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(x+1)(x+2)(x+3)(x+4) + 1
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HN
11 tháng 10 2019
(x + 1)(x + 2)(x + 3)(x + 4) - 24
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 24
= (x2 + 4x + x +4)(x2 + 3x + 2x + 12) - 24
= (x2 + 5x + 4)(x2 + 5x + 12) - 24
Đặt t = x2 + 5x + 8
Ta có: x2 + 5x + 4 = x2 + 5x + 8 - 4 (1)
x2 + 5x + 12 = x2 + 5x + 8 + 4 (2)
Thay t = x2 + 5x + 8 vào (1) và (2), ta có:
⇒ (t - 4)(t + 4) - 24
= t2 - 16 - 24
= t2 - 40
= (t - \(\sqrt{40}\))(t + \(\sqrt{40}\))
= (x2 + 5x + 8 - \(\sqrt{40}\))(x2 + 5x + 8 + \(\sqrt{40}\))
NT
1
HA
0
7 tháng 10 2021
\(-\left(x+2\right)+3\left(x^2-4\right)\)
\(=3\left(x-2\right)\left(x+2\right)-\left(x+2\right)\)
\(=\left(x+2\right)\left[3\left(x-2\right)-1\right]=\left(x+2\right)\left(3x-7\right)\)
HH
21 tháng 6 2018
Bài 1:
a) \(x^2-2xy-25+y^2\) (Sửa đề)
\(=x^2-2xy+y^2-25\)
\(=\left(x-y\right)^2-5^2\)
\(=\left(x-y-5\right)\left(x-y+5\right)\)
Vậy ...
b) \(x\left(x-1\right)+y\left(1-x\right)\)
\(=x\left(x-1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y\right)\)
Vậy ...
c) \(7x+7y-\left(x+y\right)\) (Sửa đề)
\(=7\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(7-1\right)\)
\(=6\left(x+y\right)\)
Vậy ...
d) \(x^4+y^4\)
\(=\left(x^2\right)^2+\left(y^2\right)^2\)
\(=\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left(x^2+y^2-\sqrt{2}xy\right)\left(x^2+y^2+\sqrt{2}xy\right)\)
Vậy ...
24 tháng 10 2019
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-8\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)\(=\left(x^2+7x+11-1\right)\left(x^2+7x+11+1\right)-8\)
\(=\left(x^2+7x+11\right)^2-9\)
\(=\left(x^2+7x+11-3\right)\left(x^2+7x+11+3\right)=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
HA
1
10 tháng 7 2017
1, \(x^3+8x^2+17x+10=\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
\(=x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)\)\(=\left(x+1\right)\left(x^2+7x+10\right)=\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
2. \(2x^3-3x^2+3x-1=\left(2x^3-x^2\right)-\left(2x^2-x\right)+\left(2x-1\right)\)
\(=x^2\left(2x-1\right)-x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(2x-1\right)\left(x^2-x+1\right)\)
3. \(x^4+x^2+1=\left(x^4+1\right)+x^2=\left(x^2+1\right)^2-2x^2+x^2\)\(=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4. \(81x^4+4=\left(9x^2\right)^2+2^2=\left(9x^2+2\right)^2-2.9x^2.2=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)
3 tháng 11 2017
(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
thay x^2+7x+11=y
=> (y-1)(y+1)-24=y^2-1^2-24=y^2-25=(y-5)(y+5)
= (x^2+7x+11-5)(x^2+7x+11+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)
3 tháng 11 2017
(x + 2)(x + 3)(x + 5)(x + 7) - 24
= [(x + 2)(x + 5)][(x + 3)(x + 4)] - 24
=(x2 + 7x + 10)(x2 + 7x +12) - 24
Đặt x2 + 7x + 11 = t ; ta có:
(t - 1)(t + 1) - 24
= t2 - 12 - 24
= t2 - 25
= (t - 5)(t + 5)
Thay t = x2 + 7x + 11 ta được:
(x2 + 7x + 11 - 5)(x2 + 7x +11 + 5)
= (x2 + 7x + 6)(x2 + 7x + 16)
= (x + 1)(x + 6)(x2 + 7x + 16)
Chúc bn học tốt 
5 tháng 10 2017
a) =x3-2x2+6x2-12x -12x +24
= x2(x-2)+6x(x-2)-12(x-2)
= (x-2)(x2+6x-12)
mk giải đc câu a thôi, bn zô jup mk lại vs
5 tháng 10 2017
\(a,x^3+4x^2-24x+24\)
\(=x^3+6x^2-12x-2x^2-12x+24\)
\(=\left(x^3-2x^2\right)+\left(6x^2-12x\right)-\left(12x-24\right)\)
\(=x^2\left(x-2\right)+6x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+6x-12\right)\)
\(\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
\(=\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)+1\)
\(=\left(x^2+5x+4+1\right)^2\)
\(=\left(x^2+5x+5\right)^2\)