (x+y)
g>x^2-4xy+4y^2-z^2-4zt-4t^2
= (x-2y)^2 - (z + 2t)^2
= (x-2y -x-2t) (x-2y + z +2t)
bài 2:tìm X bt
a>x.(x-1)-3x+3x=0
x (x-1) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy x=0 và x=1
b>3x.(x-2)+10-5x=0
3x(x-2) - 5 (x-2)=0
(3x-5) (x-2) =0
\(\Rightarrow\hept{\begin{cases}3x-5=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}3x=5\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=2\end{cases}}}}\)
c>x^3-5x^2+x-5=0
x^2 (x-5) + (x-5) =0
(x^2 +1)(x-5) =0
\(\Rightarrow\hept{\begin{cases}x^2+1=0\\x-5=0\end{cases}\Rightarrow\hept{\begin{cases}x^2=-1\\x=5\end{cases}\Rightarrow}\hept{\begin{cases}x\in\varphi\\x=5\end{cases}}}\)
Vậy x=5
d>x^4-2x^3+10x^2-20x=0
x^3 (x-2) + 10x(x-2) =0
(x^3 + 10x) (x-2) =0
x(x^2 + 10) (x-2) =0
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+10=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=-10\\x=2\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varphi\\x=2\end{cases}}}}\)
Vậy x=0 và x=2
NV
8 tháng 7 2015
a/\(=x^4+5x^3-2x^2-5x^3-25x^2+10x+2x^2+10x-4=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
b/ \(=x^4-7x^2+9=x^4+x^3-3x^2-x^3-x^2+3x-3x^2-3x+9=x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-3\left(x^2+x-3\right)=\left(x^2+x-3\right)\left(x^2-x-3\right)\)
NV
8 tháng 7 2015
c/ \(=4x^2-2x-6x+3=2x\left(2x-1\right)-3\left(2x-1\right)=\left(2x-1\right)\left(2x-3\right)\)
d/ \(=y^4+2xy^3+2x^2y^2-2xy^3-4x^2y^2-2x^3y+2x^2y^2+4x^3y+4x^4=y^2\left(y^2+2xy+2x^2\right)-2xy\left(y^2+2xy+2x^2\right)+2x^2\left(y^2+2xy+2x^2\right)=\left(y^2+2xy+2x^2\right)\left(y^2-2xy+2x^2\right)\)