\(10^{26}\) và \(9^{10}\)

Có: \(10>9\)

\(26>10\)

\(\Rightarrow10^{26}>9^{10}\)

C2: \(10^{26}=10^{10}.10^{16}\)

Vì: \(10^{10}>9^{10}\) 

\(\Rightarrow10^{10}.10^{16}>9^{10}\)

\(\Rightarrow10^{26}>9^{10}\)

 C1 10 ^ 26 = 100 ^ 25 = (100^5)^5 = 10000000000 ^ 5 > 81 ^ 5 = 9 ^10 => 10 ^ 26 > 9 ^ 10

C2 10 ^ 26 > 10^10 > 9^ 10  => 10 ^ 26 > 9 ^ 10 

Ta có: A=2^0+2^1+2^2+2^3+...+2^2010
         =>2A=2+2^2+2^3+2^4+...+2^2011
    =>2A-A=(2+2^2+2^3+...+2^2011)-( 1+2+2^2+2^3+...+2^2010)
    =>A= 2^2011-1

Từ đó ta suy ra A=B (=2^2011-1)
 k nha!

2A=2¹+2²+...+2²⁰¹¹

Suy ra: A=2A-A = (2¹+2²+...+2²⁰¹¹) - (2⁰+2¹+...+2²⁰¹⁰)=2²⁰¹¹-1=B

Vậy: A=B.

hihi mk cảm ơn mk sẽ lấy nick phụ k cho các bạn nhé !

mik hieu dc 3 cau roi

Bài 1:

\(2^{49}=\left(2^7\right)^7=128^7;5^{21}=\left(5^3\right)^7=125^7\\ Vì:128^7>125^7\Rightarrow2^{49}>5^{21}\)

Bài 2:

\(a,S=1+3+3^2+3^3+...+3^{99}\\ =\left(1+3+3^2+3^3\right)+3^4.\left(1+3+3^2+3^3\right)+...+3^{96}.\left(1+3+3^2+3^3\right)\\ =40+3^4.40+...+3^{96}.40\\ =40.\left(1+3^4+...+3^{96}\right)⋮40\\ b,S=1+4+4^2+4^3+...+4^{62}\\ =\left(1+4+4^2\right)+4^3.\left(1+4+4^2\right)+...+4^{60}.\left(1+4+4^2\right)\\ =21+4^3.21+...+4^{60}.21\\ =21.\left(1+4^3+...+4^{60}\right)⋮21\)

Bài 1 :

\(2^{49}=\left(2^7\right)^7=128^7\)

\(5^{21}=\left(5^3\right)^7=125^7\)

mà \(125^7< 128^7\)

\(\Rightarrow2^{49}>5^{21}\)

Bài 2 :

a) \(S=1+3+3^2+3^3+...3^{99}\)

\(\Rightarrow S=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)...+3^{96}\left(1+3+3^2+3^3\right)\)

\(\Rightarrow S=40+40.3^4+...+40.3^{96}\)

\(\Rightarrow S=40\left(1+3^4+...+3^{96}\right)⋮40\)

\(\Rightarrow dpcm\)

b) \(S=1+4+4^2+4^3+...4^{62}\)

\(\Rightarrow S=\left(1+4+4^2\right)+4^3\left(1+4+4^2\right)+...4^{60}\left(1+4+4^2\right)\)

\(\Rightarrow S=21+4^3.21+...4^{60}.21\)

\(\Rightarrow S=21\left(1+4^3+...4^{60}\right)⋮21\)

\(\Rightarrow dpcm\)

Cách 1:

Ta có: 9¹⁰ < 10¹⁰ < 10²⁰ => 9¹⁰ < 10²⁰

Cách 2: 

Ta có: 10²⁰ = (10²)¹⁰ = 100¹⁰ > 9¹⁰ => 10²⁰ > 9¹⁰

bài của tôi giống soyeon tiểu bài giảng ^^

               nhưng lãm cách 1 dễ hiểu hơn nhá

 ###

a) \(\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=32\)

\(\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=32\)

\(\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=32\)

\(\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=32\)

\(\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=32\)

\(\frac{7}{4}x.33\cdot\frac{4}{21}=32\)

đến đây thì bn tự lm đk r

b) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}\)

\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}\)

\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}\)

\(2.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}\)

\(1-\frac{2}{x}=\frac{2007}{2009}\)

\(\frac{2}{x}=\frac{2}{2009}\)

=> x = 2009

Ý bn đề vậy à ??? \(A=\frac{1}{7}+\frac{1}{13}+\frac{1}{25}+\frac{1}{49}+\frac{1}{97}....1\)

\(A=\frac{14}{98}+\frac{7}{91}+\frac{4}{100}+\frac{2}{98}+\frac{1}{97}< \frac{14}{91}+\frac{7}{91}+\frac{4}{91}+\frac{2}{91}+\frac{1}{91}=\frac{28}{91}=\frac{84}{273}< \frac{1}{3}=\frac{91}{273}\)

Vậy A < \(\frac{1}{3}\)

Hơi khó hiểu một chút nha bn

~Chúc bạn học tốt~