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1.
2x3-x2+5x+3=2x3+x2-2x2-x+6x+3=(2x+1)(x2-x+3)
2.
(x+2)(x+3)(x+4)(x+5)-24=(x2+7x+10)(x2+7x+12)-24
Đặt x2+7x+11=a
Ta cso:
(a-1)(a+1)-24=a2-1-24=a2-25=(a-5)(a+5)=(x2+7x+6)(x2+7x+16)
3.
27x3-27x2+18x-4=27x3-9x2-18x2+6x+12x-4=(3x-1)(9x2-6x+4)
\(\text{1) }2x^3-x^2+5x+3\\ =2x^3-2x^2+x^2+6x-x+3\\ =\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\\ =2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\\ =\left(2x+1\right)\left(x^2-x+3\right)\\ \)
\(\text{2) }\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+2x+5x+10\right)\left(x^2+3x+4x+12\right)-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\left(1\right)\\ \text{Đặt }x^2+7x+11=y\text{ }\text{ }\left(\text{*}\right)\\ Thay\text{ }\: \left(\text{*}\right)\text{ vào }\left(1\right)\\ \text{ }Ta\text{ }đư\text{ợc }:\\ \left(1\right)=\left(y-1\right)\left(y+1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\text{ }\text{ }\left(2\right)\\ Thay\text{ }\left(\text{*}\right)\text{ vào }\left(2\right)\\ \text{Ta lại được: }\left(2\right)=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+17\right)\\ =\left(x^2+6x+x+6\right)\left(x^2+7x+17\right)\\ =\left[\left(x^2+6x\right)+\left(x+6\right)\right]\left(x^2+7x+17\right)\\ =\left[x\left(x+6\right)+\left(x+6\right)\right]\left(x^2+7x+17\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+17\right)\\ \)
\(\text{3) }27x^3-27x^2+18x-4\\ =27x^3-18x^2-9x^2+12x+6x-4\\ =\left(27x^3-18x^2+12x\right)-\left(9x^2-6x+4\right)\\ =3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\\ =\left(3x-1\right)\left(9x^2-6x+4\right)\\ \)
a)x4+2x3+5x2+4x-12
=(x4+2x3+x2)+(4x2+4x)-12
=(x2+x)2+4(x2+x)-12
Đặt t=x2+x
=t2+4t-12=(t-2)(t+6)
=(x2+x-2)(x2+x+6)
=(x-1)(x+2)(x2+x+6)
b)(x+1)(x+2)(x+3)(x+4)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt x2+5x+4=t
t(t+2)+1=t2+2t+1
=(t+1)2=(x2+5x+4+1)2
=(x2+5x+5)2
c)(x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+5)
=(x2+8x+10)(x+2)(x+6)
d)(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4
t(t+2)-24=(t-4)(t+6)
=(x2+5x+4-4)(x2+5x+4+6)
=x(x+5)(x2+5x+10)
\(1,x^3-7x+6\)
\(=x^3+3x^2-3x^2-9x+2x+6\)
\(=x^2\left(x+3\right)-3x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x+2\right)\)
\(=\left(x+3\right)\left(x^2-2x-x+2\right)\)
\(=\left(x+3\right)\left(x-2\right)\left(x-1\right)\)
\(2,x^3-9x^2+6x+16\)
\(=x^3+x^2-10x^2-10x+16x+16\)
\(=x^2\left(x+1\right)-10x\left(x+1\right)+16\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left(x^2-2x-8x+16\right)\)
\(=\left(x+1\right)\left(x-8\right)\left(x-2\right)\)
mk ms lm hai câu thôi mà đã mệt r , bh mk lm bt mai đi học ,lúc khác lm đ cko bn
mk ghi đáp án, còn lại bạn tự biến đổi
a) \(2x^3-x^2+5x+3=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4=\left(x+1\right)\left(x+2\right)^2\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
d) \(4x^4+1=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
e) \(x^4-7x^3+14x^2-7x+1=\left(x^2-4x+1\right)\left(x^2-3x+1\right)\)
mk làm chi tiết theo yêu của của người hỏi đề:
a) \(2x^3-x^2+5x+3\)
\(=\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
b) \(x^3+5x^2+8x+4\)
\(=\left(x^3+4x^2+4x\right)+\left(x^2+4x+4\right)\)
\(=x\left(x^2+4x+4\right)+\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x^2+4x+4\right)\)
\(=\left(x+1\right)\left(x+2\right)^2\)