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a: \(=6x^3-10x^2+6x\)
b: \(=-2x^4-10x^3+6x^2\)
c: \(=-x^5+2x^3-\dfrac{3}{2}x^2\)
d: \(=2x^3+10x^2-8x-x^2-5x+4=2x^3+9x^2-13x+4\)
\(a,=\left[x^2\left(x^2-x-1\right)+x^3+x^2-3x-1\right]:\left(x^2-x-1\right)\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2x^2-2x-1\right]\\ =\left[x^2\left(x^2-x-1\right)+x\left(x^2-x-1\right)+2\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)\\ =\left[\left(x^2+x+2\right)\left(x^2-x-1\right)+1\right]:\left(x^2-x-1\right)=x^2+x+2R1\)
a: \(=\dfrac{\left(x^2+5\right)\left(x^2-5\right)+2x\left(x^2+5\right)}{x^2+5}=x^2+2x-5\)
b: \(=\dfrac{x^3-2x^2-x^2+2x+3x-6}{x-2}=x^2-x+3\)
a) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-4x^2y+2xy^2-4xy^2+2xy^2-y^3\)
\(=8x^3-8x^2y+4xy^2-y^3\)
b) \(\left(6x^5y^2-9x^4y^3+15x^3y^4\right):3x^3y^2\)
\(=2x^2-3xy+5y^2\)
a) \(=6x^3+8x^2+2x-6x^3=8x^2+2x\)
b) \(=\left[3xy\left(xy+2xy^2-4\right)\right]:3xy=xy+2xy^2-4\)
c) \(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3}{x+2}-\dfrac{5}{x-2}=\dfrac{10x+3\left(x-2\right)-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)
a, \(=6x^3+12x^2+2x-6x^3\\=12x^2+2x\)
b,
\(=xy+2xy^2-4\)
c,
\(\dfrac{10x}{x^2-4}+\dfrac{3}{x+2}-\dfrac{5}{x-2}\)
\(=\dfrac{10x}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x-6}{\left(x-2\right)\left(x+2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{10x+3x-6-5x-10}{\left(x-2\right)\left(x+2\right)}=\dfrac{8x-16}{\left(x-2\right)\left(x+2\right)}=\dfrac{8\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{8}{x+2}\)
\(a,=2xy^3\\ b,=xy+3xy^2-4\\ c,=\left(x-4\right)\left(x+4\right):\left(x-4\right)=x+4\\ d,=\left(x+5\right)^2:\left(x+5\right)=x+5\)
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)
a) Đây là phép chia ết với đa thức thương x 2 + 2x + 1.
Có thể kiểm tra lại kết quả bằng cách thực hiện nhân hai đa thức (x – 3)( x 2 + 2x +1)
b) Đa thức thương x 2 – 5.
a. 5x + 3(x2 - x - 1)
= 5x + 3x2 - 3x - 3
= 3x2 + 5x - 3x - 3
= 3x2 + 2x - 3
b. (5 - x)(5 + x) - (2x - 1)2
25 - x2 - (4x2 - 4x + 1)
= 25 - x2 - 4x2 + 4x - 1
= 25 - 1 - x2 - 4x2 + 4x
= 24 - 5x2 + 4x
a) \(5x+3\left(x^2-x-1\right)=5x+3x^2-3x-3=3x^2+2x-3\)
b) \(\left(5-x\right)\left(5+x\right)-\left(2x-1\right)^2=25-x^2-4x^2+4x-1=-5x^2+4x+24\)