(x-1)(x²+2x-6)=x³-1

<=> x³ + 2x² - 6x - x² - 2x + 6 = x³ - 1 

<=> x³ + 2x² - 6x - x² - 2x + 6 - x³ + 1 = 0 

<=> x² - 8x + 7 = 0 

<=> x² - x - 7x + 7 = 0 

<=> ( x - 1 )( x - 7 ) = 0 

Giải nốt 

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x-6\right)-\left(x^3-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x-6\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^2+2x-6\right)-\left(x^2+x+1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+2x-6-x^2-x-1\right)=0\)

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

Vậy tập nghiệm của phương trình là \(S=\left\{1;7\right\}\)

Đặt A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)

Ta có : A = \(\left(1+\frac{2}{4}\right).\left(1+\frac{2}{10}\right).\left(1+\frac{2}{18}\right).....\left(1+\frac{2}{n^2+3n}\right)\)

                 = \(\frac{6}{4}.\frac{12}{10}.\frac{20}{18}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)

                = \(\frac{3.2}{4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}.....\frac{\left(n+1\right).\left(n+2\right)}{n.\left(n+3\right)}\)

                = \(\frac{3.2.3.4.4.5....n}{2.3.4.5.6.....\left(n+2\right)}\)

                 = \(\frac{3.\left(n+1\right)}{n+2}\)

Vậy A = \(\frac{3.\left(n+1\right)}{n+2}\)

giúp với ạ 

a) Pt : \(Al+O_2\rightarrow Al_{2_{ }}O_3\)