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a: \(\left(\dfrac{1}{5}\right)^{-2}=25\)
b: \(4^{\dfrac{3}{2}}=8\)
c: \(\left(\dfrac{1}{8}\right)^{-\dfrac{2}{3}}=\left(\dfrac{1}{2}\right)^{3\cdot\dfrac{-2}{3}}=\left(\dfrac{1}{2}\right)^{-2}=4\)
d: \(\left(\dfrac{1}{16}\right)^{-0.75}=\left(\dfrac{1}{2}\right)^{4\cdot\left(-0.75\right)}=\left(\dfrac{1}{2}\right)^{-3}=8\)
\(a,1^{1,5}=1;3^{-1}=\dfrac{1}{3};\left(\dfrac{1}{2}\right)^{-2}=4\\ \Rightarrow\dfrac{1}{3}< 1< 4\\ b,2022^0=1;\left(\dfrac{4}{5}\right)^{-1}=\dfrac{5}{4};5^{\dfrac{1}{2}}=\sqrt{5}\simeq2,24\\ \Rightarrow1< \dfrac{5}{4}< \sqrt{5}\)
\(M=\left(\dfrac{1}{3}\right)^{12}\cdot\left(\dfrac{1}{3}\right)^{-15}+\left(\dfrac{2}{5}\right)^{-4}\cdot5^{-4}\cdot32\)
\(=\left(\dfrac{1}{3}\right)^{-3}+2^{-4}\cdot32\)
\(=27+\dfrac{32}{16}=27+2=29\)
Ta có:
\(u_1=\dfrac{1}{3^1-1}=\dfrac{1}{2}\\ u_2=\dfrac{2}{3^2-1}=\dfrac{1}{4}\\ u_3=\dfrac{3}{3^3-1}=\dfrac{3}{26}\)
\(\Rightarrow B\)
a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)
b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)
c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)
c/
ĐKXĐ: ...
Đặt \(cosx+\frac{2}{cosx}=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2-4\)
Pt trở thành:
\(9a+2\left(a^2-4\right)=1\)
\(\Leftrightarrow2a^2+9a-9=0\)
Pt này nghiệm xấu quá bạn :(
d/ĐKXĐ: ...
Đặt \(\frac{2}{cosx}-cosx=a\Rightarrow cos^2x+\frac{4}{cos^2x}=a^2+4\)
Pt trở thành:
\(2\left(a^2+4\right)+9a-1=0\)
\(\Leftrightarrow2a^2+9a+7=0\Rightarrow\left[{}\begin{matrix}a=-1\\a=-\frac{7}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2}{cosx}-cosx=-1\\\frac{2}{cosx}-cosx=-\frac{7}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-cos^2x+cosx+2=0\\-cos^2x+\frac{7}{2}cosx+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}cosx=-1\\cosx=2\left(l\right)\\cosx=4\left(l\right)\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pi+k2\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
b/
ĐKXĐ: ...
Đặt \(sinx+\frac{1}{sinx}=a\Rightarrow sin^2x+\frac{1}{sin^2x}=a^2-2\)
Pt trở thành:
\(4\left(a^2-2\right)+4a=7\)
\(\Leftrightarrow4a^2+4a-15=0\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-\frac{5}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx+\frac{1}{sinx}=\frac{3}{2}\\sinx+\frac{1}{sinx}=-\frac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin^2x-\frac{3}{2}sinx+1=0\left(vn\right)\\sin^2x+\frac{5}{2}sinx+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}sinx=-\frac{1}{2}\\sinx=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{\pi}{6}+k2\pi\\x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
Theo công thức tổng CSN:
\(1+\frac{2}{3}+...+\left(\frac{2}{3}\right)^n=\frac{1-\left(\frac{2}{3}\right)^{n+1}}{1-\frac{2}{3}}=3-3.\left(\frac{2}{3}\right)^{n+1}\)
\(1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^n=\frac{1-\left(\frac{1}{5}\right)^{n+1}}{1-\frac{1}{5}}=\frac{5}{4}-\frac{5}{4}\left(\frac{1}{5}\right)^{n+1}\)
\(\Rightarrow lim\frac{3-3\left(\frac{2}{3}\right)^{n+1}}{\frac{5}{4}-\frac{5}{4}\left(\frac{1}{5}\right)^{n+1}}=\frac{3}{\frac{5}{4}}=\frac{12}{5}\)
\(a,\left(\dfrac{1}{256}\right)^{-0,75}+\left(\dfrac{1}{27}\right)^{-\dfrac{4}{3}}\\ =256^{\dfrac{3}{4}}+27^{\dfrac{4}{3}}\\ =\sqrt[4]{256^3}+\sqrt[3]{27^4}\\ =145\\ b,\left(\dfrac{1}{49}\right)^{-1,5}-\left(\dfrac{1}{256}\right)^{-\dfrac{2}{3}}\\ =49^{\dfrac{3}{2}}-256^{\dfrac{2}{3}}\\ \simeq343-40,3\\ \simeq302,7\)