Thực hiện phép chia:
a) ( x3 - 3x + 2 ) : ( x - 1 )2
b) ( 3x4 - 4x3 + 1 ) : ( x - 1 )2
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a) \(\left(x^5+4x^3-6x^2\right):4x^2\)
\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)
\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
b) x^3 + x^2 - 12 x-2 x^3 - 2x^2 3x^2 - 12 3x^2 - 6x 6x - 12 x^2+3x+6 6x - 12 0
Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)
c) (-2x5 : 2x2) + (3x2 : 2x2) + (-4x^3 : 2x^2)
= \(-x^3+\dfrac{3}{2}-2x\)
d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)
\(=x-4\)
(dùng hẳng đẳng thức thứ 7)
Bài 2 :
a) 3x(x - 2) - 5x(1 - x) - 8(x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= (3x2 + 5x2 - 8x2) + (-6x - 5x) + 24
= -11x + 24
b) (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
c) (x - y)2 + (x + y)2 - 2(x - y)(x + y)
= (x - y)2 - 2(x - y)(x + y) + (x + y)2
= [(x - y) + x + y)2 = [x - y + x + y] = (2x)2 = 4x2
Bài 1 :
a]= \(\frac{1}{4}\)x3 + x - \(\frac{3}{2}\).
b] => [x3 + x2 -12 ] = [ x2 +3 ][x-2] + [-6]
c]= -x3 -2x +\(\frac{3}{2}\).
d] = [ x3 - 64 ] = [ x2 + 4x + 16][ x- 4].
a: \(\dfrac{20a^4b^5c^2}{\left(-5ab^2c\right)^2}\)
\(=\dfrac{20a^4b^5c^2}{25a^2b^4c^2}\)
\(=\dfrac{4}{5}a^2b\)
b: \(\dfrac{\left(-15x^2y^3\right)^7}{\left(15xy^3\right)^6}-\dfrac{32x^{18}y^5}{\left(-4x^5y\right)^2}\)
\(=\dfrac{-15^7\cdot x^{14}\cdot y^{21}}{15^6\cdot x^6\cdot y^{18}}-\dfrac{32x^{18}y^5}{16x^{10}y^2}\)
\(=-15x^8y^3-2x^8y^3\)
c: \(\dfrac{-\dfrac{1}{3}x^5y^2}{-2xy}-\dfrac{x^2+2x+1}{x+1}\)
\(=\dfrac{2}{3}x^3y-x-1\)
Mình xp giúp được mỗi câu đầu thôi nha ;-;;;; 2 câu sau mình chưa học, bạn thông cảm ;-;;;.
`a,` \(\text{P(x) =}\)\(2x^3-3x+x^5-4x^3+4x-x^5+x^2-2\)
`P(x)= (2x^3 - 4x^3)-(3x-4x) +(x^5-x^5) +x^2-2`
`P(x)= -2x^3- (-x)+0+x^2-2`
`P(x)=-2x^3+x+x^2-2`
`Q(x)= x^3-x^2+3x+1+3x^2`
`Q(x)= x^3- (x^2-3x^2) +3x+1`
`Q(x)=x^3- (-2x^2)+3x+1`
\(a,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x\left(x+5\right)}{x\left(x+6\right)}=\dfrac{x+5}{x+6}\\ b,=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)
\(c,=\dfrac{3x^2+2x+1+x^2-2x+1-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)
\(a,\Rightarrow x\left(x+3\right)-\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left(x+3\right)\left(x-x+3\right)=0\\ \Rightarrow3\left(x+3\right)=0\Rightarrow x=-3\\ b,A:B=\left(2x^2-x+4x-2\right):\left(2x-1\right)\\ =\left[x\left(2x-1\right)+2\left(2x-1\right)\right]:\left(2x-1\right)\\ =x+2\)
1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)
2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)
3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)
4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)
Bài 3:
b: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)+10=0\)
\(\Leftrightarrow6x^2+12-6x^2+12x-6=0\)
hay \(x=-\dfrac{1}{2}\)
Bài 2:
a: \(x^3+3x^2+3x+1=\left(x+1\right)^3\)
b: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)
( 4 x 4 – 4 x 3 + 3x – 3) : (x – 1) = 4 x 3 + 3
Đáp án cần chọn là: D