A>5²⁰¹

Vì khi tính một vài số của A thì đã lớn hơn 5²⁰¹

Ta có:

\(A=5+5^2+5^3+5^4+...+5^{200}\)

\(5A=5.\left(5+5^2+5^3+...+5^{200}\right)\)

\(5A=5^2+5^3+5^4+...+5^{201}\)

\(5A-A=\left(5^2+5^3+5^4+...+5^{200}+5^{201}\right)-\left(5+5^2+5^3+5^4+...+5^{200}\right)\)

\(4A=5^2+5^3+5^4+...+5^{200}+5^{201}-5-5^2-5^3-5^4-...-5^{200}\)

\(4A=\left(5^2-5^2\right)+\left(5^3-5^3\right)+\left(5^4-5^4\right)+...+\left(5^{200}-5^{200}\right)+5^{201}-5\)

\(4A=0+0+0+...+0+5^{201}-5\)

\(4A=5^{201}-5\)

\(A=\frac{5^{201}-5}{4}\)

Vì \(5^{201}-5< 5^{201}\)

\(\Rightarrow\frac{5^{201}-5}{4}< \frac{5^{201}}{4}< 5^{201}\)

hay \(A< 5^{201}\)

Vậy \(A< 5^{201}\)

1.\(x=\dfrac{5}{2}\)

2.\(y=\dfrac{10}{2}=5\)

3.\(-3y=15\Leftrightarrow y=-5\)

4/\(9t=-11\Leftrightarrow t=-\dfrac{11}{9}\)

A = 5 + 5² + 5³ + 5⁴ + ... + 5²⁰⁰

5A = 5² + 5³ + 5⁴ + 5⁵ + ... + 5²⁰¹

5A - A = (5² + 5³ + 5⁴ + 5⁵ + ... + 5²⁰¹) - (5 + 5² + 5³ + 5⁴ + ... + 5²⁰⁰)

4A = 5²⁰¹ - 5 < 5²⁰¹

=> A < 5²⁰¹

tối mik giải cho nhé, giờ bận

Bài 1:

a: Sửa đề: 1/3^200

1/2^300=(1/8)^100

1/3^200=(1/9)^100

mà 1/8>1/9

nên 1/2^300>1/3^200

b: 1/5^199>1/5^200=1/25^100

1/3^300=1/27^100

mà 25^100<27^100

nên 1/5^199>1/3^300

1853567804232223

a) (-13).5 < 0

b) 200 > 200 . (-3)

c) (-17) . 2 < -17

d) (-11) . 8 < -11.

Lời giải:
$M=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{2014}{5^{2014}}$

$5M=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{2014}{5^{2013}}$

$\Rightarrow 4M=5M-M=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}-\frac{2014}{5^{2014}}$
$4M+\frac{2014}{5^{2014}}=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2013}}$

$5(4M+\frac{2014}{5^{2014}})=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2012}}$

$\Rightarrow 4(4M+\frac{2014}{5^{2014}})=5-\frac{1}{5^{2013}}$

$M=\frac{5}{16}-\frac{1}{16.5^{2013}-\frac{2014}{4.5^{2014}}$