a, 2^2x+1 = 8³.16⁵

=> 2^2x+1 = (2³)³ . (2⁴)⁵

=> 2^2x+1 = 2⁹ . 2²⁰

^=> 2^2x+1 = 2²⁹

=> 2x+1 = 29 ( vì cơ số 2 > 0 )

=> 2x = 29 - 1

=> 2x = 28

=> x = 28 : 2

=> x = 14

Vậy x = 14

Ta có : \(6.2^{12}+2^{13}=3.2.2^{12}+2^{13}=3.2^{13}+2^{13}=2^{13}\left(3+1\right)=2^{13}.4=2^{13}.2^2=2^{15}=\left(2^3\right)^5=8^5\)\(3^{10}=\left(3^2\right)^5=9^5\)

Vì 8 < 9

=> 8⁵ < 9⁵ ( vì 5 > 0 )

=> 6.2¹² + 2¹³ < 3¹⁰

Vậy 6.2¹² + 2¹³ < 3¹⁰

Đặt \(A=\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\)

\(A>\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}+\frac{3}{14}=\frac{3}{14}.5=\frac{15}{14}>1\)

\(A< \frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{3}{10}.5=\frac{15}{10}=\frac{3}{2}< 2\)

Vậy \(1< A< 2\)

=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}\)

mà \(\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}+\frac{3}{15}=\frac{15}{15}=1\)

=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}>1\)(1)

mà \(\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}+\frac{3}{10}=\frac{15}{10}< \frac{20}{10}=2\)

=>\(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\)(2)

Từ (1) và (2) => \(1< \frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}< 2\)

\(a)\) Chưa biết -_- 

\(b)\) \(12^2-8^2=4^2.3^2-4^2.2^2=4^2\left(3^2-2^2\right)>4^2\)

Vậy \(12^2-8^2>4^2\)

\(c)\) \(7^8+7^9=7^8\left(1+7\right)=7^8.8< 7^8.49=7^8.7^2=7^{10}\)

Vậy \(7^8+7^9< 7^{10}\)

Chúc bạn học tốt ~ 

a)  13^2 = (10+3)^2 = 10^2 + 3^2 + 2.10.3 > 10^2 + 3^2

b) 12^2 - 8^2 = (8 + 4)^2 - 8^2 = 8^2 + 2.4.8 + 4^2 - 8^2 = 2.4.8 + 4^2 > 4^2

c) 7^10 = 7.7^9 = (6+1).7^9 = 6.7^9 + 7^9 = 6.7^9 + 7.7^8 > 7^8 + 7^9

Okok cảm ơn nhiều :)

a ,13² >10² +3²

^{vì 169>109}

b,12^{2 _} 8² > 4² 

vì 80>16

c,7⁸ +7⁹ < 7¹⁰

vì 46118408< 282475249

Lời giải:
Xét tỉ số:

\(\frac{5^{10}+12^{10}}{13^{10}}=(\frac{5}{13})^{10}+(\frac{12}{13})^{10}< (\frac{5}{13})^2+(\frac{12}{13})^2=1\)

$\Rightarrow 5^{10}+12^{10}< 13^{10}$

Theo đầu bài ta có:
\(\hept{\begin{cases}A=\frac{10^{12}-1}{10^{13}-1}\Rightarrow10A=\frac{10^{13}-10}{10^{13}-1}=\frac{\left(10^{13}-1\right)-9}{10^{13}-1}=1-\frac{9}{10^{13}-1}\\B=\frac{10^{10}+1}{10^{11}+1}\Rightarrow10B=\frac{10^{11}+10}{10^{11}+1}=\frac{\left(10^{11}+1\right)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}\end{cases}}\)
Do \(1-\frac{9}{10^{13}-1}< 1< 1+\frac{9}{10^{11}+1}\Rightarrow10A< 10B\Rightarrow A< B\)

a.-3/4 < 2/4

b.-8/13 > -12/13

c.-3/7 < 4/5

d.12/-18 = -2/3