Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(=4x^2-12x+9\)
b) \(=4x^2+2x+\dfrac{1}{4}\)
c) \(=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)
1) \(\left(x+1\right)^2=x^2+2x+1\)
2) \(\left(2x+1\right)^2=4x^2+4x+1\)
3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)
4) \(\left(2x+3\right)^2=4x^2+12x+9\)
5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)
6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)
7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)
8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)
9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)
10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)
Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)
a, 81 - y2 = 92 - y2 = ( 9 - y ).( 9 + y )
b, ( x - 3y )3 = x3 - 3x2.3y + 3x.( 3y )2 - ( 3y )3 = x3 - 9x2y + 27xy2 - 27y3
c, ( 2x + 2 )3 = ( 2x )3 + 3.( 2x )2.2 + 3.2x.22 + 23 = 8x3 + 24x2 + 24x + 8
a, 81 - y2 = 92 - y2 = ( 9 - y ).( 9 + y )
b, ( x - 3y )3 = x3 - 3x2 .3y + 3x.( 3y )2 - ( 3y )3 = x3 - 9x2y + 27xy2 - 27y3
c, ( 2x + 2 )3 = ( 2x )3 + 3.( 2x )2 .2 + 3.2x.22 + 23 = 8x3 + 24x2 + 24x + 8
ok thế là xong
a)
x3+3x2+3x+1
b)8x3+18x2+54x+27
c)x3+3636x2+3636x+1780360128
d)x6-6x2+12x-8
e)
8x3-36x2y+54xy2-27y3
a, \(\left(2x-1\right)^2=\left(2x\right)^2-2.2x.1+1^2=4x^2-4x+1\) 1
b, \(\left(3x+1\right)^2=\left(3x\right)^2+2.3x.1+1^2\) \(=9x^2+6x+1\)
c, \(\left(\frac{1}{2}x+1\right)^3\) = \(\left(\frac{1}{2}x\right)^3+3.\left(\frac{1}{2}x\right)^2.1+3.\frac{1}{2}x.1^2+1^3\)
= \(\frac{1}{8}x^3+\frac{3}{4}x^2+\frac{3}{2}x+1\)
d, \(-x^2+100\) = \(x^2+10^2\)