bài 1: phân tích các đa thức sau thành nhân tử.
a, x^2-81
b,4x^2-25
c, x^4-y^4
d, x^2+6xy+9y^2
e,6x-9-x^2
f, x^2 -4x^2 +4y^2 +4xy
g, (a+b)^3 + (a-b)^3
h, (3x+1)^2-(x+1)^2
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Những câu hỏi liên quan
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x-y\right)\left(x+y\right)\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left[3\left(x-y\right)+2\left(x+y\right)\right]\left[3\left(x-y\right)-2\left(x+y\right)\right]\)
\(=\left(3x-3y+2x+2y\right)\left(3x-3y-2x-2y\right)\)
\(=\left(5x-y\right)\left(x-5y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
26 tháng 10 2021
\(a,=x\left(4x^2-1\right)=x\left(2x-1\right)\left(2x+1\right)\\ b,=2\left(3x-2\right)-x\left(3x-2\right)=\left(2-x\right)\left(3x-2\right)\\ c,=x\left(y-1\right)-\left(y-1\right)=\left(x-1\right)\left(y-1\right)\\ d,=\left(y+2\right)^2-4x^2=\left(y+2-2x\right)\left(y+2+2x\right)\\ e,=x^2-x-2x+2=\left(x-1\right)\left(x-2\right)\)
26 tháng 10 2021
a) \(4x^3-x^2=x^2\left(4x-1\right)\)
b) \(6x-4+x\left(2-3x\right)=2\left(3x-2\right)-x\left(3x-2\right)=\left(2-x\right)\left(3x-2\right)\)
c) \(xy+1-x-y=\left(xy-x\right)-\left(y-1\right)=x\left(y-1\right)-\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)
d) \(y^2-4x^2+4y+4=\left(y^2+4y+4\right)-4x^2=\left(y+2\right)^2-\left(2x\right)^2=\left(y-2x+2\right)\left(y+2x+2\right)\)
e) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
TL
20 tháng 8 2021
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
22 tháng 12 2023
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
17 tháng 12 2016
mình chỉ phân tích thôi
a) 6x(4-x)+x-4
=6x(4-x)-(4-x)
=(6x-1)(4-x)
c) 25x^2-10x+1-16z^2
=(5x-1)^2-16z^2
=(5x-1-4z)(5x-1+4z)
ban xem lại đề bài câu b đi chắc là sai đó
còn các câu trên bạn tự làm nhé
TK
17 tháng 12 2016
Thực hiện phép tính:
a) (2x-3y)(4x2+6xy+9y2)
=8x3-27y3
b) (6x3+3x2+4x+2):(3x2+2)
=(3x2+2)(2x+1):(3x2+2)
=2x+1
c) (x+2)2+(3-x)-2(x+3)(x-3)
=x2+4x+4+3-x-2x2+18
=-x2+4x+25
8 tháng 4 2022
Bài 2:
a: 6x(4-x)+(x-4)
=6x(4-x)-(4-x)
=(4-x)(6x-1)
b: \(=x^2-1+y\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(1+y\right)=\left(y+1\right)\left(x+1\right)\left(x-1\right)\)
c: \(=\left(5x-1\right)^2-\left(4z\right)^2\)
=(5x-1-4z)(5x-1+4z)
a) \(x^2-81=\left(x-9\right)\left(x+9\right)\)
b) \(4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
c) \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
d) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
e) \(6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)
f) \(x^2-4x^2+4y^2+4xy=\left(x^2+4xy+4y^2\right)-4x^2=\left(x+2y\right)^2-4x^2\\ =\left(x+2y+2x\right)\left(x+2y-2x\right)=\left(3x+2y\right)\left(2y-x\right)\)
g) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)=2a\left(a^2+3b^2\right)\)
h) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)