phân tích thành nhân tử: (x+6)(x+7)+(x+4)(x+7)+(x+4)(x+5)
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NL
Nguyễn Lê Phước Thịnh
CTVHS
26 tháng 8 2023
a: =4(x-2)(x+1)+4(x-2)^2+(x+1)^2
=(2x-4)^2+2*(2x-4)(x+1)+(x+1)^2
=(2x-4+x+1)^2=(3x-3)^2=9(x-1)^2
b: =x^7(x^2-1)-x^5(x+1)+x^3(x+1)+(x^2-1)
=(x+1)[x^7(x-1)-x^5+x^3+x-1]
=(x+1)[x^7(x-1)-x^3(x-1)(x+1)+(x-1)]
=(x+1)(x-1)(x^7-x^4-x^3+1)
=(x+1)(x-1)(x^3-1)(x^4-1)
=(x+1)(x-1)^2*(x^2+x+1)(x^2+1)(x-1)(x+1)
=(x+1)^2*(x-1)^3*(x^2+1)(x^2+x+1)
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
HT
1
NT
1
11 tháng 9 2018
\(x^9-x^7+x^6-x^5-x^4+x^3-x^2+1\)
\(=\left(x^9-x^7\right)+\left(x^6-x^5\right)-\left(x^4-x^3\right)-\left(x^2-1\right)\)
\(=x^7\left(x^2-1\right)+x^5\left(x-1\right)-x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^7-1\right)+\left(x-1\right)\left(x^5-x^3\right)\)
\(=\left(x^2-1\right)\left(x^7-1\right)+x^3\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[x^7-1+x^3\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(x^7+x^4-x^3-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left[x^4\left(x^3+1\right)-\left(x^3+1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(x^3+1\right)\left(x^4-1\right)\)
\(=\left(x+1\right)^3\left(x-1\right)^2\left(x^2+1\right)\left(x^2-x+1\right)\)
Tham khảo nhé~
NL
Nguyễn Lê Phước Thịnh
CTVHS
8 tháng 10
1: \(3x\left(x-1\right)+7x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(7x^2+3x\right)\)
=x(7x+3)(x-1)
2: \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left(3x+15\right)^2-\left(x-7\right)^2\)
=(3x+15+x-7)(3x+15-x+7)
=(4x+8)(2x+22)
\(=4\left(x+2\right)\cdot2\cdot\left(x+11\right)=8\left(x+2\right)\left(x+11\right)\)
3: \(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
=(x-y)(x-y-z)
4: \(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-2\cdot\left(x^2+2x\right)-3\)
\(=\left(x^2+2x-3\right)\left(x^2+2x+1\right)=\left(x+3\right)\left(x-1\right)\left(x+1\right)^2\)
5: \(9x^2y^2+15x^2y-21xy^2\)
\(=3xy\cdot3xy+3xy\cdot5x-3xy\cdot7y\)
=3xy(3xy+5x-7y)
6: \(x^3-2x^2y+xy^2-9xy^4\)
\(=x\left(x^2-2xy+y^2-9y^4\right)\)
\(=x\left\lbrack\left(x-y\right)^2-\left(3y^2\right)^2\right\rbrack=x\left(x-y-3y^2\right)\left(x-y+3y^2\right)\)
7: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
=x(x+4)-3(x+4)
=(x+4)(x-3)
8: (x+1)(x+3)(x+5)(x+7)+15
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+15\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
ta có
[(x+6)(x+7)+(4+x)(x+7)]+(x+4)(x+5)
=[(x+7)(x+6+x+4)]+(x+4)(X+5)
=(x+7)(2x+10)+(x+4)(x+5)
=(x+7)2(x+5)+(x+4)(x+5)
=(x+5)(2x+14+x+4)
=(x+5)(3x+18)
(x+6).(x+7)+(x+4).(x+7)+(x+4).(x+5)
= x^2+13x+42+x^2+11x+28+x^2+9x+20
= 3x^2+33x+90
= 3.(x^2+11x+30)
=3.[(x^2+5x)+(6x+30)]
= 3.(x+5).(x+6)
Tk mk nha