Phân tích đa thức thành nhân tử
x^4-x^3-x+1
9(x-5)^2-(x-7)^2
K
Khách
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C
1
1 tháng 11 2020
M = x9 - x7 + x6 - x5 - x4 + x3 - x2 + 1
= ( x9 - x7 ) + ( x6 - x4 ) - ( x5 - x3 ) - ( x2 - 1 )
= x7( x2 - 1 ) + x4( x2 - 1 ) - x3( x2 - 1 ) - ( x2 - 1 )
= ( x2 - 1 )( x7 + x4 - x3 - 1 )
= ( x - 1 )( x + 1 )[ x4( x3 + 1 ) - ( x3 + 1 ) ]
= ( x - 1 )( x + 1 )( x3 + 1 )( x4 - 1 )
= ( x - 1 )( x + 1 )( x + 1 )( x2 - x + 1 )( x2 - 1 )( x2 + 1 )
= ( x + 1 )2( x - 1 )( x2 - x + 1 )( x - 1 )( x + 1 )( x2 + 1 )
= ( x + 1 )3( x - 1 )2( x2 + 1 )( x2 - x + 1 )
MD
0
NH
4
DN
6 tháng 11 2016
Ta có:
\(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^8\right)+\left(x^8-x^7\right)-\left(x^6-x^5\right)-\left(2x^5-2x^4\right)-\left(x^4-x^3\right)+\left(x^2-x\right)+\left(x-1\right) \)
\(=x^8.\left(x-1\right)+x^7.\left(x-1\right)-x^5.\left(x-1\right)-2x^4.\left(x-1\right)-x^3\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^8+x^7-x^5-2x^4-x^3+x+1\right)\)
HM
0
NH
1
25 tháng 11 2017
Ta có:\(9\left(x+5\right)^2-4\left(x-7\right)^2=9\left(x^2+10x+25\right)-4\left(x^2-14x+49\right)\)
\(=9x^2+90x+225-4x^2+56x-196=5x^2+146x+29\)
\(=\left(5x^2+145x\right)+\left(x+29\right)=\left(x+29\right)\left(5x+1\right)\)
24 tháng 9 2021
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
12 tháng 10 2021
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
HT
3
3 tháng 10 2019
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)(1)
Đặt \(x^2+5x+4=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-15=t^2+2t+1-16\)
\(=\left(t+1\right)^2-4^2=\left(t+5\right)\left(t-3\right)\)
\(=\left(x^2+5x+9\right)\left(x^2+5x+1\right)\)
3 tháng 10 2019
b) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5+x-9\right)\left(2x+5-x+9\right)\)
\(=\left(3x-4\right)\left(x+14\right)\)
BN
1
AH
Akai Haruma
Giáo viên
27 tháng 6 2023
Bạn nên viết lại đa thức bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để được hỗ trợ tốt hơn.
x4 - x3 - x + 1
= x3.(x-1) - ( x -1)
= (x-1).(x3 -1)
=(x-1)(x-1)(x2+x+1)
9.(x-5)2 -( x-7)2
={ 3.(x-5)}2 - (x-7)2
= [ 3.(x-5) - x + 7][3.(x-5) + x-7]
= (3x -15 - x + 7)( 3x - 15 + x -7)
= (2x -8)(4x - 22)