Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(2x+1\right)^3\)
\(=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1+1\)
\(=8x^3+12x^2+6x+1\)
b) \(\left(x-3\right)^3\)
\(=x^3-3.x^2.3+3.x.3^2-3^3\)
\(=x^3-9x^2+27x-27\)
Bài 2:
a: \(x^3+15x^2+75x+125=\left(x+5\right)^3\)
b: \(1-15y+75y^2-125y^3=\left(1-5y\right)^3\)
c: \(8x^3+4x^2y+\dfrac{3}{2}xy^2+8y^3=\left(2x+2y\right)^3\)
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
\(a,=x^3+3x^2+3x+1\\ b,=8x^3+36x^2+54x+27\\ c,=x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}\\ d,=x^6-6x^4+12x^2-8\\ e,=8x^3-36x^2y+54xy^2-27y^3\)
a, \(\left(2x-1\right)^2=\left(2x\right)^2-2.2x.1+1^2=4x^2-4x+1\) 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, \(-x^2+100\) = \(x^2+10^2\)
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
4:
a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2
b: 8(7^2+1)(7^4+1)(7^8+1)
=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^16-1)<7^16-1
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
mik chỉ biết bài 5 thôi !
\(1,=-\left(y^2+12y+36\right)=-y^2-12y-36\)
\(2,=-\left(16-8y+y^2\right)=-16+8y-y^2\)
\(3,=-\left(\dfrac{4}{9}+\dfrac{4}{3}x+x^2\right)=-\dfrac{4}{9}-\dfrac{4}{3}x-x^2\)
\(4,=-\left(x^2-3x+\dfrac{9}{4}\right)=-x^2+3x-\dfrac{9}{4}\)
\(5,-\left(2+3y\right)^2=-\left(4+12y+9y^2\right)=-4-12y-9y^2\)
.... mấy ý còn lại bn tự lm nhé, tương tự thhooi
1) \(-\left(y+6\right)^2=-y^2-12y-36\)
2) \(-\left(4-y\right)^2=-y^2+8y-16\)
3) \(-\left(x+\dfrac{2}{3}\right)^2=-x^2-\dfrac{4}{3}x-\dfrac{4}{9}\)
4) \(-\left(x-\dfrac{3}{2}\right)^2=-x^2+3x-\dfrac{9}{4}\)
5) \(-\left(3y+2\right)^2=-9y^2-12y-4\)
6) \(-\left(2y-3\right)^2=-4y^2+12y-9\)
7) \(-\left(5x+2y\right)^2=-25x^2-20xy-4y^2\)
8) \(-\left(2x-\dfrac{3}{2}\right)^2=-4x^2+6x-\dfrac{9}{4}\)
a) \(\left(3x-2\right)^2=\left(3x\right)^2-2.3x.2+2^2=9x^2-12x+4\)
b) \(\left(\dfrac{x}{3}+y^3\right)^2=\left(\dfrac{x}{3}\right)^2+2\dfrac{x}{3}y^3+\left(y^3\right)^2=\dfrac{x^2}{9}+\dfrac{2}{3}xy^3+y^6\)
c) \(9x^2-225=\left(3x\right)^2-\left(15\right)^2=\left(3x-15\right)\left(3x+15\right)\)
d) \(\left(2x-3y\right)^3=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^2-\left(3y\right)^3=8x^3-3.4x^2.3y+6x.9y^2-27y^3=8x^3-36x^2y+54xy^2-27y^3\)
e) \(\left(2x^2+\dfrac{3}{2}\right)^3=\left(2x^2\right)^3+3\left(2x^2\right)^2\dfrac{3}{2}+3.2x^2\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3=8x^6+3.4x^4.\dfrac{3}{2}+6x^2.\dfrac{9}{4}+\dfrac{27}{8}=8x^6+18x^4+\dfrac{27}{2}x^2+\dfrac{27}{8}\)
f) \(\left(-2xy^2+\dfrac{1}{2}x^3y\right)^3=\left(-2xy^2\right)+3\left(-2xy^2\right)^2\dfrac{1}{2}x^3y+3\left(-2xy^2\right)\left(\dfrac{1}{2}x^3y\right)^2+\left(\dfrac{1}{2}x^3y\right)^3=-8x^3y^6+3.4x^2y^4.\dfrac{1}{2}x^3y-6xy^2.\dfrac{1}{4}x^6y^2+\dfrac{1}{8}x^9y^3=-8x^3y^6+6x^5y^5-\dfrac{3}{2}x^7y^4+\dfrac{1}{8}x^9y^3\)
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) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)
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