`a,-x^3/8 + 3/(4x^2) - 3/(2x) +1`

`=-(x^3/8 - 3/(4x^2) + 3/(2x) - 1)`

`=-(x/2 - 1)^3`

`b,x^6 - 3/(2x^{4} y) + 3/(4x^{2}y^{2}) - 1/(8y^{3})`

`=(x^3 - 1/(2y))^{3}`

a) \(x^2+2x+1\)

\(=\left(x+1\right)^2\)

b) \(9-24x+16x^2\)

\(=\left(3-4x\right)^2\)

c) \(4x^2+\dfrac{1}{4}+2x\)

\(=4x^2+2x+\dfrac{1}{4}\)

\(=\left(2x+\dfrac{1}{2}\right)^2\)

này mình có vài câu không làm được, xin lỗi bạn nha

\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)

Bài 1: 

a) \(a^2-6a+9=\left(a-3\right)^2\)

b) \(\dfrac{1}{4}x^2+2xy^2+4y^4=\left(\dfrac{1}{2}x+2y^2\right)^2\)

Bài 2:

a)  \(\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\)

\(\Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\)

b) \(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)

a) \(x^4+4x^2+4=\left(x^2+2\right)^2\)

b) \(\left(2y-x\right)^2+2\left(2y-x\right)+1=\left(2y-x+1\right)^2\)

c) \(\left(2a-4b\right)^2+4a-8b+1=\left(2a-4b\right)^2+2\cdot\left(2a-4b\right)\cdot1+1^2=\left(2a-4b+1\right)^2\)

Bài 5:

a: \(x^3-1-\left(x^2+2x\right)\left(x-2\right)=5\)

=>\(x^3-1-\left(x^3-2x^2+2x^2-4\right)=5\)

=>\(x^3-1-x^3+4=5\)

=>3=5(vô lý)

b: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)

=>\(x^3+3x^2+3x+1-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)

=>\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)

=>12x-4=-10

=>12x=-6

=>x=-6/12=-1/2

BÀi 3:

a: \(A=\left(a+b\right)^3-\left(a-b\right)^3\)

\(=a^3+3a^2b+3ab^2+b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\)

\(=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3=6a^2b+2b^3\)

b:Sửa đề: \(A=\left(u-v\right)^3+3uv\left(u-v\right)\)

\(=u^3-3u^2v+3uv^2-v^3+3u^2v-3uv^2=u^3-v^3\)

c: \(C=6\left(c-d\right)\left(c+d\right)+2\left(c-d\right)^2-\left(c-d\right)^3\)

\(=6\left(c^2-d^2\right)+2\left(c^2-2cd+d^2\right)-c^3+3c^2d-3cd^2+d^3\)

\(=6c^2-6d^2+2c^2-4cd+2d_{}^2-c^3+3c^2d-3cd^2+d^3\)

\(=-c^3+3c^2d-3cd^2+d^3+8c^2-4cd-4d^2\)

Bài 2:

a: \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

b: \(m^3+9m^2n+27mn^2+27n^3\)

\(=m^3+3\cdot m^2\cdot3n+3\cdot m\cdot\left(3n\right)^2+\left(3n\right)^3\)

\(=\left(m+3n\right)^3\)

a. (3a + 1)³

= 27a³ + 27a² + 9a + 1

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: \(m^3+9m^2n+27mn^2+27n^3=\left(m+3n\right)^3\)

`a, a^2 + 10ab + 25b^2 = (a+5b)^2`

`b, 1 + 9a^2 - 6a = (3a-1)^2`

a) \(a^2+10ab+25b^2\)

\(=a^2+2\cdot5b\cdot a+\left(5b\right)^2\)

\(=\left(a+5b\right)^2\)

b) \(1+9a^2-6a\)

\(=1-6a+9a^2\)

\(=\left(1+3a\right)^2\)