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\(\left(\dfrac{1}{4}\right)^{2n}=\left(\dfrac{1}{8}\right)^2\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2.2n}=\left(\dfrac{1}{2}\right)^{3.2}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{4n}=\left(\dfrac{1}{2}\right)^6\)
\(\Rightarrow4n=6\)
\(\Rightarrow n=\dfrac{6}{4}=\dfrac{3}{2}\)
Quãng đường đó thực tế dài là :
\(30.120000=3600000=36.10^5=3,6.10^6\left(cm\right)\)
b, \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{35}\)
a) Vì \(-45< -16\) nên \(\left(-\dfrac{45}{17}\right)^{15}< \left(\dfrac{-16}{17}\right)^{15}\)
b) Vì \(21< 23\) nên \(\left(-\dfrac{8}{9}\right)^{21}< \left(-\dfrac{8}{9}\right)^{23}\)
c) \(27^{40}=3^{3^{40}}=3^{120}\)
\(64^{60}=8^{2^{60}}=8^{120}\)
Vì \(3< 8\) nên \(3^{120}< 8^{120}\) hay \(27^{40}< 64^{60}\)
A=\(\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...\)
A-\(\frac{1}{5}=\frac{1}{5^2}+\frac{1}{5^3}+...\)
5 -\(\left(A-\frac{1}{5}\right)=5.\left(\frac{1}{5^2}+\frac{1}{5^3}+...1\right)\)
10.\(\left(A-\frac{1}{5}\right)=A\)
\(10A-5=A\)
\(10.A-A=5\)
\(9A=5\)
\(A=\frac{5}{9}\)
T hk pk dung hay sai nha, ma m cu lam ik, co j len t sua
a, \(\frac{8}{2^n}=2\Rightarrow2.2^n=8\)
\(\Rightarrow2^{n+1}=2^3\)
\(\Rightarrow n+1=3\)
\(\Rightarrow n=2\)
d,\(\left(2n-3\right)^2=9\)
\(\left(2n-3\right)^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}2n-3=-3\\2n-3=3\end{cases}\Rightarrow\orbr{\begin{cases}2n=-3+3\\2n=3+3\end{cases}\Rightarrow}\orbr{\begin{cases}2n=0\\2n=6\end{cases}\Rightarrow}\orbr{\begin{cases}n=0\\n=3\end{cases}}}\)
Vậy n=0; n= 3
`(2x - 1)^2 = 36`
`(2x - 1)^2 = 6^2` hoặc `(2x - 1)^2 = (-6)^2`
`2x - 1 = 6` hoặc `2x - 1 = -6`
`2x = 6 + 1` hoặc `2x = -6 + 1`
`2x = 7` hoặc `2x = -5`
`x = 7 : 2` hoặc `x = -5 : 2`
`x = 3,5` hoặc `x = -2,5`
\(\left(2x-1\right)^2=36\)
\(\Rightarrow\left(2x-1\right)^2=6^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(A=\frac{x^2-2x+1}{x+1}=\frac{x^2-2x-3+4}{x+1}=\frac{\left(x+1\right)\left(x-3\right)+4}{x+1}=x-3+\frac{4}{x+1}\inℤ\)
mà \(x\inℤ\)nên \(\frac{4}{x+1}\inℤ\)do đó \(x+1\inƯ\left(4\right)=\left\{-4,-2,-1,1,2,4\right\}\)
\(\Leftrightarrow x\in\left\{-5,-3,-2,0,1,3\right\}\).
1/2 mũ n = 1/2 mũ 6
Vậy x = 6
\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{8}\right)^2\)
\(=>\left(\dfrac{1}{2}\right)^n=\left[\left(\dfrac{1}{2}\right)^3\right]^2\)
\(=>\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^6\)
\(\Rightarrow n=6\)