Giúp mình với mình cần gấp

Giúp mình câu a thôi mình giải đc câu b rồi

E = \(6x+\sqrt{9x^2-12x+4}\)

E = \(6x+\sqrt{\left(3x-2\right)^2}\)

E = \(6x+\left|3x-2\right|\)

E = \(6x+3x-2\)

E = \(9x-2\)

F = \(5x-\sqrt{x^2+4x+4}\)

F = \(5x-\sqrt{\left(x+2\right)^2}\)

F = \(5x-\left|x+2\right|\)

F = \(5x-x+2\)

F = \(4x+2\)

\(7x\left(x-1\right)=x-1\)

\(\Leftrightarrow7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\7x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{7}\end{cases}}}\)

\(7x\left(x-1\right)=x-1\)

\(\Leftrightarrow7x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x-1\right)=0\)

\(\Rightarrow\hept{\begin{cases}x-1=0\Rightarrow x=1\\7x-1=0\Rightarrow x=\frac{1}{7}\end{cases}}\)

Vậy \(x=1\) hoặc \(x=\frac{1}{7}\)

a) a³b³ - 1 

= (ab)³ - 1

= ( ab - 1 ) ( a²b² + ab + 1 )

thank nha

\(\left(n+3\right)^2-\left(n-1\right)^2\)

\(=\left(n+3-n+1\right)\left(n+3+n-1\right)\)

\(=4\left(2n+2\right)\)

\(=8n+8\)

\(=8\left(n+1\right)⋮8\left(đpcm\right)\)

\(\left(n+6\right)^2-\left(n-6\right)^2\)

\(=\left(n+6-n+6\right)\left(n+6+n-6\right)\)

\(=0\cdot2n\)

\(=0⋮24\)

 a.                                                  Giải :

             Ta có:

                                   (n+3)² - (n-1)² = [(n+3)(n+3)-(n-1)(n-1)]

                                        =[(n²+9)-(n²-1)]=n²-n²+9-1

                                       =9-1=8(đpcm)