Tìm số tự nhiên n, biết :
a) \(\dfrac{16}{2^n}\)=2
b) \(\dfrac{\left(-3\right)^n}{81}\)=-27
c) 8n:2n=4
Tìm số tự nhiên n, biết :
a) \(\dfrac{16}{2^n}=2\)
b) \(\dfrac{\left(-3\right)^n}{81}=-27\)
c) \(8^n:2^n=4\)
a)
b,
\(\dfrac{\left(-3\right)^n}{81}=-27\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^4}=-27\Rightarrow\left(-3\right)^{n-4}=\left(-3\right)^3\Rightarrow n-4=3\Rightarrow n=7\)
c,\(8^n:2^n=4\Rightarrow4^n=4\Rightarrow n=1\)
=> (-3)n-4 = (-3)3
=> n - 4 = 3 => n = 7
c) 8n : 2n = 4
4n = 4.
Tìm các số tự nhiên m,n biết :
a) \(\left(-\dfrac{1}{5^{ }}\right)^n\) =\(-\dfrac{1}{125}\)
b)\(\left(-\dfrac{2}{11^{ }}\right)^m=\dfrac{4}{121}\)
c)\(7^{2n}+7^{2n+2}=2450\)
c)\(7^{2n}+7^{2n+2}=2450\)
⇒\(7^{2n}+7^{2n}.7^2=2450\)
⇒\(7^{2n}.50=2450\)
⇒\(7^{2n}=49\)\(=7^2\)
⇒2n=2
⇒n=1
a)\(\left(-\dfrac{1}{5}\right)^n=-\dfrac{1}{125}\) b)\(\left(-\dfrac{2}{11}\right)^m=\dfrac{4}{121}\)
\(\left(-\dfrac{1}{5}\right)^n=\left(-\dfrac{1}{5}\right)^3\) \(=\left(-\dfrac{2}{11}\right)^m=\left(-\dfrac{2}{11}\right)^2\)
⇒n=3 ⇒m=2
1.
a, \(^{^2}\left(x-2\right)=9\) b,\(^{^3}\left(3x-1\right)=-8\) c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\) d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\) e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\) f,\(\left(\dfrac{1}{2}\right)^{2x-1}=8\)
2.tìm số tự nhiên n biết
a, \(3^{n-1}=27\) b, \(3^{n-1}=\dfrac{1}{243}\) c, \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\) d, \(\left(-\dfrac{1}{3}\right)^{n-5}=\dfrac{1}{81}\) e,\(2^{-1}.2^n+4.2^n=9.2^5\)
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!Biết biểu thức \(A=\dfrac{2}{3}+\dfrac{4}{3^2}+\dfrac{6}{3^3}+...+\dfrac{2n}{3^n}\) được tính theo công thức \(A=\dfrac{a.\left(3^n-b\right)-cn}{c.3^n}\) với a,b,c là các số tự nhiên và nguyên tố cùng nhau. Tính abc
Lời giải:
$A=\frac{2}{3}+\frac{4}{3^2}+\frac{6}{3^3}+...+\frac{2n}{3^n}$
$3A=2+\frac{4}{3}+\frac{6}{3^2}+....+\frac{2n}{3^{n-1}}$
$3A-A=2+\frac{2}{3}+\frac{2}{3^2}+....+\frac{2}{3^{n-1}}-\frac{2n}{3^n}$
$2A=2+\frac{2}{3}+\frac{2}{3^2}+....+\frac{2}{3^{n-1}}-\frac{2n}{3^n}$
$A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}-\frac{n}{3^n}$
$3A=3+1+\frac{1}{3}+....+\frac{1}{3^{n-2}}-\frac{n}{3^{n-1}}$
$3A-A=3-\frac{1}{3^{n-1}}-\frac{n}{3^{n-1}}+\frac{n}{3^n}$
$2A=3-\frac{n+1}{3^{n-1}}+\frac{n}{3^n}$
$2A=\frac{3^{n+1}-2n-3}{3^n}$
$A=\frac{3.3^n-2n-3}{2.3^n}$
$\Rightarrow a=3; b=1; c=2\Rightarrow abc=6$
Tìm x ∈ N biết :
a) \(8< 2^x\le2^9.2^{-5}\)
b)\(27< 81^3:3^x< 243\)
c)\(\left(\dfrac{2}{5}\right)^x>\left(\dfrac{5}{2}\right)^{-3}.\left(\dfrac{-3}{5}\right)^2\)
\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
a) Tìm số tự nhiên n biết:
\(\dfrac{4}{3\cdot5}+\dfrac{8}{5\cdot9}+\dfrac{12}{9\cdot15}+....+\dfrac{32}{n\cdot\left(n+16\right)}=\dfrac{16}{25}\)
b) Chứng tỏ rằng:
\(\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2018}>4\)
a) \(2\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+...+\dfrac{16}{n\left(n+16\right)}\right)=\dfrac{16}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{1}{3}-\dfrac{1}{n+16}=\dfrac{8}{25}\)
\(\dfrac{n+13}{3\left(n+16\right)}=\dfrac{8}{25}\)
\(24n+384=25n+325\)
\(25n-24n=384-325\)
\(n=59\)
b) Sai đề nha
\(\left\{{}\begin{matrix}\dfrac{2018}{2019}< 1\\\dfrac{2019}{2020}< 1\\\dfrac{2020}{2021}< 1\\\dfrac{2021}{2022}< 1\end{matrix}\right.\)
\(\Rightarrow\dfrac{2018}{2019}+\dfrac{2019}{2020}+\dfrac{2020}{2021}+\dfrac{2021}{2022}< 4\)
chị ơi hình như chị nhầm rồi P/s cuối phải là 1/n.(n+6)thì phải
Tìm các giới hạn sau:
\(a,lim\dfrac{\sqrt[3]{8n^3+2n}}{-n+3}\)
\(b,lim\dfrac{\left(2n\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n-1\right)\left(3-2n\right)}\)
\(a,lim\dfrac{^3\sqrt{8n^3+2n}}{-n+3}\)
\(=lim\dfrac{^3\sqrt{8+\dfrac{2}{n^2}}}{-1+\dfrac{3}{n}}=\dfrac{^3\sqrt{8}}{-1}=\dfrac{2}{-1}=-2\)
\(\lim\dfrac{\left(2n\sqrt{n}+1\right)\left(\sqrt{n}+3\right)}{\left(n-1\right)\left(3-2n\right)}=\lim\dfrac{\left(2+\dfrac{1}{n\sqrt{n}}\right)\left(1+\dfrac{3}{\sqrt{n}}\right)}{\left(1-\dfrac{1}{n}\right)\left(\dfrac{3}{n}-2\right)}=\dfrac{2.1}{1.\left(-2\right)}=-1\)
2) tìm số tự nhiên n biết:
\(\left(\dfrac{1}{3}\right)^n=\dfrac{1}{27};\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)
\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{27}\right)\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
\(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
a, \(\left(\dfrac{1}{3}\right)^n=\dfrac{1}{27}\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
Vì \(\dfrac{1}{3}\ne-1,\dfrac{1}{3}\ne0;\dfrac{1}{3}\ne1\) nên \(n=3\)
Vậy........
b, \(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
Vì \(\dfrac{3}{5}\ne-1,\dfrac{3}{5}\ne0;\dfrac{3}{5}\ne1\) nên \(n=4\)
Vậy..........
Chúc bạn học tốt!!!
Tìm số tự nhiên n, biết:
a)\(\frac{16}{2n}=2\)
b)\(\frac{\left(-3^n\right)}{81}=-27\)