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T
5 tháng 8 2018
\(2^{50}=\left(2^5\right)^{10}=32^{10}\)
\(5^{20}=\left(5^2\right)^{10}=25^{10}\)
Suy ra: 250 > 520
b)
\(9^{200}=\left(9^2\right)^{100}=81^{100}\)
Suy ra: 99100 > 81100
21 tháng 10 2021
\(\left(2x-1\right)^2+\left(y-3\right)^8+\left(z-5\right)^{20}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\y-3=0\\z-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=3\\z=5\end{matrix}\right.\)
29 tháng 6 2017
a,\(\left(\dfrac{3}{7}+\dfrac{1}{2}\right)^2\)
\(=\left(\dfrac{13}{14}\right)^2\)
\(=\dfrac{169}{196}\)
b,\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=\left(\dfrac{-1}{12}\right)^2\)
\(=\dfrac{1}{144}\)
c,\(\dfrac{5^4.20^4}{25^5.4^5}\)
\(=\dfrac{100^4}{100^5}\)
\(=\dfrac{1}{100}\)
d,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4\)
\(=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=\left(\dfrac{\left(-10\right)}{3}.\dfrac{\left(-6\right)}{5}\right)^4.\left(\dfrac{-10}{3}\right)\)
\(=4^4.\left(\dfrac{-10}{3}\right)\)
\(=256.\left(\dfrac{-10}{3}\right)\)
\(=\dfrac{-2560}{3}\)
28 tháng 5 2022
Bài 8:
a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)
b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)
KD
23 tháng 9 2016
a.\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
=\(\left(\frac{6}{14}+\frac{7}{14}\right)^2\)
=\(\left(\frac{13}{14}\right)^2\)
=\(\frac{13^2}{14^2}\)
=\(\frac{169}{196}\)
b.\(\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
=\(\left(\frac{9}{12}-\frac{10}{12}\right)^2\)
=\(\left(\frac{-1}{12}\right)^2\)
=\(\frac{-1^2}{12^2}\)
=\(\frac{1}{144}\).
c.Phần C bn viết lại đề bài đi,mk ko hiểu
d.\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)
=\(\frac{-10^5}{3^5}.\left(\frac{-6^4}{5^4}\right)\)
=\(\frac{-100000}{243}.\frac{1296}{625}\)
=\(\frac{-2560}{3}\)
Không biết đúng ko nữa
VT
1
15 tháng 10 2023
\(25A=\dfrac{5^{502}+25}{5^{502}+1}=1+\dfrac{24}{5^{502}+1}\)
\(25B=\dfrac{5^{602}+25}{5^{602}+1}=1+\dfrac{24}{5^{602}+1}\)
\(5^{502}+1< 5^{602}+1\)
=>\(\dfrac{24}{5^{502}+1}>\dfrac{24}{5^{602}+1}\)
=>25A>25B
=>A>B
12 tháng 2 2022
Bài 1:
a: \(=5^2\left(5^3-5^2+1\right)=5^2\cdot101⋮101\)
b: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮11\)
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) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)
\(5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)
\(4B=5^{2010}-1\)
\(B=\frac{5^{2010}-1}{4}\)
\(B=1+5+5^2+...+\)\(5^{2009}\)
\(\Rightarrow5B=5+5^2+5^3+...+5^{2010}\)
\(\Rightarrow5B-B=4B=5^{2010}-1\)
\(\Rightarrow B=\frac{5^{2010}-1}{4}\)