a)Ta có : \(\hept{\begin{cases}3^{100}=\left(3^2\right)^{50}=9^{50}\left(1\right)\\2^{150}=\left(2^3\right)^{50}=8^{50}\left(2\right)\end{cases}}\)

Mà 9 > 8 => 9⁵⁰ > 8⁵⁰ => 3¹⁰⁰ > 2¹⁵⁰

Vậy 3¹⁰⁰ > 2¹⁵⁰

b) Ta có : \(\hept{\begin{cases}27^5=\left(3^3\right)^5=3^{15}\left(3\right)\\243^3=\left(3^5\right)^3=3^{15}\left(4\right)\end{cases}}\)

Từ (3) và (4) => 3¹⁵ = 3¹⁵ hay 27⁵ = 243³

Vậy 27⁵ = 243³ ( nên sửa lại 245 --> 243 nhá)

c) Ta có : \(81^{75}=\left(3^4\right)^{75}=3^{300}=\left(3^3\right)^{100}=27^{100}\)

Mà 27 < 30 => 27¹⁰⁰ < 30¹⁰⁰ hay 81⁷⁵ < 30¹⁰⁰

Vậy 81⁷⁵ < 30¹⁰⁰

a. 

\(3^{100}=\left(3^2\right)^{50}=9^{50}\) 

\(2^{150}=\left(2^3\right)^{50}=8^{50}\) 

\(9^{50}>8^{50}\) 

\(\Rightarrow3^{100}>2^{150}\) 

b. 

\(27^5=\left(3^3\right)^5=3^{15}\) 

\(243^3=\left(3^5\right)^3=3^{15}\) 

\(3^{15}=3^{15}\) 

\(\Rightarrow27^5=243^3\) 

c. 

\(81^{75}=\left(3^4\right)^{75}=3^{300}=\left(3^3\right)^{100}=27^{100}\) 

\(27^{100}< 30^{100}\Rightarrow81^{75}< 30^{100}\) 

e) 3ⁿ⁺² + 5.3^{n + 1} = 216

=> 3ⁿ . 3² + 5.3ⁿ . 3¹ = 216

=> 3ⁿ . 9 + 15.3ⁿ = 216

=> 3ⁿ ( 9 + 15) = 216

=> 3ⁿ . 24 = 216

=> 3ⁿ = 9

=> n = 2

f) 5^{n + 1} - 5^{n - 1} = 125⁴ . 2³ . 37

=> 5ⁿ . 5 - 5ⁿ  . 1/5 = 125⁴ . 2³ . 37

=> 5ⁿ ( 5 - 1/5) = 125⁴ . 2³ . 37

=> 5ⁿ . 24/5 = 125⁴ . 2³ . 37

=> n không thỏa mãn

a,\(3^{n+2}+5.3^{n+1}=216\)

\(3^n.9+5.3^n.3=216\)

\(3^n\left(9+5.3\right)=216\)

\(3^n.24=216\)

\(3^n=9\Rightarrow n=2\)

c) 3^{n + 2} - 3^{n + 1} = 1458

=> 3ⁿ . 3² - 3ⁿ . 3¹ = 1458

=> 3ⁿ (3² - 3) = 1458

=> 3ⁿ . 6 = 1458

=> 3ⁿ = 243

=> n = 5

d) 2^{n - 1} + 4.2ⁿ = 9.2⁵

=> 2ⁿ . 2¹ + 4.2ⁿ = 9.2⁵

=> 2ⁿ (2 + 4) = 9.2⁵

=> 2ⁿ . 6 = 9.2⁵

=> 2ⁿ = \(\frac{9\cdot2^5}{6}=48\)

=> không tìm được x

a. 

\(5^n+5^{n+2}=650\) 

\(5^n\left(1+5^2\right)=650\) 

\(5^n\left(1+25\right)=650\) 

\(5^n\cdot26=650\) 

\(5^n=650:26\) 

\(5^n=25\) 

\(5^n=5^2\Rightarrow n=2\) 

b. 

\(3^{n+3}+5\cdot3^n=864\) 

\(3^n\left(3^3+5\right)=864\) 

\(3^n\left(27+5\right)=864\) 

\(3^n\cdot32=864\) 

\(3^n=864:32\) 

\(3^n=27\) 

\(3^n=3^3\Rightarrow n=3\)

a) 5ⁿ + 5ⁿ⁺² = 650

=> 5ⁿ + 5ⁿ . 5² = 650

=> 5ⁿ (1 + 5²) = 650

=> 5ⁿ . 26 = 650

=> 5ⁿ = 25

=> n = 2

b) 3^{n+ 3} + 5.3ⁿ = 864

=> 3ⁿ . 3³ + 5.3ⁿ = 864

=> 3ⁿ(3³ + 5) = 864

=> 3ⁿ . 32 = 864

=> 3ⁿ = 27

=> n = 3

Phần C đề thiếu

\(D=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}\)

\(\Rightarrow3D=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)

\(\Rightarrow3D-D=(1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}})-\)\((\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}})\)

\(\Rightarrow2D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow6D=3+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)

\(\Rightarrow6D-2D=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}\)

\(\Rightarrow4D=3-\frac{203}{3^{100}}\)

\(\Rightarrow D=\frac{3}{4}-\frac{\frac{203}{3^{100}}}{4}< \frac{3}{4}\left(đpcm\right)\)

sửa rồi nhá bn

a/

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)

\(A=2A-A=1-\frac{1}{2^{100}}< 1\)

b/

\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2018}}\)

\(2B=3B-B=1-\frac{1}{3^{2019}}\Rightarrow B=\frac{1}{2}-\frac{1}{2.3^{2019}}< \frac{1}{2}\)

a. 

\(\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{12}\right)+\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{12}\right)\) 

\(=\left(\frac{5}{4}+\frac{3}{5}+\frac{3}{8}-\frac{7}{5}\right):\left(-\frac{11}{12}\right)\)  

\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{12}\right)\) 

\(=\frac{33}{40}:\left(-\frac{11}{12}\right)\) 

\(=\frac{33}{40}\cdot\left(-\frac{12}{11}\right)\) 

\(=\frac{-9}{10}\)  

b. 

\(\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{15}\right)+\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\) 

\(=\left(\frac{3}{8}-\frac{7}{5}+\frac{5}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)  

\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{15}\right)\) 

\(=\frac{33}{40}:\left(-\frac{11}{15}\right)\) 

\(=\frac{33}{40}\cdot\left(-\frac{15}{11}\right)\) 

\(=\frac{-9}{8}\)

a) \(\left|-\frac{2}{11}+\frac{3}{22}x\right|-\frac{1}{2}=\frac{5}{7}\)

=> \(\left|-\frac{2}{11}+\frac{3}{22}x\right|=\frac{17}{14}\)

=> \(\orbr{\begin{cases}-\frac{2}{11}+\frac{3}{22}x=\frac{17}{14}\\-\frac{2}{11}+\frac{3}{22}x=-\frac{17}{14}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{215}{21}\\x=-\frac{53}{7}\end{cases}}\)

b) \(-\frac{7}{8}x-5\frac{3}{4}=3\)

=> \(-\frac{7}{8}x-\frac{23}{4}=3\)

=> \(-\frac{7}{8}x=3+\frac{23}{4}=\frac{35}{4}\)

=> \(x=\frac{35}{4}:\left(-\frac{7}{8}\right)=\frac{35}{4}\cdot\left(-\frac{8}{7}\right)=-10\)

c) \(2x+\left(-\frac{2}{7}\right)-7=-11\)

=> \(2x-\frac{2}{7}-7=-11\)

=> \(2x=-11+7+\frac{2}{7}=-\frac{26}{7}\)

=> \(x=\left(-\frac{26}{7}\right):2=-\frac{13}{7}\)

d) \(\frac{3}{7}+x:\frac{14}{15}=\frac{1}{2}\)

=> \(x:\frac{14}{15}=\frac{1}{2}-\frac{3}{7}=\frac{1}{14}\)

=> \(x=\frac{1}{14}\cdot\frac{14}{15}=\frac{1}{15}\)

\(\frac{x+5}{x+3}< 1\)

=> \(\frac{x+3+2}{x+3}< 1\)

=> \(1+\frac{2}{x+3}< 1\)

=> \(\frac{2}{x+3}< 0\)

=> x + 3 < 0

=> x < -3

Vậy x < -3 để \(\frac{x+5}{x+3}< 1\)

x + y = 3y

=> x= 3y-y

\(\frac{1}{x}\)= \(\frac{1}{3y-y}\)

\(\frac{1}{x}\)+\(\frac{1}{y}\)=\(\frac{1}{3y-y}\)+\(\frac{1}{y}\)

= \(\frac{y}{y\left(3y-y\right)}+\frac{3y-y}{y\left(3y-y\right)}\)=\(\frac{y+3y-y}{3y^2-y^2}\)=\(\frac{3y}{y^2\left(3-1\right)}=\frac{3}{2y}\)

Ta có x+y=3y

\(\Rightarrow x=2y\)

Thay vào ta có

\(\frac{1}{2y}+\frac{1}{y}=\frac{1}{2y}+\frac{2}{2y}=\frac{3}{2y}\)