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a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
1)x2-8x-9
= x^2 - 9x +x -9
= x(x+1) - 9 (x+1)
= (x-9) (x+1)
2)x2+3x-18
3)x3-5x2+4x
=x^3 - 4x^2 - x^2 + 4x
= x^2 (x-1) - 4x(x-1)
= (x^2 - 4x) (x-1)
= x(x-4)(x-1)
4)x3-11x2+30x
5)x3-7x-6
6)x16-64
\(=\left(x^8\right)^2-8^2\)
\(=\left(x^8-8\right)\left(x^8+8\right)\)
7)x3-5x2+8x-4
8)x2-3x+2
= x^2 - 2x - x +2
= x(x-1) -2(x-1)
= (x-2)(x-1)
1) \(\left(x-9\right)\left(x+1\right)\) 2) \(\left(x-3\right)\left(x+6\right)\) 3) \(x\left(x-4\right)\left(x-1\right)\)
4) \(x\left(x-6\right)\left(x-5\right)\) 5)\(\left(x-3\right)\left(x+1\right)\left(x+2\right)\) 6) ........
7) \(\left(x-1\right)\left(x-2\right)\left(x-2\right)\) 8) \(\left(x-2\right)\left(x-1\right)\)
1) \(x^6-x^4-9x^3+9x^2\)
\(=x^2\left(x^4-x^2-9x+9\right)\)
\(=x^2\left[x^2\left(x^2-1\right)-9\left(x-1\right)\right]\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
2) \(x^4-4x^3+8x^2-16x+16\)
\(=x^2\left(x^2+4\right)-4x\left(x^2+4\right)+4\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
3) \(x^4-25x^2+20x-4=x^4+5x^3-2x^2-5x^3-25x^2+10x+2x^2+10x-4\)
\(=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
4) \(5x\left(x-2y\right)+2\left(2y-x\right)^2\)\(=5x\left(x-2y\right)+2\left(x-2y\right)^2=\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
5) \(x^2\left(x^2-6\right)-x^2+9=x^4-7x^2+9\)
\(=x^4+x^3-3x^2-x^3-x^2+3x-3x^2-3x+9\)
\(=x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-3\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x^2-x-3\right)\)
6) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)^2+\left(y-4\right)^3=\left(y-4\right)^2\left(7x+y-4\right)\)
7) \(x^3+2x^2-6x-27=x^3-3x^2+5x^2-15x+9x-27\)
\(=x^2\left(x-3\right)+5x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2+5x+9\right)\)
a, sửa (x3+y3+z3) thành (x+y+z)3
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+3x^2\left(y+z\right)+3x\left(y+z\right)^2+\left(y+z\right)^3-a^3-b^3-c^3\)
\(=3\left(y+z\right)\left[x^2+x\left(y+z\right)\right]+y^3+3y^2z+3yz^2+z^3-y^3-z^3\)
\(=3\left(y+z\right)\left(x^2+xy+xz\right)+3yz\left(y+z\right)\)
\(=3\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
b, \(x^3+x^2z+y^2z-xyz+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+z\left(x^2-xy+y^2\right)\)
\(=\left(x+y+z\right)\left(x^2-xy+y^2\right)\)
c, không phân tích được
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3\left(x+y\right)z\left(x+y+z\right)\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
bài 2 nè
a+b+c = 0
=>(a+b+c)^3 = 0
a^3 + b^3 + c^3 + 3(a+b)(b+c)(a+c) = 0
vì a+b = -c
a+c = -b
b+c = -a
thay vào => a^3 + b^3 + c^3 - 3abc = 0
=> a^3 + b^3 + c^3 = 3abc
Phân tích các đa thức thành nhân tử:
a. x6-y6
b.x35+x34+x33+.......+x2+x+1
c.x2-6xy+9y2-9
e.(x-9)(x-7)+1
Ta có : x35 + x34 + x33 +.......+ x2 + x + 1
= (x35 + x34 ) + (x33 + x32) +.......+ (x3 + x2) + (x + 1)
= x34(x + 1) + x32(x + 1) + .... + x2(x + 1) + (x + 1)
= (x + 1) ( x34 + x32 + ..... + x2 + 1)
Ta có : (x - 9)(x - 7) + 1
= x2 - 16x + 63 + 1
= x2 - 16x + 64
= x2 - 2.x.8 + 82
= (x - 8)2
a) = (x3 - 8) - (x2 - 4) = (x - 2).(x2 + 2x + 4) - (x - 2)(x+2) = (x - 2).(x2 + 2x + 4 - x - 2) = (x - 2).(x2 + x + 2)
b) = (x7 - x) + (x2 + x+ 1) = x(x6 - 1) + (x2 + x+ 1) = x(x3 - 1)(x3+ 1) + (x2 + x+ 1) = x(x3 + 1).(x - 1).(x2 + x+ 1) + (x2 + x+ 1)
= (x2 + x+ 1) .[(x3 + 1).(x2 - x) + 1] = (x2 + x+ 1) .(x5 - x4 + x2 - x + 1)
c) = (x3 + 3x2) + (3x2 + 9x) + (2x + 6)
= x2.(x + 3) + 3x.(x + 3) + 2(x + 3)
= (x2 + 3x+2)(x+3) = (x2 + 2x + x+ 2)(x+3) = (x+1)(x+2)(x+3)
\(x^9-x^7+x^6-x^5-x^4+x^3-x^2+1\)
\(=\left(x^9-x^7\right)+\left(x^6-x^5\right)-\left(x^4-x^3\right)-\left(x^2-1\right)\)
\(=x^7\left(x^2-1\right)+x^5\left(x-1\right)-x^3\left(x-1\right)-\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^7-1\right)+\left(x-1\right)\left(x^5-x^3\right)\)
\(=\left(x^2-1\right)\left(x^7-1\right)+x^3\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left[x^7-1+x^3\left(x-1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(x^7+x^4-x^3-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left[x^4\left(x^3+1\right)-\left(x^3+1\right)\right]\)
\(=\left(x+1\right)\left(x-1\right)\left(x^3+1\right)\left(x^4-1\right)\)
\(=\left(x+1\right)^3\left(x-1\right)^2\left(x^2+1\right)\left(x^2-x+1\right)\)
Tham khảo nhé~
cho hỏi sao
(x+1)(x−1)(x3+1)(x4−1)
=(x+1)3(x−1)2(x2+1)(x2−x+1)
được vậy