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14 tháng 9 2021

c)\(7^{2n}+7^{2n+2}=2450\)

\(7^{2n}+7^{2n}.7^2=2450\)

\(7^{2n}.50=2450\)

\(7^{2n}=49\)\(=7^2\)

⇒2n=2

⇒n=1

14 tháng 9 2021

a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\)                   b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)

\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\)                    \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)

⇒n=3                                          ⇒m=2

18 tháng 4 2017

a) =>

b) =>

27 tháng 6 2017

a) (12)m=132

\(\Rightarrow\left(\dfrac{1}{2}\right)^m=\left(\dfrac{1}{2}\right)^5\Rightarrow m=5\)

b)

\(\Rightarrow\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\Rightarrow n=3\)

22 tháng 10 2017

a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)

\(\Rightarrow2n-1=3\)

\(\Rightarrow2n=4\)

\(\Rightarrow n=2\)

22 tháng 10 2017

a) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow2^{-\left(2n-1\right)}=2^{-3}\)

\(\Rightarrow2^{-2n+1}=2^{-3}\)

\(\Rightarrow-2n+1=-3\)

\(\Rightarrow-2n=-4\)

\(\Rightarrow n=-2\)

Vậy ...

b) \(\left(\dfrac{7}{5}\right)^n=\dfrac{343}{125}\)

\(\Rightarrow\left(\dfrac{7}{5}\right)^n=\left(\dfrac{7}{5}\right)^3\)

\(\Rightarrow n=3\)

Vậy ....

25 tháng 7 2023

Bài 3 :

\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}\)

\(\dfrac{1}{2!}=\dfrac{1}{2.1}=1-\dfrac{1}{2}< 1\)

\(\dfrac{1}{3!}=\dfrac{1}{3.2.1}=1-\dfrac{1}{2}-\dfrac{1}{3}< 1\)

\(\dfrac{1}{4!}=\dfrac{1}{4.3.2.1}< \dfrac{1}{3!}< \dfrac{1}{2!}< 1\)

.....

\(\)\(\dfrac{1}{2023!}=\dfrac{1}{2023.2022....2.1}< \dfrac{1}{2022!}< ...< \dfrac{1}{2!}< 1\)

\(\Rightarrow\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)

25 tháng 7 2023

Bạn xem lại đề 2, phần mẫu của N

a: =>x(1/2+1/4+1/2017)=x(1/3+1/5+1/2017)

=>x=0

b: =>1/-3=-7/21

e: a/b=2/7

nên a=2/7b

=>b=7/2a

b/c=14/15

=>b=14/15c

\(\Leftrightarrow\)7/2a=14/15c

=>a/c=4/15

13 tháng 7 2017

a)(1/2)m=1/32

(1/2)m=(1/2)5

=>m=5

24 tháng 12 2023

a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)

b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)

\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)

\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)

\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)

\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)

\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)

15 tháng 9 2017

1/2m = 1/32

1/2m = 1/25

=> m =5