Ta có: \(5^{n+1}-4.5^n=5^n.5-4.5^n=\left(5-4\right)5^n=5^n\)

CHÚC BẠN HỌC TỐT.........

a) 3x\(^n\) (6x\(^{n-3}\)+1) - 2x\(^n\) ( 9x\(^{n-3}\) - 1)

= 18x\(^{n-2}\) + 3x\(^n\) - 18x\(^{n-2}\) + 2x\(^n\)

= 5x\(^n\)

b) 5\(^{n+1}\) - 4.5\(^n\)

= 5\(^n\) . ( 5-4) = 5\(^n\)

\(3x^n.\left(6x^{n-3}+1\right)-2x^n.\left(9x^{n-3}-1\right)\)

\(=18x^{n+n-3}+3x^n-18x^{n+n-3}+2x^n\)

\(=18x^{2n-3}+3x^n-18^{2n-3}+2x^n\)

\(=3x^n+2x^n=x^n\left(3+2\right)=5x^n\)

a) Ta có: \(x^{n-1}\left(x+y\right)-y\left(x^{n-1}+y^{n-1}\right)\)

\(=x^n+x^{n-1}\cdot y-x^{n-1}\cdot y-y\cdot y^{n-1}\)

\(=x^n-y^n\)

a) \(x^{n-1}\left(x+y\right)-y\left(x^{n-1}+y^{n-1}\right)\)

\(=x^{n-1}x+x^{n-1}y-x^{n-1}y-y^{n-1}y\)

\(=x^n-y^n\)

b) \(6x^n\left(x^2-1\right)+2x^3\left(3x^{n+1}+1\right)\)

\(=6x^nx^2-6x^n+2x^33x^{n+1}+2x^3\)

\(=6x^{n+2}-6x^n+6x^{3+n+1}+2x^3\)

\(=6x^{n+2}-6x^n+6x^{n+4}+2x^3\)

Đề có sai ko vậy bạn ???

\(3x^n\left(6x^{n-3}+1\right)-2x^n\left(9x^{n-3}-1\right)\)

\(=3x^n\cdot6x^{n-3}+3x^n-2x^n\cdot9x^{n-3}+2x^n\)

\(=18x^{2n-3}+3x^n-18^{2x-3}+2x^n\)

\(=3x^n+2x^n\)

\(=5x^n\)

\(=18x^{2n-3}+3x^n-18x^{2n-3}+2x^n=5x^n\)

\(3x^n\left(6x^{n-3}+1\right)-2x^n\left(9x^{n-3}-1\right)\)

\(=18x^{2n-3}+3x^n-18x^{2n-3}+2x^n\)

\(=3x^n+2x^n\)

\(=5x^n\)

\(ĐKXĐ:x\ne\pm3\)

\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{x^2+9}{\left(x+3\right)\left(x^2+9\right)}:\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)

\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)

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