Question

So sánh :

a) \(\frac{-13}{38}\)và \(\frac{29}{-88}\)

b) 3³⁰¹ và 5¹⁹⁹

c) Cho P = \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n+1\right)\left(2n+3\right)}\). So sánh P với 1

Answer 1

a,

\(-\frac{13}{38}=-1--\frac{25}{38}=-1+\frac{25}{38}\)

\(\frac{29}{-88}=-\frac{29}{88}=-1--\frac{59}{88}=-1+\frac{59}{88}\)

Vì \(\frac{25}{38}< \frac{59}{88}\Rightarrow-\frac{13}{38}< \frac{29}{-88}\)

b,

Ta có:

3³⁰¹ > 3³⁰⁰ = [3³]¹⁰⁰ = 27¹⁰⁰

5¹⁹⁹ < 5²⁰⁰ = [5²]¹⁰⁰ = 25¹⁰⁰

Mà 27¹⁰⁰ > 25¹⁰⁰ => 3³⁰¹ > 5¹⁹⁹

c,

\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left[2n+1\right]\left[2n+3\right]}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)

\(=1-\frac{1}{2n+3}< 1\)

Vậy P < 1

Answer 2

\(5^{199}=\left(5^{\frac{199}{301}}\right)^{301}\)

\(5^{\frac{199}{301}}< 3^1\)

\(\Leftrightarrow5^{199}< 3^{301}\)

Answer 3

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)

\(=1-\frac{1}{2n+3}< 1\)

Answer 4

b) Ta có :

3³⁰¹ > 3³⁰⁰ = ( 3³ ) ¹⁰⁰ = 27¹⁰⁰

5¹⁹⁹ < 5²⁰⁰ = ( 5² ) ¹⁰⁰ = 25¹⁰⁰

Vì 25¹⁰⁰ < 27¹⁰⁰ nên 3³⁰¹ > 5¹⁹⁹