a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)

b: =(1-2x)(1+2x)

c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)

d: =(x+3)^3

e: \(=\left(2x-y\right)^3\)

f: =(x+2y)(x^2-2xy+4y^2)

a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)

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

c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)

a.\(x^3-6x^2+12x-8=0\Rightarrow\)\(\left(x-2\right)^3=0\Rightarrow x=2\)

b.\(x^3+9x^2+27x+27=0\Rightarrow\left(x+3\right)^3=0\)\(\Rightarrow x=-3\)

c. \(8x^3-12x^2+6x-1=0\)

\(\Rightarrow\left(2x-1\right)^3=0\)

\(\Rightarrow x=\frac{1}{2}\)

xin lỗi vì ko giúp đc zì !!! Tại ....... e ms lớp 6 à !!!! 

a) \(100x^2-\left(x^2+25\right)^2\)

\(=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)( Áp dụng hằng đẳng thức số 3 )

b) ko khai phân tích dc bạn ạ

c) 

dễ thế này tự suy nghĩ đi biểu k đc học những cái cơ bản ak!

a, \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)

b, \(27x^3+27x^2+9x+1=0\Leftrightarrow27x^3+1+27x^2+9x=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2-3x+1\right)+9x\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(9x^2+2>0\right)=0\Leftrightarrow x=-\frac{1}{3}\)

c, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=\frac{2}{3};x=-1\)

d, \(\left(x+1\right)^3-25\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\Leftrightarrow x=-1;x=-6;x=4\)

Câu b đề sai nha bạn.

a) \(x^3+9x^2+27x+27=\left(x+3\right)^3\)

b) \(3\sqrt{3x^3}+18x^2+12\sqrt{3x}+8=\left(\sqrt{3x}+2\right)^3\)

c) \(\dfrac{1}{4}-x^2=\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)

\(a,=\left(x+2\right)^3=100^3=1000000\\ b,=\left(x-3\right)^3=10^3=1000\\ c,Sửa:x^3-3x^2+3x-1=\left(x-1\right)^3=10^3=1000\\ d,Sửa:x^3+15x^2+75x+125=\left(x+5\right)^3=20^3=8000\)

Bài 4:

a) Ta có: \(x^3+6x^2+12x+8\)

\(=x^3+2x^2+4x^2+8x+4x+8\)

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

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

\(=\left(x+2\right)^3\)

b) Ta có: \(x^3-3x^2+3x-1\)

\(=x^3-x^2-2x^2+2x+x-1\)

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

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

\(=\left(x-1\right)^3\)

c) Ta có: \(1-9x+27x^2-27x^3\)

\(=1-3x-6x+18x^2+9x^2-27x^3\)

\(=\left(1-3x\right)-6x\left(1-3x\right)+9x^2\left(1-3x\right)\)

\(=\left(1-3x\right)\left(1-6x+9x^2\right)\)

\(=\left(1-3x\right)^3\)

d) Ta có: \(x^3+\frac{3}{2}x^2+\frac{3}{4}x+\frac{1}{8}\)

\(=x^3+3\cdot x^2\cdot\frac{1}{2}+3\cdot x\cdot\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3\)

\(=\left(x+\frac{1}{2}\right)^3\)

e) Ta có: \(27x^3-54x^2y+36xy^2-8y^3\)

\(=\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot2y+3\cdot3x\cdot\left(2y\right)^2-\left(2y\right)^3\)

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