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\(a,\)\(x^3-13x-12\)
\(=x^3-x-12x-12\)
\(=x\left(x^2-1\right)-12\left(x+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x+4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
a) \(x^3-13x-12\)
\(=x^3+x^2-x^2-x-12x-12\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-12\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-12\right)\)
\(=\left(x+1\right)\left(x^2-4x+3x-12\right)\)
\(=\left(x+1\right)\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=\left(x+1\right)\left(x-4\right)\left(x+3\right)\)
b) \(2x^4+3x^3-9x^2-3x+2\)câu này hình như sai đề rồi, bạn xem lại nhen
c) \(x^4-3x^3-6x^2+3x+1\)câu này cx thế, bạn xem lại nha
\(x^2+2xy+7x+7y+y^2+10\)
\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y\right)^2+7\left(x+y\right)+\frac{49}{4}-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}\right)^2-\frac{9}{4}\)
\(=\left(x+y+\frac{7}{2}-\frac{3}{2}\right)\left(x+y+\frac{7}{2}+\frac{3}{2}\right)\)
\(=\left(x+y+2\right)\left(x+y+5\right)\)
b)Ta có: x2y+xy2+x+y=2010
<=>xy.x+xy.y+x+y=2010
<=>11x+11y+x+y=2010
<=>12(x+y)=2010
<=>x+y=167,5
=>(x+y)2=28056,25
<=>x2+y2+2xy=28056,25
<=>x2+y2=28034,25
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
/ (4x−2)(10x+4)(5x+7)(2x+1)+17=0(4x−2)(10x+4)(5x+7)(2x+1)+17=0
⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0⇔(4x−2)(5x+7)(10x+4)(2x+1)+17=0
⇔(20x2+18x−14)(20x2+18x+4)+17=0⇔(20x2+18x−14)(20x2+18x+4)+17=0
Đặt t= 20x2+18x+4(t≥0)20x2+18x+4(t≥0) ta có:
(t-18).t +17=0
⇔t2−18t+17=0⇔t2−18t+17=0
⇔(t−17)(t−1)=0⇔(t−17)(t−1)=0
⇔[t=17(tm)t=1(tm)⇔[t=17(tm)t=1(tm) ⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0⇔[20x2+18x+4=1720x2+18x+4=1⇔[20x2+18x−13=020x2+18+3=0
⇔[(20x+9−341−−−√)(20x+9+341−−−√)=0(20x+9−21−−√)(20x+9+21−−√)=0⇔[(20x+9−341)(20x+9+341)=0(20x+9−21)(20x+9+21)=0
⇔⎡⎣⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢x=−9+341−−−√20x=−9−341−−−√20x=−9+21−−√20x=−9−21−−√20
\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)
\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)
\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)
Đặt ....
a, \(3x^3-4x^2+5x-4\)
\(=3x^3-3x^2-x^2+x+4x-4\)
\(=3x^2\left(x-1\right)-x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(3x^2-x+4\right)\left(x-1\right)\)
b, \(4x^3-3x^2+5x-21\)
\(=4x^3-7x^2+4x^2-7x+12x-21\)
\(=x^2\left(4x-7\right)+x\left(4x-7\right)+3\left(4x-7\right)\)
\(=\left(x^2+x+3\right)\left(4x-7\right)\)
c, \(3x^3+8x^2+14x+15\)
\(=3x^3+5x^2+3x^2+5x+9x+15\)
\(=x^2\left(3x+5\right)+x\left(3x+5\right)+3\left(3x+5\right)\)
\(=\left(x^2+x+3\right)\left(3x+5\right)\)
Bài này dùng phương pháp nhẩm nghiệm (tối ưu nhất với đa thức bậc ba)
Chúc bạn học tốt.
\(64x^4+y^4\)
\(=\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
\(x^5+x-1\)
\(=x^5+x^2-\left(x^2-x+1\right)\)
\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
a. \(=x^3+2^3+1^3-x^3\)
\(=\left(x^3-x^3\right)+8+1\)
\(=0+8+1\)
\(=9\)
Bài 1 :
a) ( x + 2 )( x2 - 2x + 4 ) + (1 - x)(1+x+ + x2 )
= ( x3 - 8 ) + ( 1 - x3 )
= x3 - 8 + 1 - x3
= 7
b) 7x( 4x - 2) - ( x - 3)( x+1 ) + 16x
= 28x2 - 14x - x2 - x + 3x + 3 + 16x
= 27x2 + 3
a) \(7\left(3x-2\right)+y\left(3x-2\right)=\left(3x-2\right)\left(7+y\right)\)
b) \(x\left(y-x\right)-3\left(x-y\right)=x\left(y-x\right)+3\left(y-x\right)=\left(y-x\right)\left(x+3\right)\)
c) \(x^2-6xy+9y^2=\left(x-3y\right)^2\)
a. (3x - 2)(7+y)
b. (x - y)(x - 3)
c. (x-3y)2