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a) Đặt A=(x+2)(x+3)(x+4)(x+5)-24
= (x+2)(x+5)(x+3)(x+4)-24
= (x^2+7x+10)(x^2+7x+12)-24
Đặt x^2+7x+11 = a thay vào A ta được :
A=(a-1)(a+1)=a^2-25 = a^2 - 5^2 = (a-5)(a+5) ( 2)
Thế a vào (2) ta được :
A=(x^2+7x+11-5)(x^2+7x+11+5)
= (x^2+7x+6)(x^2+7x+16)
b) = (x2+8x+7)(x2+8x+15)+15
Đặt X=x2+8x+11
f(x) = (X-4)(X+4)+15
= X2-16+15
= X2-12
= (X-1)(X+1)
=> f(x)= (x2+8x+11-1)(x2+8x+11+1)
f(x) = (x2+8x+10)(x2+8x+12)
Đến đây là vẫn còn phân tích được nhưng không dùng phương pháp đặt biến phụ:
f(x) = (x2+8x+10)(x2+8x+12)
= (x2+8x+10)[(x2+2x)+(6x+12)]
= (x2+8x+10)[x(x+2)+6(x+2)]
= (x+2)(x+6)(x2+8x+10)
d) 2x4 - 3x3 - 7x2 + 6x + 8 = (x - 2)(2x3 + x2 - 5x - 4)
Ta lại có 2x3 + x2 - 5x - 4 là đa thức có tổng hệ số của các hạng tử bậc lẻ và bậc chẵn bằng nhau nên có một nhân tử là x+1 nên 2x3 + x2 - 5x - 4 = (x+1)(2x2-x-4)
Vậy 2x4 - 3x3 - 7x2 + 6x + 8 = (x-2)(x+1)(2x2-x-4)
a) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x-1\right)\left(x+2\right)\right].\left[x\left(x+1\right)\right]=24\)
\(=\left(x^2+2x-x-2\right)\left(x^2+x\right)=24\)
\(=\left(x^2+x-2\right)\left(x^2+x\right)=24\)
\(=\left[\left(x^2+x-1\right)-1\right].\left[\left(x^2+x-1\right)+1\right]=24\)
\(=\left(x^2+x-1\right)^2-1=24\)
\(=\left(x^2+x-1\right)^2=25\)
xin lỗi mk chỉ làm được đến đây thôi cậu làm tiếp nhé
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, x3+ 6x2+11x+6
= x3 + 2x2 + 4x2 + 8x + 3x + 6
= x2(x + 2) + 4x(x + 2) + 3(x + 2)
= (x + 2)(x2 + 4x + 3)
2, x4+3x3-7x2-27x-18
= x4 + 3x3 - 9x2 + 2x2 - 27x -18
= (x4 - 9x2) + (3x3 - 27x) + (2x2 - 18)
= x2(x2 - 9) + 3x(x2 - 9) + 2(x2 - 9)
= (x2 - 9)(x2 + 3x + 2)
= (x + 3)(x - 3)(x2 + 3x + 2)
3, x3-8x2+x+42
= x3 - 7x2 - x2 + 7x - 6x + 42
= (x3 - 7x2) - (x2 - 7x) - (6x - 42)
= x2(x - 7) - x(x - 7) - 6(x - 7)
= (x - 7)(x2 - x - 6)
4, x4+5x3-7x2-41x-30
= x4 + x3 + 4x3 - 4x2 - 11x2 - 11x - 30x - 30
= (x4 + x3) + (4x3 - 4x2) - (11x2 + 11x) - (30x + 30)
= x3(x + 1) + 4x2(x + 1) - 11x(x + 1) - 30(x + 1)
= (x3 + 4x2 - 11x - 30)(x + 1)
5, x5+x-1
= x5 - x4 + x3 + x4 - x3 + x2 - x2+ x -1
= x3(x2 - x + 1)+ x2(x2 - x + 1)- (x2 - x + 1)
= (x2 - x + 1)(x3 + x2 - 1)
6, x5-x4-1
= x5 - x3 - x2 - x4 + x2 + x + x3 - x - 1
= x2(x3 - x - 1) - x(x3 - x - 1) + (x3 - x - 1)
= (x2 - x + 1)(x3 - x - 1)
1, x 3+ 6x 2+11x+6
= x 3 + 2x 2 + 4x 2 + 8x + 3x + 6
= x 2 ﴾x + 2﴿ + 4x﴾x + 2﴿ + 3﴾x + 2﴿
= ﴾x + 2﴿﴾x 2 + 4x + 3﴿
2, x 4+3x 3‐7x 2‐27x‐18
= x 4 + 3x 3 ‐ 9x 2 + 2x 2 ‐ 27x ‐18
= ﴾x 4 ‐ 9x 2 ﴿ + ﴾3x 3 ‐ 27x﴿ + ﴾2x 2 ‐ 18﴿
= x 2 ﴾x 2 ‐ 9﴿ + 3x﴾x 2 ‐ 9﴿ + 2﴾x 2 ‐ 9﴿
= ﴾x 2 ‐ 9﴿﴾x 2 + 3x + 2﴿
=﴾x + 3﴿﴾x ‐ 3﴿﴾x 2 + 3x + 2﴿
3, x 3‐8x 2+x+42
= x 3 ‐ 7x 2 ‐ x 2 + 7x ‐ 6x + 42
= ﴾x 3 ‐ 7x 2 ﴿ ‐ ﴾x 2 ‐ 7x﴿ ‐ ﴾6x ‐ 42﴿
= x 2 ﴾x ‐ 7﴿ ‐ x﴾x ‐ 7﴿ ‐ 6﴾x ‐ 7﴿
= ﴾x ‐ 7﴿﴾x 2 ‐ x ‐ 6﴿
4, x 4+5x 3‐7x 2‐41x‐30
= x 4 + x 3 + 4x 3 ‐ 4x 2 ‐ 11x 2 ‐ 11x ‐ 30x ‐ 30
= ﴾x 4 + x 3 ﴿ + ﴾4x 3 ‐ 4x 2 ﴿ ‐ ﴾11x 2 + 11x﴿ ‐ ﴾30x + 30﴿
= x 3 ﴾x + 1﴿ + 4x 2 ﴾x + 1﴿ ‐ 11x﴾x + 1﴿ ‐ 30﴾x + 1﴿
= ﴾x 3 + 4x 2 ‐ 11x ‐ 30﴿﴾x + 1﴿
5, x 5+x‐1
= x 5 ‐ x 4 + x 3 + x 4 ‐ x 3 + x 2 ‐ x 2+ x ‐1
= x 3 ﴾x 2 ‐ x + 1﴿+ x 2 ﴾x 2 ‐ x + 1﴿‐ ﴾x 2 ‐ x + 1﴿
= ﴾x 2 ‐ x + 1﴿﴾x 3 + x 2 ‐ 1﴿ 6, x 5‐x 4‐1
= x 5 ‐ x 3 ‐ x 2 ‐ x 4 + x 2 + x + x 3 ‐ x ‐ 1
= x 2 ﴾x 3 ‐ x ‐ 1﴿ ‐ x﴾x 3 ‐ x ‐ 1﴿ + ﴾x 3 ‐ x ‐ 1﴿
= ﴾x 2 ‐ x + 1﴿﴾x 3 ‐ x ‐ 1﴿
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
1 ) x3 - 2x2 + x
= x( x2 - 2x + 1 )
= x ( x-1)2
2) 4x3 - 25x
= x ( 4x2 - 25)
= x( 2x-5) ( 2x +5)
11) \(x^2-y^2-4x+4\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
13) \(x^4+4=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-4x^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
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khi bạn tích tui
tui không tích lại bạn đâu
THANKS
a, 3x3-8x2+8x-5
= x2(3x-5)-x(3x-5)+3x-5
=(3x-5)(x2-x+1)
b, 4x3-3x2+5x-21
= x2(4x-7) +x(4x-7)+3(4x-7)
=(4x-7)(x2+x+3)
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) \(\left(x-2\right)\left(x-1\right)\)