phân tích thành nhân tử
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
HN
2
Những câu hỏi liên quan
XO
16 tháng 10 2020
(x - 5)2 - 4(x - 3)2 + 2(2x - 1)(x - 5) + (2x - 1)2
= [(x - 5)2 + 2(2x - 1)(x - 5) + (2x - 1)2) - [2(x - 3)]2
= (x - 5 + 2x - 1)2 - (2x - 6)2
= (3x - 6)2 - (2x - 6)2
= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)
16 tháng 10 2020
( x - 5 )2 - 4( x - 3 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2
= [ ( x - 5 )2 + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )2 ] - 22( x - 3 )2
= ( x - 5 + 2x - 1 )2 - ( 2x - 6 )2
= ( 3x - 6 )2 - ( 2x - 6 )2
= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )
= x( 5x - 12 )
6 tháng 2 2022
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
MH
17 tháng 10 2023
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)-24\)
\(=\left(x^2+7x+11\right)^2-25\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
T
17 tháng 10 2023
\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)
Đặt \(y=x^2+7x+10\), khi đó biểu thức trở thành:
\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(x^2+7x+10+1-5)(x^2+7x+10+1+5)\\=(x^2+7x+6)(x^2+7x+16)\\=(x^2+x+6x+6)(x^2+7x+16)\\=[x(x+1)+6(x+1)](x^2+7x+16)\\=(x+1)(x+6)(x^2+7x+16)\\Toru\)
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: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
HA
1
28 tháng 8 2016
Ta có :
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^4+x^3+x^2-x^3+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^4+x^3+x^2\right)-\left(x^3-1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^2+x+1\right)x^2-\left(x-1\right)\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left[3\left(x^2-x+1\right)-\left(x^2+x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2+2-4x\right)\)
\(=2\left(x^2+x+1\right)\left(x^2+1-2x\right)\)
\(=2\left(x^2+x+1\right)\left(x-1\right)^2\)
CP
1
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 4 2021
\(\Leftrightarrow x^3-\left(m-1\right)x^2-\left(m-1\right)x-2x^2+2\left(m-1\right)x+2m-2=0\)
\(\Leftrightarrow x\left(x^2-\left(m-1\right)x-m+1\right)-2\left(x^2-\left(m-1\right)x-m+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-\left(m-1\right)x-m+1\right)=0\)
TL
2
HV
27 tháng 12 2016
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\\ =3\left(x^4-x+x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^4-x\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1^2\right)\)
\(=3\left[x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)+\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3+x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(4x^2-2x+2\right)\\ =2\left(x^2+x+1\right)\left(x^2-x+1\right)\)
27 tháng 12 2016
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)\)
\(=3\left[\left(x^4+x^3+x^2\right)-\left(x^3-x^2-x\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left[x^2\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left[3\left(x^2-x+1\right)-\left(x^2+x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)
\(=2\left(x^2+x+1\right)\left(x^2-2x+1\right)\)
\(=2\left(x^2+x+1\right)\left(x-1\right)^2\)
HL
phân tích đa thức thành nhân tử \(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
1
15 tháng 6 2017
\(x^2\cdot\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x+4\right)^2-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left(x^4+x^2+1\right)-\left(x^4+x^2+1+2x^3+2x^2+2x\right)\)
\(=2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)\)
\(=2\left(x^4+x^2+1-x^3-x^2-x\right)\)
\(=2\left(x^4-x^3-x+1\right)\)
\(=2\left(x^3\left(x-1\right)-\left(x-1\right)\right)\)
\(=2\left(x-1\right)\left(x^3-1\right)\)
\(=2\left(x-1\right)^2\left(x^2+x+1\right)\)
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
\(=3\left[x^4+2x^2+1-x^2\right]-\left(x^2+x+1\right)^2\)
\(=3\left[\left(x^2+1\right)^2-x^2\right]-\left(x^2+x+1\right)^2\)
\(=3\left(x^2+x+1\right)\left(x^2-x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)
\(=2\left(x-1\right)^2\cdot\left(x^2+x+1\right)\)