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a) \(VT=12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}\)
\(VP=18^{16}=2^{16}\cdot3^{32}\)
=> VT=VP
b) \(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^5}\)
(đề sai)
c) \(\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{1}{4}\)
\(VT=\frac{9^3}{\left(3^4-3^3\right)^2}=\frac{3^6}{\left[3^3\left(3-1\right)\right]^2}=\frac{1}{2^2}=\frac{1}{4}=VP\)
a) \(\left(-\frac{3}{4}\right)^{3x-1}=\frac{-27}{64}\)
\(\Leftrightarrow\left(-\frac{3}{4}\right)^{3x-1}=\left(-\frac{3}{4}\right)^3\)
\(\Leftrightarrow3x-1=3\)
\(\Leftrightarrow3x=4\)
\(\Leftrightarrow x=\frac{4}{3}\)
b) Đề sai ! Sửa :
\(\left(\frac{4}{5}\right)^{2x+5}=\frac{256}{625}\)
\(\Leftrightarrow\left(\frac{4}{5}\right)^{2x+5}=\left(\frac{4}{5}\right)^4\)
\(\Leftrightarrow2x+5=4\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
c) \(\frac{\left(x+3\right)^5}{\left(x+5\right)^2}=\frac{64}{27}\)
\(\Leftrightarrow\left(x+3\right)^3=\left(\frac{4}{3}\right)^3\)
\(\Leftrightarrow x+3=\frac{4}{3}\)
\(\Leftrightarrow x=-\frac{5}{3}\)
d) \(\left(x-\frac{2}{15}\right)^3=\frac{8}{125}\)
\(\Leftrightarrow\left(x-\frac{2}{15}\right)^3=\left(\frac{2}{15}\right)^3\)
\(\Leftrightarrow x-\frac{2}{15}=\frac{2}{15}\)
\(\Leftrightarrow x=\frac{4}{15}\)
Ta có : 3x + 3x + 2 = 810
=> 3x(1 + 32) = 810
=> 3x.10 = 810
=> 3x = 81
=> 3x = 34
=> x = 4
ta có \(3^3+3^x+2=810\)
=>\(3^x\left(1+3^2\right)=810\)
=>\(3^x.10=810\)
=>\(3^x=81\)
=>\(3^x=3^4\)
=>x=4
Vậy x=4
3. \(\left(\frac{1}{2^5}\right)^{25}=\left(\frac{1^5}{2^5}\right)^{25}=\left[\left(\frac{1}{2}\right)^5\right]^{25}=\left(\frac{1}{2}\right)^{125}\)
\(\left(\frac{1}{3^{25}}\right)^5=\left(\frac{1^{25}}{3^{25}}\right)^5=\left[\left(\frac{1}{3}\right)^{25}\right]^5=\left(\frac{1}{3}\right)^{125}\)
Vì \(\frac{1}{2}>\frac{1}{3}\Rightarrow\left(\frac{1}{2^5}\right)^{25}>\left(\frac{1}{3^{25}}\right)^5\)
1. \(3^{800}=\left(3^8\right)^{100}=6561^{100}\)
\(5^{500}=\left(5^5\right)^{100}=3125^{100}\)
Vì \(6561>3125\Rightarrow3^{800}>5^{500}\)
2. \(\left(-2\right)^{3000}=\left[\left(-2\right)^3\right]^{1000}=\left(-8\right)^{1000}\)
\(\left(-3\right)^{2000}=\left[\left(-3\right)^2\right]^{1000}=9^{1000}\)
Vì \(-8< 9\Rightarrow\left(-2\right)^{3000}< \left(-3\right)^{2000}\)
\(\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3\right]^2.\left(\frac{25}{4}\right)^2\)
\(=\left[\left(\frac{2}{5}\right)^3.\frac{25}{4}\right]^2\)
\(=\left[\frac{8}{125}.\frac{25}{4}\right]^2\)
\(=\left(\frac{2}{5}\right)^2\)
\(=\frac{4}{25}\)
\(15\frac{1}{5}:\left(\frac{-5}{7}\right)-25\frac{1}{5}.\left(\frac{-7}{5}\right)\)
\(=15\frac{1}{5}.\frac{-7}{5}-25\frac{1}{5}.\frac{-7}{5}\)
\(=\frac{-7}{5}\left(15\frac{1}{5}-25\frac{1}{5}\right)\)
\(=\frac{-7}{5}.\left(-10\right)\)
\(=14\)
\(\frac{\left(5^4-5^3\right)}{125^4}-\frac{64}{125}\)
\(=\frac{\left(625-125\right)}{500}-\frac{64}{125}\)
\(=\frac{500}{500}-\frac{64}{125}\)
\(=0-0,51\)
\(=-0,51\)
Ta có:
\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{\left(5^3\right)^3.\left(5-1\right)^3}{\left(5^3\right)^5}=\frac{5^9.4^3}{5^{15}}=\frac{4^3}{5^6}=\frac{64}{5^6}\) (1)
\(\frac{64}{25^3}=\frac{64}{\left(5^2\right)^3}=\frac{64}{5^6}\) (2)
Từ (1) và (2) =>\(\frac{\left(5^4-5^3\right)^3}{125^5}=\frac{64}{25^3}\)
Bằng nhau