Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,\left(6x+1\right)\left(x+2\right)-2x\left(3x-5\right)\)
\(=6x^2+12x+x+2-6x^2+10x\)
\(=23x+2\)
a) (6x + 1)(x + 2) - 2x(3x - 5)
= 6x2 + 12x + x + 2 - 6x2 + 10x
= (6x2 - 6x2) + (12x + x + 10x) + 2
= 23x + 2
b) (2x - 1)2 - (2x - 3)(2x + 3)
= 4x2 - 4x + 1 - 4x2 + 9
= (4x2 - 4x2) - 4x + (1 + 9)
= -4x + 10
c) (2x - 3)3 - (3x + 1)(5 - 4x) - 16x2
= 8x3 - 36x2 + 54x - 15x + 12x2 - 5 + 4x - 16x2
= 8x3 - (36x2 - 12x2 + 16x2) + (54x - 15x + 4x) - 5
= 8x3 - 40x2 + 43x - 5
d) (3x + 2) - (x - 5) - x(3x - 13)
= 3x + 2 - x + 5 - 3x2 + 13x
= (3x - x + 13x) + (2 + 5) - 3x2
= 15x + 7 - 3x2
a) ( x - 2 )3 - x( x + 1 )( x - 1 ) + 6x( x - 3 )
= x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 - 18x
= x3 - 6x - 8 - x3 + x
= -5x - 8
b) ( x + 1 )3 - ( x - 1 )3 - 6( x - 1 )2
= x3 + 3x2 + 3x + 1 - ( x3 - 3x2 + 3x - 1 ) - 6( x2 - 2x + 1 )
= x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 + 12x - 6
= 12x - 4
c) ( 2x + 1 )( 4x2 - 2x + 1 ) + ( 2 - 3x )( 4 + 6x + 9x2 ) - 9
= ( 2x )3 + 13 + 23 - ( 3x )3 - 9
= 8x3 + 1 + 8 - 27x3 - 9
= -19x3
d) ( x + 1 )3 + ( x - 1 )3 + x3 - 3x( x - 1 )( x + 1 )
= x3 + 3x2 + 3x + 1 + x3 - 3x2 + 3x - 1 + x3 - 3x( x2 - 1 )
= 3x3 + 6x - 3x2 + 3x
= 9x
a, nhân ra để đc 1 hdt sau đó đưa phân tử chunglaf đc mâk
b,nhân ra cx đc mà dùng hdt cx đc
hok tốt
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
a) ( x2 + 1/x + 1/9 ).( x - 1/3 ) - ( x- 1/3 )3
\(=\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\left(x-\frac{1}{3}\right)\left[x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2x}{3}-\frac{1}{9}\right]\)
\(=\left(x-\frac{1}{3}\right)\left[\frac{2x}{3}+\frac{1}{x}\right]\)