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\(8A=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\)

\(8B=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)

\(\text{Vì }\frac{7}{8^{19}+1}>\frac{7}{8^{24}+1}\)

\(\Rightarrow8A>8B\)

\(\Rightarrow A>B\)

\(\text{Câu B làm tương tự nhé}\)

27 tháng 5 2018

a) \(A=2^{24}=\left(2^3\right)^8=8^8.\)(1)

\(B=3^{16}=\left(3^2\right)^8=9^8\)(2)

Từ (1) và (2) \(\Rightarrow A< B\)

Vậy \(A< B.\)

b) \(B=\left(0,3\right)^{30}=\left(0,3^2\right)^{15}=0,09^{15}\)(1)

\(A=\left(0,1\right)^{15}\)(2)

Từ (1) và (2) \(\Rightarrow A>B\)

Vậy \(A>B.\)

c) \(A=\left(\frac{-1}{4}\right)^8=\left(\frac{1}{4}\right)^8=\left[\left(\frac{1}{2}\right)^2\right]^8=\left(\frac{1}{2}\right)^{16}\)(1)

\(B=\left(\frac{1}{8}\right)^5=\left[\left(\frac{1}{2}\right)^3\right]^5=\left(\frac{1}{2}\right)^{15}\)(2)

Từ (1) và (2) \(\Rightarrow A>B\)

Vậy \(A>B.\)

d) \(A=102^7=102^6.102\)(1)

\(B=9^{13}=9^{12}.9=\left(9^2\right)^6.9=81^6.9\)(2)'

Từ (1) và (2) \(\Rightarrow A>B\)

Vậy \(A>B.\)

e) \(8A=8\frac{8^{18}+1}{8^{19}+1}=\frac{8^{19}+8}{8^{19}+1}=1+\frac{7}{8^{19}+1}\)(1)

\(8B=8\frac{8^{23}+1}{8^{24+1}}=\frac{8^{24}+8}{8^{24}+1}=1+\frac{7}{8^{24}+1}\)(2)

Từ (1) và (2) \(\Rightarrow8A>8B\Rightarrow A>B\)

Vậy \(A>B.\)

f) \(A=\frac{5^5}{5+5^2+5^3+5^4}=\frac{5^4}{1+5+5^2+5^3}=\frac{625}{156}>\frac{468}{156}=3.\)(1)

\(B=\frac{3^5}{3+3^2+3^3+3^4}=\frac{3^4}{1+3+3^2+3^3}=\frac{81}{40}< \frac{120}{40}=3.\)(2)

Từ (1) và (2) \(\Rightarrow A>B\)

Vậy \(A>B.\)

27 tháng 5 2018

a, ta có A=2^24=64^4

             B=3^16=81^4

Vì 64^4<81^4

Vậy 2^24<3^36

b, ta có A=0,1^15

             B=0,3^30=0,09^15

Vì 0,1^15< 0,09^15

Vậy 0,1^15<0,3^30

20 tháng 9 2019

\(a,\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[2^4-4^2\right]\)

\(=\left[2^{17}+16^2\right]\cdot\left[9^{15}-3^{15}\right]\cdot\left[16-16\right]\)

\(=\left[2^{17}+16^2\right]\left[9^{15}-3^{15}\right]\cdot0=0\)

\(b,\left[8^{2017}-8^{2015}\right]\cdot\left[8^{2014}\cdot8\right]\)

\(=8^{2015}\left[8^2-1\right]\cdot8^{2015}\)

\(=8^{2015}\cdot63\cdot8^{2015}=8^{4030}\cdot63\)sửa lại câu b , có vấn đề rồi

\(c,\frac{2^8+8^3}{2^5\cdot2^3}=\frac{2^8+\left[2^3\right]^3}{2^5\cdot2^3}=\frac{2^8+2^9}{2^8}=\frac{2^8\left[1+2\right]}{2^8}=3\)

2.a, \(2^6=\left[2^3\right]^2=8^2\)

Mà 8 = 8 nên 82 = 82 hay 26 = 82

b, \(5^3=5\cdot5\cdot5=125\)

\(3^5=3\cdot3\cdot3\cdot3\cdot3=243\)

Mà 125 < 243 nên 53 < 35

c, 26 = [23 ]2 = 82

Mà 8 > 6 nên 82 > 62 hay 26 > 62

d, 7200 = [72 ]100 = 49100

6300 = \(\left[6^3\right]^{100}\)= 216100

Mà 49 < 216 nên 49100 < 216100 hay 7200 < 6300

19 tháng 7 2019

bài 2

làm câu B;C nha

B)

\(27^3=\left(3^3\right)^3=3^9\)

\(9^5=\left(3^2\right)^5=3^{10}\)

vì \(10>9\)

\(=>9^5>27^3\)

C)

\(\left(\frac{1}{8}\right)^6=\left(\frac{1}{2^3}\right)^6=\frac{1^6}{2^{18}}=\frac{1}{2^{18}}\)

\(\left(\frac{1}{32}\right)^4=\left(\frac{1}{2^5}\right)^4=\frac{1^4}{2^{20}}=\frac{1}{2^{20}}\)

vì \(2^{18}< 2^{20}\)

\(=>\frac{1}{2^{18}}>\frac{1}{2^{20}}\)

\(=>\left(\frac{1}{8}\right)^6>\left(\frac{1}{32}\right)^4\)

19 tháng 7 2019

\(\text{A.}\frac{32^3.9^5}{8^3.6^6}=\frac{\left(2^5\right)^3.\left(3^2\right)^5}{\left(2^3\right)^3.\left(2.3\right)^6}=\frac{2^{15}.3^{10}}{2^9.2^6.3^6}=\frac{3^{10}}{3^6}=3^4=81\)

\(\text{B.}\frac{\left(5^5-5^4\right)^3}{50^6}=\frac{2500^3}{50^6}=\frac{\left(50^2\right)^3}{50^6}=\frac{50^6}{50^6}=1\)

Bài 2:

\(\text{A.Ta có:}\)

\(5^6=\left(5^3\right)^2=125^2\)

\(\left(-2\right)^{14}=2^{14}=\left(2^7\right)^2=128^2\)

Vì \(125< 128\)

\(\Rightarrow125^2< 128^2\)

\(\Rightarrow5^6< \left(-2\right)^{14}\)

\(\text{B.Ta có:}\)

\(9^5=\left(3^2\right)^5=3^{10}\)

\(27^3=\left(3^3\right)^3=3^9\)

Vì \(9< 10\)

\(\Rightarrow3^9< 3^{10}\)

\(\Rightarrow27^3< 9^5\)

\(\text{C.Ta có:}\)

\(\left(\frac{1}{8}\right)^6=\left[\left(\frac{1}{2}\right)^3\right]^6=\left(\frac{1}{2}\right)^{18}\)

\(\left(\frac{1}{32}\right)^4=\left[\left(\frac{1}{2}\right)^5\right]^4=\left(\frac{1}{2}\right)^{20}\)

Vì \(18< 20\)

\(\Rightarrow\left(\frac{1}{2}\right)^{18}< \left(\frac{1}{2}\right)^{20}\)

\(\Rightarrow\left(\frac{1}{8}\right)^6< \left(\frac{1}{32}\right)^4\)

16 tháng 10 2022

a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)

=>4^x=4^8

=>x=8

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)

=>2^x=2^11

=>x=11

c: =>1/6*6^x+6^x*36=6^15(1+6^3)

=>6^x=6*6^15

=>x=16

d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)

=>x=9

3 tháng 9 2016

Câu 1:

a) 2225 và 3150

         Ta có:2225=(29)25=51225

                  3150=(36)25=72925

       Vì 51225<72925

                 Suy ra: 2225<3150

3 tháng 9 2016

Câu 2:

a)\(25^3:5^2=\left(5^2\right)^3:5^2=5^6:5^2=5^4\)

b)\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)

c)\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3+\frac{1}{4}:2=3+\frac{1}{8}=\frac{25}{8}\)

Câu 3:

a)\(9.3^3.\frac{1}{81}.3^2=3^2.3^3.3^2.\left(\frac{1}{3^4}\right)=3^7:3^4=3^3\)

b)\(4.2^5:\left(2^3.\frac{1}{16}\right)=2^2.2^5:\left(2^3.\frac{1}{2^4}\right)=2^7:\frac{1}{2}=2^8\)

c)\(3^2.2^5.\left(\frac{2}{3}\right)^2=288.\frac{4}{9}=2^7\)

d)\(\left(\frac{1}{3}\right)^3.\frac{1}{3}.9^2=\left(\frac{1}{3}\right)^4.\left(3^2\right)^2=3^4.\left(\frac{1}{3}\right)^4=3^4:3^4=1\)

 

15 tháng 6 2018
https://i.imgur.com/7Dfkl7S.jpg