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chép nhầm đề. Làm lại:
\(C=\frac{\left(3.4.2^4\right)^2}{5.2^5.4^2-16^2}=\frac{3^2.2^4.2^8}{5.2^5.2^4-2^8}=\frac{9.2^{12}}{5.2^9-2^8}=\frac{9.2^{12}}{2^8.\left(5.2-1\right)}=\frac{9.2^4}{9}=\frac{2^4}{1}=2^4=16\)
Gấp lắm thì làm vậy
\(C=\frac{\left(3.4.2^2\right)^2}{5.2^5.4^2-16^2}=\frac{3^2.2^4.2^4}{5.2^5.2^4-2^8}=\frac{3^2.2^8}{5.2^9-2^8}=\frac{3^2.2^8}{2^8.\left(5.2-1\right)}=\frac{9}{9}=1\)
\(\dfrac{1}{2}A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{2023}\)
\(A-\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(\dfrac{1}{2}A=\left(\dfrac{1}{2}\right)^{2023}-1\)
\(A=\dfrac{1}{2^{2022}}-2\)
12A=12+(12)2+(12)3+(12)4+...+(12)202312A=12+(12)2+(12)3+(12)4+...+(12)2023
A−12A=(12)2023−1A−12A=(12)2023−1
12A=(12)2023−112A=(12)2023−1
A=122022−2
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}=\frac{3^2.2}{10}=\frac{18}{10}\)
\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{2^{35}\left(11-2\right)}\)
\(=\frac{3^2.2^{36}}{2^{35}.9}\)
\(=\frac{3^2.2^{36}}{2^{35}.3^2}\)
\(=2\)
~~~Hok tốt~~~
\(\frac{\left(3.4.2^4\right)^2}{5.2^5.4^2-16^2}\)
\(=\frac{\left(3.2^2.2^4\right)^2}{5.2^5\left(2^2\right)^2-\left(2^4\right)^2}\)
\(=\frac{\left(3.2^6\right)^2}{5.2^9-2^8}\)
\(=\frac{3^2.2^{12}}{2^8\left(5.2-1\right)}\)
\(=\frac{9.2^4}{9}\)
\(=16\)
\(=\frac{3^2.2^4.2^8}{5.2^5.2^4-2^8}\)
\(=\frac{3^22^{12}}{2^8\left(5.2-1\right)}\)\(=2^4\)