a.       P = 0

 b . a = 0

Với giá trị a là 1 số tự nhiên thì P>0.

Ta có:

\(P=2a^{2n+1}-3a^{2n}+5a^{2n+1}-7a^{2n}+3a^{2n+1}\)

\(P=\left(2a^{2n+1}+5a^{2n+1}+3a^{2n+1}\right)+\left(-3a^{2n}-7a^{2n}\right)\)

Suy ra: \(P=10a^{2n+1}+\left(-10a\right)^{2n}\)

Mà \(2n⋮2\)còn \(2n+1\)ko chia hết cho 2

Do đó: \(a>0\)thì P>0

Nhầm cái chỗ suy ra:

\(P=10a^{2n+1}+\left(-10\right)a^{2n}\)

Bổ sung thêm

Vì 2n+1 > 2n

a^2n =x ; x>=0 mọi a; n thuộc n

\(P=2.a.x-3x+5.a.x-7x+3.a.x\)

\(P=10.a.x-10x=10x\left(a-1\right)\)

\(P>0\Rightarrow\left\{{}\begin{matrix}x>0\\a>1\end{matrix}\right.\) ; a>1 => a>0 => kết luân a>1

\(p=2a^{2n+1}+5a^{2n+1}-3a^{2n}-7a^{2n}+3a^{2n1}\)

\(p=\left(2a^{2n+1}+5a^{2n+1}+3a^{2n+1}\right)+\left(-3a^{2n}-7a^{2n}\right)\)

\(\Rightarrow P=10a^{2n+1}+\left(-10a\right)^{2n}\)

Mà \(2n⋮2\)còn \(2n+1⋮2̸\)

Do đó \(a>2\)thì\(P>0\)

cHÚC BẠN HỌC TÔT ~!!!

\(P=10a^{2n+1}-10a^{2n}>0\Leftrightarrow10a^{2n+1}>10a^{2n}\Leftrightarrow10a^{2n}.a>10a^{2n}\Leftrightarrow\hept{\begin{cases}a>0\\a>1\end{cases}\Leftrightarrow a>1}\)

ĐKXĐ: \(a\ne\frac{3}{2},a\ne-\frac{3}{2}\)

a, \(P=\left(\frac{a-1}{2a-3}-\frac{3a}{4a+6}+\frac{7a-2a^2-1}{18-8a^2}\right):\frac{1}{6-4a}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}+\frac{7a-2a^2-1}{2\left(9-4a^2\right)}\right):\frac{-1}{4a-6}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}-\frac{7a-2a^2-1}{2\left(4a^2-9\right)}\right):\frac{-1}{2\left(2a-3\right)}\)

\(=\left(\frac{a-1}{2a-3}-\frac{3a}{2\left(2x+3\right)}-\frac{7a-2a^2-1}{2\left(2a-3\right)\left(2a+3\right)}\right)\left[-2\left(2a-3\right)\right]\)

\(=\left[\frac{2\left(a-1\right)\left(2a+3\right)-3a\left(2a-3\right)-\left(7a-2a^2-1\right)}{2\left(2a-3\right)\left(2a+3\right)}\right]\left[-2\left(2a-3\right)\right]\)

\(=\frac{4a-5}{2\left(2a-3\right)\left(2a+3\right)}\left[-2\left(2a-3\right)\right]\)

\(=-\frac{\left(4a-5\right)}{2a+3}=\frac{5-4a}{2a+3}\)