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Ta có:
\(\left(\frac{1}{2}\right)^{225}=\left[\left(\frac{1}{2}\right)^9\right]^{25}=\left(\frac{1}{516}\right)^{25}\)
\(\left(\frac{1}{3}\right)^{100}=\left[\left(\frac{1}{3}\right)^4\right]^{25}=\left(\frac{1}{81}\right)^{25}\)
\(\frac{1}{516}< \frac{1}{81}\Rightarrow\left(\frac{1}{516}\right)^{25}< \left(\frac{1}{81}\right)^{25}\Rightarrow\left(\frac{1}{2}\right)^{225}< \left(\frac{1}{3}\right)^{100}\)
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)
\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)
\(\Leftrightarrow\frac{38}{x+3}=16\)
\(\Leftrightarrow x+3=2,375\)
\(\Leftrightarrow x=-0,625\)
\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)
\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)
\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)
\(\Leftrightarrow\frac{38}{x+3}=14\)
\(\Leftrightarrow\left(x+3\right)14=38\)
\(\Leftrightarrow14x+42=38\)
\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)
Vậy \(x=-\frac{2}{7}\)
\(\frac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(=>\left(\frac{1}{2}+4\right)\cdot2^n=\frac{9}{2}\cdot2^6\)
\(=>\frac{9}{2}\cdot2^n=\frac{9}{2}\cdot2^6\)
\(=>2^n=2^6\)
\(=>n=6\)
=> 6x - 3 - 5 - 15x = 44
=> -9x - 8 = 44
=> -9x = 52
=> x = \(\frac{-52}{9}\)
nhớ
3(2x-1)-5(1+3x)=44
\(\Leftrightarrow\)6x-3-5-15x=44
\(\Leftrightarrow\)-11x=52
\(\Leftrightarrow\)x=-52/11
a)\(\left(\frac{1}{5}\right)^5\).\(5^5\)=\(\frac{1}{3125}\).3125=1
(1/5.5)5
=5/55
=15
=1