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ND
0
DT
5
29 tháng 6 2018
a) Ta có: \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
Mà \(\frac{1}{32}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\Rightarrow m=5\)
b)Ta có: \(\frac{343}{125}=\left(\frac{7}{5}\right)^3\)
Mà \(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\Rightarrow n=3\)
29 tháng 6 2018
\(a)\) \(\left(\frac{1}{2}\right)^m=\frac{1}{32}\)
\(\Leftrightarrow\)\(\left(\frac{1}{2}\right)^m=\frac{1^5}{2^5}\)
\(\Leftrightarrow\)\(\left(\frac{1}{2}\right)^m=\left(\frac{1}{2}\right)^5\)
\(\Leftrightarrow\)\(m=5\)
Vậy \(m=5\)
\(b)\) \(\frac{343}{125}=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\)\(\frac{7^3}{5^3}=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\)\(\left(\frac{7}{5}\right)^3=\left(\frac{7}{5}\right)^n\)
\(\Leftrightarrow\)\(n=3\)
Vậy \(n=3\)
Chúc bạn học tốt ~
DT
5
S
12 tháng 12 2017
Ta có : \(\frac{2^7.3^4}{3^3.2^5}=\frac{2^2.3}{1.1}=4.3=12\)
k cho mk nha!
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
ND
Nguyễn Đức Trí
VIP
27 tháng 7 2023
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
TN
1
14 tháng 9 2017
\(\left(\frac{1}{2}\right)^{40}=\left(\frac{1}{2}\right)^{10\cdot4}=\left(\frac{1}{16}\right)^{10}\)
Mà ta có
\(\left(\frac{1}{32}\right)^{10}< \left(\frac{1}{16}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{2}\right)^{40}>\left(\frac{1}{32}\right)^{10}\)
BM
2
27 tháng 8 2021
(\(\frac{1}{5}\))2 .n = (\(\frac{1}{125}\))3 - n
<=> \(\frac{1}{25}\)n +n = \(\frac{1}{5^9}\)
<=> \(\frac{26}{25}\)n = \(\frac{1}{5^9}\)
<=> n = \(\frac{1}{5^9}\): \(\frac{26}{25}\)= \(\frac{1}{2031250}\)
BM
0
\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
\(=\left(\frac{6}{14}+\frac{7}{14}\right)^2\)
\(=\left(\frac{13}{14}\right)^2\)
\(=\frac{13^2}{14^2}\)
\(=\frac{169}{196}\)