9⁶⁴ - 1 = (9³² + 1)(9³² - 1) = ... = (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9² + 1)(9 + 1)(9 - 1) > (9³² + 1)(9¹⁶ + 1)(9⁸ + 1)(9⁴ + 1)(9²​ + 1)(9 + 1)

9⁶⁴=(9³²​+1).(9³²-1)

=(9³²+1)(9¹⁶+1)(9¹⁶-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁸-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9⁴-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9²-1)

=(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)(9-1)

Vì (9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)(9-1)>(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)

=>9⁶⁴-1>(9³²+1)(9¹⁶+1)(9⁸+1)(9⁴+1)(9²+1)(9+1)

\(9^{64}-1\)

\(=\left(9^{32}+1\right)\left(9^{32}-1\right)\)

\(=\left(9^{32}+1\right)\left(9^{16}+1\right)\left(9^{16}-1\right)\)

\(...\)

\(=\left(9^{32}+1\right)\left(9^{16}+1\right)\left(9^8+1\right)\left(9^4+1\right)\left(9^2+1\right)\left(9+1\right)\left(9-1\right)\)

\(\Rightarrow9^{64}-1>\left(9+1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)

\(B=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{81}\right)\left(1-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{99}{100}\)

\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{9.11}{10.10}=\left(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{9}{10}\right).\left(\dfrac{3}{2}.\dfrac{4}{3}...\dfrac{11}{10}\right)=\dfrac{1}{10}.\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)

\(B=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)...\left(1-\dfrac{1}{9}\right)\left(1+\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\\ B=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{8}{9}\cdot\dfrac{9}{10}\right)\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{10}{9}\cdot\dfrac{11}{10}\right)\\ B=\dfrac{1}{10}\cdot\dfrac{11}{2}=\dfrac{11}{20}>\dfrac{11}{21}\)

`A = 3/4 xx 8/9 xx ... xx 99/100`

`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`

`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`

`= 1/10 xx 11`

`= 11/10`.

Ta có: `11/10 > 1`

`11/19 < 1`.

`=> A > 11/19`.

Ta có : 

\(\left(\frac{1}{32}\right)^7=\frac{1^7}{32^7}=\frac{1}{\left(2^5\right)^7}=\frac{1}{2^{5.7}}=\frac{1}{2^{35}}\)

\(\left(\frac{1}{16}\right)^9=\frac{1^9}{16^9}=\frac{1}{\left(2^4\right)^9}=\frac{1}{2^{4.9}}=\frac{1}{2^{36}}\)

Vì \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\) ( cùng tử, mẫu nào bé hơn thì phân số đó lớn hơn ) nên \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

Vậy \(\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

Chúc bạn học tốt ~ 

Ta có  :     \(\left(\frac{1}{32}\right)^7=\left(\frac{1}{2^5}\right)^7=\frac{1}{2^{35}}\)

                 \(\left(\frac{1}{16}\right)^9=\left(\frac{1}{2^4}\right)^9=\frac{1}{2^{36}}\)

DO :  \(\frac{1}{2^{35}}>\frac{1}{2^{36}}\)\(\Rightarrow\left(\frac{1}{32}\right)^7>\left(\frac{1}{16}\right)^9\)

Tk mk nha !!! 

> nha bn

Ta có :

\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)

ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)

             = 3/2^2.8/3^2 ... 99/10^2

             = 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2

             = 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10

             = 1/10 . 11/2 = 11/20 < 11/19

              Vậy M < 11/19

\(\text{a) }\left(-\frac{1}{16}\right)^{100}=\frac{\left(-1\right)^{100}}{16^{100}}=\frac{1}{16^{100}}\)

\(\left(-\frac{1}{2}\right)^{500}=\frac{\left(-1\right)^{500}}{2^{500}}=\frac{1}{\left(2^5\right)^{100}}=\frac{1}{32^{100}}\)

Ta co 

\(16^{100}< 32^{100}\)

\(\Rightarrow\frac{1}{16^{100}}>\frac{1}{32^{100}}\)

\(\Rightarrow\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)

a. 

Ta có:

\(\left(-\frac{1}{16}\right)^{100}=\frac{\left(-1\right)^{100}}{16^{100}}=\frac{1}{16^{100}}\)

\(\left(-\frac{1}{2}\right)^{500}=\frac{\left(-1\right)^{500}}{2^{500}}=\frac{1}{\left(2^5\right)^{100}}=\frac{1}{32^{100}}\)

Vì \(\frac{1}{16^{100}}>\frac{1}{32^{100}}\Rightarrow\left(-\frac{1}{16}\right)^{100}>\left(-\frac{1}{2}\right)^{500}\)

b.

Ta có:

\(\left(-32\right)^9=\left[-\left(2^5\right)\right]^9=-\left(2^{45}\right)\)

\(\left(-16\right)^{13}=\left[-\left(2^4\right)\right]^{13}=-\left(2^{52}\right)\)

Vì \(-\left(2^{45}\right)>-\left(2^{52}\right)\Rightarrow\left(-32\right)^9>\left(-16\right)^{13}\)

#Chúc bạn học tốt!#

Baì này mình mới làm lúc sáng bạn vào câu hỏi tương tự có đấy

\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)

\(\Rightarrow2A=3^{32}-1\)

\(\Rightarrow A=\frac{3^{32}-1}{2}< 3^{32}-1=C\)

\(B=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{81}\right).\left(1-\frac{1}{100}\right)\)

\(B=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{80}{81}.\frac{99}{100}\)

\(B=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{8.10}{9.9}.\frac{9.11}{10.10}\)

\(B=\frac{1.2.3...8.9}{2.3.4...9.10}.\frac{3.4.5...10.11}{2.3.4...9.10}\)

\(B=\frac{1}{10}.\frac{11}{2}\)

\(B=\frac{11}{20}>\frac{11}{21}\)