Bài 1 :

a)25^4 và 125^3

Ta có

25^4=(5^2)^4=5^8

125^3=(5^3)^3=5^9

Vì 5^9 > 5^8 nên 125^3 > 25^4

b) \(6^{38}>6^{37}>5^{37}\)

c) \(243^4=3^{20};27^6=3^{18}< 3^{20}\)

d) \(7^{48}>7^{72}>6^{72}\)

2|x|=16-10

2|x|=6

|x|=3

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

\(10+2|x|=2.\left(3^2-1\right)\)

\(\Leftrightarrow10+2|x|=16\)

\(\Leftrightarrow2|x|=6\)

\(\Leftrightarrow|x|=3\)

\(\Leftrightarrow x\pm\sqrt{3}\)

Ta có :3²ⁿ =(3²)ⁿ = 9ⁿ

           2³ⁿ =(2³)ⁿ = 8ⁿ

Vì 9 > 8 =>9ⁿ > 8ⁿ =>3²ⁿ > 2³ⁿ

Vậy 3²ⁿ >2³ⁿ

Chúc bạn học tốt !!!!

3²n và 2²n

3²n = 9n

2²n = 8n

Vì 9n > 8n nên 3²n >2²n

*good luck*

(5²⁰¹⁰ - 5²⁰⁰⁸) : 5²⁰⁰⁸ = 5²⁰¹⁰ : 5²⁰⁰⁸ - 5²⁰⁰⁸ : 5²⁰⁰⁸ = 5² - 1 = 25 - 1 = 24

(7²⁰⁰⁵ + 7²⁰⁰⁴) : 7²⁰⁰⁴ = 7²⁰⁰⁵ : 7²⁰⁰⁴ + 7²⁰⁰⁴ : 7²⁰⁰⁴ = 7 + 1 = 8

[(5² .2³ - 7².2) : 2].6 - 7.2⁵ = (5². 2³ : 2 - 7².2:2).6 - 7.2⁵ = (5². 2² - 7²).6 - 7.2⁵ = (25.4 - 49).6 - 7. 32 = (100 - 49).6 - 224 = 51.6 - 224 = 306 - 224 = 82 

(5²⁰¹⁰ - 5²⁰⁰⁸) :5²⁰⁰⁸

= 5²⁰¹⁰ :5²⁰⁰⁸ -5²⁰⁰⁸ :5²⁰⁰⁸

= 5² - 1

= 25 - 1

=24

(7²⁰⁰⁵ +7²⁰⁰⁸) :7²⁰⁰⁴

= 7²⁰⁰⁵ :7²⁰⁰⁴ +7²⁰⁰⁸ :7²⁰⁰⁴

= 7 +7⁴

= 7 + 2401

= 2408 

\(\left(3,25-\frac{2}{3}x\right):\left(\frac{-7}{4}\right)=-3\)

\(\left(\frac{14}{3}-\frac{2}{3}x\right):\left(\frac{-7}{4}\right)=-3\)

\(\frac{14}{3}-\frac{2}{3}x=\left(-3\right).\left(\frac{-7}{4}\right)\)

\(\frac{14}{3}-\frac{2}{3}x=\frac{21}{4}\)

\(-\frac{2}{3}x=\frac{21}{4}-\frac{14}{3}\)

\(-\frac{2}{3}x=\frac{7}{12}\)

\(x=\frac{7}{12}:\left(-\frac{2}{3}\right)\)

\(x=-\frac{7}{8}\)

\(PT\Leftrightarrow3,25-\frac{2}{3}x=\left(-3\right).\left(\frac{-7}{4}\right)\)

\(\Leftrightarrow3,25-\frac{2}{3}x=\frac{21}{4}\)

\(\Leftrightarrow\frac{2}{3}x=3,25-\frac{21}{4}\)

\(\Leftrightarrow\frac{2}{3}x=-2\)

\(\Leftrightarrow x=\left(-2\right).\frac{3}{2}\)

\(\Leftrightarrow x=-3\)

\(PT\Leftrightarrow\left(x-2\right).\left(x+7\right)=\left(x+9\right).\left(x-1\right)\)

      \(\Leftrightarrow x^2+5x-14=x^2+8x-9\)

      \(\Leftrightarrow-3x=5\)

      \(\Leftrightarrow x=\frac{-5}{3}\left(TMĐK\right)\)

\(\frac{x-2}{x-1}=\frac{x+9}{x+7}\)

=> \(\frac{\left(x-1\right)-1}{x-1}=\frac{\left(x+7\right)+2}{x+7}\)

=> \(1-\frac{1}{x-1}=1+\frac{2}{x+7}\)

=> \(-\frac{1}{x-1}=\frac{2}{x+7}\)

=> \(-1.\left(x+7\right)=2\left(x-1\right)\)

=> \(-x-7=2x-2\)

=> \(-x-2x=-2+7\)

=> \(-3x=5\)

=> \(x=5:\left(-3\right)=-\frac{5}{3}\)

\(\left(\frac{-2x}{5}+1\right):\left(-5\right)=-\frac{1}{25}\)

\(\Leftrightarrow\frac{-2x+5}{5}=-\frac{1}{25}\cdot\left(-5\right)=\frac{1}{5}\)

\(\Leftrightarrow-2x+5=1\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\)