Ta có : \(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)

           \(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}\)     

           \(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)

            ...

            \(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow\)K<\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

K<\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

K<\(\frac{1}{3}-\frac{1}{100}< \frac{1}{3}\)

\(\Rightarrow K< \frac{1}{3}\)  (1)

Ta có : \(\frac{1}{4^2}=\frac{1}{4.4}=\frac{1}{16}\)

            \(\frac{1}{5^2}=\frac{1}{5.5}>\frac{1}{5.6}\)

            \(\frac{1}{6^2}=\frac{1}{6.6}>\frac{1}{6.7}\)

             ...

             \(\frac{1}{99^2}=\frac{1}{99.99}>\frac{1}{99.100}\)

             \(\frac{1}{100^2}=\frac{1}{100.100}>\frac{1}{100.101}\)

\(\Rightarrow K>\frac{1}{16}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}+\frac{1}{100.101}\)

K>\(\frac{1}{16}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)

K>\(\frac{1}{16}+\frac{1}{5}-\frac{1}{101}>\frac{1}{5}\)  (2)

Từ (1) và (2)

\(\Rightarrow\frac{1}{5}< K< \frac{1}{3}\)

Vậy \(\frac{1}{5}< K< \frac{1}{3}.\)

a,\(5^3.2-100:4+2^3.5\)

= 125 . 2 - 25 + 8 . 5

= 250 - 25 + 40

= 265

b, \(6^2:9+50.2-3^3.3\)

= 36 : 9 + 100 - 27 . 3

= 4 + 100 - 81

= 23

bạn kia ko làm các ý khác à

b) \(5^3\cdot2-100:4+2^3\cdot5\)

\(=125\cdot2-25+8\cdot5\)

\(=250-25+40\)

\(=225+40=265\)

c) \(6^2:9+50\cdot2+3^3-3\)

\(=36:9+100+27-3\)

\(=4+100+27-3\)

\(=104+27-3=131-3=128\)

d) \(3^2\cdot5+2^3\cdot10-81:3\)

\(=9\cdot5+8\cdot10-27\)

\(=45+80-27\)

\(=125-27=98\)

e) \(5^{13}:5^{10}-25\cdot2^2\)

\(=5^{13-10}-5^2\cdot2^2\)

\(=5^3-\left(5\cdot2\right)^2\)

\(=125-10^2\)

\(=125-100=25\)

f) \(20:2^2+5^9:5^8\)

\(=20:4+5^{9-8}\)

\(=5+5^1=5+5=10\)

g) \(100:5^2+7\cdot3^2\)

\(=10^2:5^2+7\cdot9\)

\(=\left(10:5\right)^2+63\)

\(=2^2+63=4+63=67\)

h) \(84:4+3^9:3^7+5^0\)

\(=21+3^{9-7}+1\)

\(=21+3^2+1\)

\(=21+9+1=30+1=31\)

i) \(29-\left[16+3\cdot\left(51-49\right)\right]\)

\(=29-\left[16+3\cdot2\right]\)

\(=29-\left[16+6\right]\)

\(=29-22=7\)

j) \(\left(15^{19}:5^{17}+3\right)\cdot0:7\)

\(=\left[\left(3\cdot5\right)^{19}:5^{17}+3\right]\cdot0\)

Vì số nào nhân cho 0 cũng bằng 0 nên giá trị biểu thức trên bằng 0

k) \(7^9:7^7-3^2+2^3\cdot5\)

\(=7^{9-7}-9+8\cdot5\)

\(=7^2-9+40\)

\(=49-9+40=40+40=80\)

l) \(1200:2+6^2\cdot2^1+18\)

\(=600+36\cdot2+18\)

\(=600+72+18\)

\(=600+\left(72+18\right)=600+90=690\)

m) \(5^9:5^7+70:14-20\)

\(=5^{9-7}+5-20\)

\(=5^2+5-20\)

\(25+5-20=30-20=10\)

Những câu sau mình làm sau nhé bạn!!!!!!!

A = 2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 

2A = 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100

2A - A = ( 2^4 + 2^5 + 2^6 + 2^7 + 2^8 + .... + 2^90 + 2^100 ) - (  2^3 + 2^4+ 2^5+ 2^6 + 2^7 + ... + 2^90 ) 

A = 2^100 - 2^3 

B = 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 

5B = 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 

5B - B = ( 5 + 5^2 + 5^3 + 5^4 + 5^5 + .... + 5^50 + 5^51 ) - ( 1 + 5 + 5^2 + 5^3 + 5^4 + .... + 5^50 )

4B = 5^51 - 1 

B = 5^51 - 1 / 4

a) \(12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\)

\(=12\cdot\dfrac{4}{9}+\dfrac{4}{3}\)

\(=\dfrac{12\cdot4}{9}+\dfrac{4}{3}\)

\(=\dfrac{16}{3}+\dfrac{4}{3}\)

\(=\dfrac{16+4}{3}\)

\(=\dfrac{20}{3}\)

b) \(\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}\cdot\left(-\dfrac{1}{2}\right)^2\right]\)

\(=\dfrac{9}{4}-\left(\dfrac{1}{2}:2-9\cdot\dfrac{1}{4}\right)\)

\(=\dfrac{9}{4}-\left(\dfrac{1}{4}-9\cdot\dfrac{1}{4}\right)\)

\(=\dfrac{9}{4}-\dfrac{1}{4}\cdot\left(1-9\right)\)

\(=\dfrac{9}{4}+\dfrac{8}{4}\)

\(=\dfrac{17}{4}\) 

c) \(\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\)

\(=-\dfrac{1}{12}:\dfrac{5}{11}+\dfrac{1}{12}\)

\(=\dfrac{1}{12}\cdot-\dfrac{11}{5}+\dfrac{1}{12}\)

\(=\dfrac{1}{12}\cdot\left(-\dfrac{11}{5}+1\right)\)

\(=\dfrac{1}{12}\cdot-\dfrac{6}{5}\)

\(=-\dfrac{1}{10}\) 

d) \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{4}{9}:\left(2+\dfrac{2}{3}\right)-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}+\dfrac{4}{9}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{9}{10}+\dfrac{1}{6}\)

\(=-\dfrac{11}{15}\) 

e) \(\dfrac{3^7\cdot8^6}{6^6\cdot\left(-2\right)^{12}}\)

\(=\dfrac{3^7\cdot\left(2^3\right)^6}{2^6\cdot3^6\cdot2^{12}}\)

\(=\dfrac{3^7\cdot2^{18}}{2^{6+12}\cdot3^6}\)

\(=\dfrac{2^{18}\cdot3^7}{2^{18}\cdot3^6}\)

\(=3^{7-6}\)

\(=3\)

\(a,12\cdot\left(-\dfrac{2}{3}\right)^2+\dfrac{4}{3}\\ =12\cdot\dfrac{4}{9}+\dfrac{4}{3}\\ =\dfrac{16}{3}+\dfrac{4}{3}\\ =\dfrac{20}{3}\\ b,\left(\dfrac{3}{2}\right)^2-\left[0,5:2-\sqrt{81}.\left(-\dfrac{1}{2}\right)^2\right]\\ =\dfrac{9}{4}-\left(\dfrac{1}{2}\cdot\dfrac{1}{2}-9\cdot\dfrac{1}{4}\right)\\ =\dfrac{9}{4}-\left(\dfrac{1}{4}-\dfrac{9}{4}\right)\\ =\dfrac{9}{4}-\left(-\dfrac{8}{4}\right)\\ =\dfrac{17}{4}\)

\(c,\left(-\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{5}{11}+\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\\ =\left(-\dfrac{9}{12}+\dfrac{8}{12}\right)\cdot\dfrac{11}{5}+\left(-\dfrac{3}{12}+\dfrac{4}{12}\right)\\ =-\dfrac{1}{12}\cdot\dfrac{11}{5}+\dfrac{1}{12}\\ =-\dfrac{11}{60}+\dfrac{1}{12}\\ =-\dfrac{1}{10}\)

\(d,\dfrac{-1^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left(-\dfrac{5}{6}\right)\\ =-\dfrac{1}{15}+\dfrac{4}{9}\cdot\dfrac{3}{8}+\dfrac{5}{6}\\ =-\dfrac{1}{15}+\dfrac{1}{6}+\dfrac{5}{6}\\ =\dfrac{1}{10}+\dfrac{5}{6}\\ =\dfrac{14}{15}\)

`e,` Không hiểu đề á c: )

a) 3A = 3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101
=> 3A - A = (3 + 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^ 101) - (1 + 3 + 3 ^2 + 3 ^ 3 + ... + 3 ^100)
=> 2A = 3^101 - 1 => A = (3^101 - 1)/2
b) 4B = 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101
=> 4B - B = (4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 + 4^ 101) - (1 + 4 + 4 ^ 2 + 4 ^3 + 4 ^ 4 + ... + 4 ^ 100 )
=> 3B = 4^101 - 1 => B = ( 4^101 - 1)/2
c) xem lại đề ý c xem quy luật như thế nào nhé.
d) 3D = 3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151
=> 3D - D = (3^101 + 3^ 102 + 3^ 103 + ... + 36 150 + 3^ 151) - (3 ^100 + 3 ^ 101 + 3 ^ 102 + .... + 3 ^ 150)
=> 2D = 3^ 151 - 3^100 => D = ( 3^ 151 - 3^100)/2

a) Có A=\(1+3+3^2+3^3+....+3^{100}\)

\(\Rightarrow\)3A =\(3\left(1+3+3^2+3^3+...+3^{100}\right)\)=\(3+3^2+3^3+3^4+...+3^{101}\)

\(\Rightarrow2A=3+3^2+3^3+....+3^{101}-1-3-3^2-3^3-....-3^{100}=3^{101}-1\)\(\Rightarrow A=\dfrac{3^{101}-1}{2}\)

Bài b/c/d : bn cứ lm tương tự.

1. 5³ = 5.5.5 = 125

2. 2⁷ = 2.2.2.2.2.2.2 = 128

3. 4⁴ = 4.4.4.4 = 256

4. 7³ = 7.7.7 = 343

6. 3⁵ = 243

7. 2⁶ =  64

8. 3⁴ =  81

9. 8³ =  512

11. 13² = 169

12. 11² = 121

13. 14² = 196

14. 15² = 225

16. 17² = 289

17. 18² = 324

18. 19² = 361

19. 20² = 400

21. 10⁴ = 10000

22. 10⁵ = 100000

23. 10⁶ = 1000000

24. 10⁷ = 10000000

bạn làm như bạn gia hân là đúng nhé

giải giúp mình bài 2 trên với ạ

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\\ A< \frac{9}{10}< 1\Rightarrow A< 1\)

Ta có 1/2²<1/1.2

         1/3²<1/2.3

         1/4²<1/3.4

         ................

        1/8²<1/7.8

=>B<1/1.2+1/2.3+1/3.4+...+1/7.8

=>B<1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8

=>B<1-1/8

Vậy B < 1

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