Thực hiện phép tính :
b) (x3 - 4x2 + x + 6) : ( x - 2)
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a: A=x^5-32
Khi x=3 thì A=3^5-32=243-32=211
b: B=x^8-x^7+x^6-x^5+x^4-x^3+x^2-x+x^7-x^6+x^5-x^4+x^3-x^2+x-1
=x^8-1
=2^8-1=255
a) 2x.(3x2 – 5x + 3)
=2x3-10x2+6x
b(-2x-1).( x2 + 5x – 3 ) – (x-1)3
=-2x3 - 10x2 + 6x - x2 - 5x + 3 - x3 + 3x2 - 3x + 1
= -3x3 - 8x2 - 2x + 4
d) (6x5y2 – 9x4y3 + 15x3y4) : 3x3y2
=2x2-3xy+5y2
⇔
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].
Bài 2:
a: =x(x^2-25)
=x(x-5)(x+5)
b: =x(x-2y)+3(x-2y)
=(x-2y)(x+3)
c: =(2x-3)(4x^2+6x+9)+2x(2x-3)
=(2x-3)(4x^2+8x+9)
\(=\dfrac{x^2\left(x-2\right)+3\left(x-2\right)}{x-2}=x^2+3\)
\(\dfrac{x^3-2x^2+3x-6}{x-2}=\dfrac{x^2\left(x-2\right)+3\left(x-2\right)}{x-2}\\ =\dfrac{\left(x^2+3\right)\left(x-2\right)}{x-2}=x^2+3\)
\(\left(x^3-2x^x+3x-6\right):\left(x-2\right)\)
\(=x^3:\left(x-2\right)-2x^x:\left(x-2\right)+3x:\left(x-2\right)-6:\left(x-2\right)\)
\(=x^3:x-x^3:2-2x^x:x+2x^x:2+3x:x-3x:2-6:x+6:2\)
\(=x^2-\dfrac{x^3}{2}-2x^{x-1}+x^x+3-\dfrac{3x}{2}-\dfrac{6}{x}+3\)
\(=x^2-\dfrac{x^3+3x}{2}-2x^{x-1}+x^x+6-\dfrac{6}{x}\)
Mk làm chi tiết từng bc một nên hơi dài
~ học tốt ~
\(\frac{x^3-4x^2+x+6}{x-2}\)
\(=\frac{x^3-2x^2-2x^2+x+6}{x-2}\)
\(=\frac{\left(x^3-2x^2\right)-\left(2x^2-x-6\right)}{x-2}\)
\(=\frac{x^2\left(x-2\right)-\left(2x^2-4x+3x-6\right)}{x-2}\)
\(=\frac{x^2\left(x-2\right)-[\left(2x^2-4x\right)+\left(3x-6\right)]}{x-2}\)
\(=\frac{x^2\left(x-2\right)-[2x\left(x-2\right)+3\left(x-2\right)]}{x-2}\)
\(=\frac{x^2\left(x-2\right)-\left(x-2\right)\left(2x+3\right)}{x-2}\)
\(=\frac{\left(x-2\right)\left(x^2-2x-3\right)}{x-2}\)
\(=\frac{\left(x-2\right)\left(x^2+x-3x-3\right)}{x-2}\)
\(=\frac{\left(x-2\right)[\left(x^2+x\right)-\left(3x-3\right)]}{x-2}\)
\(=\frac{\left(x-2\right)[x\left(x-1\right)-3\left(x-1\right)}{x-2}\)
\(=\frac{\left(x-2\right)\left(x-1\right)\left(x-3\right)}{x-2}\)
\(=\left(x-1\right)\left(x-3\right)\)
\(\frac{x^3-4x^2+x+6}{x-2}\)
\(=\frac{x^3-2x^2-2x^2+x+6}{x-2}\)
\(=\frac{x^2\left(x-2\right)-\left(x-2\right)\left(2x+3\right)}{x-2}\)
\(=\frac{\left(x-2\right)\left(x^2-2x-3\right)}{x-2}\)
\(=x^2-2x-3\)
@TrầnMinhPhong.
Đến đoạn này là được rồi .