Tìm n, biết:
a) 8^n+1= 8^2
b) 7^n= 343
c) 16 phần 2n= 2
d) 3^2:3^n= 3^5
e) 9.3^4.3^n= 3^7
g) (n-2)^2= 1
h) (n-1)^3=8
k) 3^2.3^n=3^5
Tìm số nguyên n biết: 1/9.27^n=3^n. b)3^-2.3^4.3^n=3^7. c) 2^n-1.2^n+4.2^n=9.2^5. d)32^-n.16^n=2048. Các bạn giải hộ mình nhé
a)1/9.27^n=3^n
3^n=3^n
=>n={0;1;2;3;...}
a) n= 2;3;5;7;...(n là số nguyên)
Tìm các số tự nhiên n
\(5^{n+3}-5^{n+1}=5^{12}.120\)
\(2^{n+1}+4.2^n=3.2^7\)
\(3^{n+2}-3^{n+1}=486\)
\(3^{2n+3}-3^{2n+2}=2.3^{10}\)
a) \(5^{n+3}-5^{n+1}=5^{12}.120\Leftrightarrow5^{n+1}.\left(5^2-1\right)=5^{12}.5.24\)
\(\Leftrightarrow24.5^{n+1}=5^{13}.24\Leftrightarrow5^{n+1}=5^{13}\Leftrightarrow n+1=13\Leftrightarrow n=12\)
b) \(2^{n+1}+4.2^n=3.2^7\)
\(\Leftrightarrow2^n\left(2+4\right)=3.2^7\Leftrightarrow6.2^n=3.2^7\Leftrightarrow2^n=2^6\Leftrightarrow n=6\)
c) \(3^{n+2}-3^{n+1}=486\)
\(\Leftrightarrow3^{n+1}.\left(3-1\right)=486\Leftrightarrow2.3^{n+1}=486\Leftrightarrow3^{n+1}=243\)
\(\Leftrightarrow3^n=243:3=81=3^3\Leftrightarrow n=3\)
d) \(3^{2n+3}-3^{2n+2}=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.\left(3-1\right)=2.3^{10}\)
\(\Leftrightarrow3^{2n+2}.2=2.3^{10}\Leftrightarrow3^{2n+2}=3^{10}\Leftrightarrow2n+2=10\Leftrightarrow2n=8\Leftrightarrow n=4\)
1. CMR: ∀ n∈\(N^{\cdot}\)
a) \(A=5^n+2.3^{n-1}+1\text{⋮}8\)
b) \(B=3^{n+2}+4^{2n+1}\text{⋮}13\)
c) \(C=6^{2n}+3^{n+2}+3^n\text{⋮}11\)
d) \(D=1^n+2^n+5^n+8^n\text{⋮}8\)
2. \(CMR:\) \(1^{2002}+2^{2002}+...+2002^{2002}\text{⋮}11\)
3. a) cho a,b ∈Z, t/m:\(a^2+b^2\text{⋮}7\). \(CMR:a\text{⋮}7;b\text{⋮}7\)
b) \(CMR:\) Nếu \(a^2+b^2\text{⋮}21\) thì \(a^2+b^2\text{⋮}441\) (a,b ∈Z)
\(1,\)
\(a,\) Với \(n=1\Leftrightarrow5+2\cdot1+1=8⋮8\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow5^k+2\cdot3^{k-1}+1⋮8\)
Với \(n=k+1\)
\(5^n+2\cdot3^{n-1}+1=5^{k+1}+2\cdot3^k+1\\ =5^k\cdot5+2\cdot3^k+1\\ =5^k\cdot2+2\cdot3^k+5^k\cdot3+1\\ =2\left(5^k+3^k\right)+5^k+2\cdot5^{k-1}+1+2\cdot3^{k-1}-2\cdot3^{k-1}\\ =2\left(5^k+3^k\right)+\left(5^k+2\cdot3^{k-1}+1\right)-2\left(3^{k-1}+5^{k-1}\right)\)
Vì \(5^k+3^k⋮\left(5+3\right)=8;5^{k-1}+3^{k-1}⋮\left(5+3\right)=8;5^k+2\cdot3^{k-1}+1⋮8\) nên \(5^{k+1}+2\cdot3^k+1⋮8\)
Theo pp quy nạp ta được đpcm
\(b,\) Với \(n=1\Leftrightarrow3^3+4^3=91⋮13\left(đúng\right)\)
Giả sử \(n=k\left(k\ge1\right)\Leftrightarrow3^{k+2}+4^{2k+1}⋮13\)
Với \(n=k+1\)
\(3^{n+2}+4^{2n+1}=3^{k+3}+4^{2k+3}\\ =3^{k+2}\cdot3+16\cdot4^{2k+1}\\ =3^{k+2}\cdot3+3\cdot4^{2k+1}+13\cdot4^{2k+1}\\ =3\left(3^{k+2}+4^{2k+1}\right)+13\cdot4^{2k+1}\)
Vì \(3^{k+2}+4^{2k+1}⋮13;13\cdot4^{2k+1}⋮13\) nên \(3^{k+3}+4^{2k+3}⋮13\)
Theo pp quy nạp ta được đpcm
\(1,\)
\(c,C=6^{2n}+3^{n+2}+3^n\\ C=36^n+3^n\cdot9+3^n\\ C=\left(36^n-3^n\right)+\left(3^n\cdot9+2\cdot3^n\right)\\ C=\left(36^n-3^n\right)+3^n\cdot11\)
Vì \(36^n-3^n⋮\left(36-3\right)=33⋮11;3^n\cdot11⋮11\) nên \(C⋮11\)
\(d,D=1^n+2^n+5^n+8^n\)
Vì \(1^n+2^n+5^n⋮\left(1+2+5\right)=8;8^n⋮8\) nên \(D⋮8\)
\(2,\)
Ta thấy:\(1+2+...+2002=\left(2002+1\right)\left(2002-1+1\right):2=2003\cdot2002:2⋮11\left(2002⋮11\right)\)
Do đó \(1^{2002}+2^{2002}+...+2002^{2002}⋮1+2+...+2002⋮11\)
tìm n
8:n-5
40:2n+5
16:3-2n
n+10:n+4
n+7:n-7
n+3:8-n
3n-5:n-3
3n+2:2n+1
7-2n:n-1
giải hộ mình voi ghi cả cách giải ra cho minhd nha
dài kinh,bài này chắc làm đến tối! bn ơi,bn cho từng câu một thôi!đau đầu lắm!
a ) 8 chia hết cho n - 5
nên n - 5 thuộc Ư( 8 ) = { 1 ; 2 ; 4 ; 8 }
=> n thuộc { 6 ; 7 ; 9 ; 13 }
Mik làm mẫu 1 câu thôi , còn lại bạn tự làm nhé !
Tìm số nguyên n biết: 1/9.27^n=3^n. b)3^-2.3^4.3^n=3^7. c) 2^n-1.2^n+4.2^n=9.2^5. d)32^-n.16^n=2048. Các bạn giải hộ mình nhé
Toán lớp 6
a)1/9.27^n=3^n
3^n=3^n
=>n={0;1;2;3...}
Tích nha ^_^ !!!
Tìm n thuộc Z biết:
a)\(\frac{1}{9}.27^n=3^n\)
b) \(3^2.3^4.3^n=3^7\)
c) \(2^{-1}.2^n+4.2^n=9.2^5\)
d) \(32^{-n}.16^n=2048\)
a) Ta có: \(\frac{1}{9}\cdot27^n=3^n\)
\(\Leftrightarrow\frac{1}{3^2}\cdot\left(3^3\right)^n=3^n\)
\(\Leftrightarrow3^{3n}=3^{n+2}\)
\(\Rightarrow3n=n+2\)
\(\Rightarrow n=1\)
b) Ta có: \(3^2.3^4.3^n=3^7\)
\(\Rightarrow3^n=3\)
\(\Rightarrow n=1\)
c) Ta có: \(2^{-1}.2^n+4.2^n=9.2^5\)
\(\Leftrightarrow2^n\cdot\frac{9}{2}=9.2^5\)
\(\Rightarrow2^n=2^6\)
\(\Rightarrow n=6\)
d) Ta có: \(32^{-n}.16^n=2048\)
\(\Leftrightarrow\frac{1}{2^{5n}}\cdot2^{4n}=2^{11}\)
\(\Leftrightarrow2^{4n}=2^{5n+11}\)
\(\Rightarrow4n=5n+11\)
\(\Rightarrow n=-11\)
Tính các giới hạn sau:
a) \(\lim\limits\dfrac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}\)
b) \(\lim\limits\dfrac{1}{\sqrt{3n+2}-\sqrt{2n+1}}\)
c) \(\lim\limits\dfrac{4.3^n+7^{n+1}}{2.5^n+7^n}\)
a.
\(A=\lim\frac{\sqrt[3]{n^6-7n^3-5n+8}}{n+12}=\lim \frac{\sqrt[3]{\frac{n^6-7n^3-5n+8}{n^3}}}{\frac{n+12}{n}}=\lim \frac{\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}}{1+\frac{12}{n}}\)
Ta thấy:
\(\lim\sqrt[3]{n^3-7-\frac{5}{n^2}+\frac{8}{n^3}}=\infty \)
\(\lim (1+\frac{12}{n})=1\)
Suy ra $A=\infty$
b.
\(B=\lim\frac{1}{\sqrt{3n+2}-\sqrt{2n+1}}=\lim \frac{1}{\frac{3n+2-(2n+1)}{\sqrt{3n+2}+\sqrt{2n+1}}}=\lim \frac{\sqrt{3n+2}+\sqrt{2n+1}}{n+1}\)
\(=\lim \frac{\sqrt{\frac{3n+2}{n}}+\sqrt{\frac{2n+1}{n}}}{\frac{n+1}{\sqrt{n}}}=\lim \frac{\sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}}}{\sqrt{n}+\frac{1}{\sqrt{n}}}\)
Ta thấy:
\(\lim( \sqrt{3+\frac{2}{n}}+\sqrt{2+\frac{1}{n}})=\sqrt{3}+\sqrt{2}>0\)
\(\lim (\sqrt{n}+\frac{1}{\sqrt{n}})=\infty\)
$\Rightarrow B=\infty$
c.
\(C=\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{4(\frac{3}{7})^n+7}{2(\frac{5}{7})^n+1}\)
Ta thấy:
\(\lim [4(\frac{3}{7})^n+7]=4.0+7=7\) với $|\frac{3}{7}|<1$
\(\lim [2(\frac{5}{7})^n+1]=2.0+1=1\) với $|\frac{5}{7}|<1$
$\Rightarrow C=\frac{7}{1}=7$
1. Tìm a,b,c biết:
a) a/b = 8/5; b/c = 2/7 và a+b+c= 61
b) ab = 1/2; bc= 2/3; ac = 3/4
c) 3a=2b; 5b = 7c và 3a + 5c - 7b= 60
2. tìm các số nguyên n sao cho:
1) 5^n + 5^n+2 = 650
2) 32^-n .16^n = 1024
3) 3^-1 .3^n+ 5. 3^n-1 = 162
4) 125. 5\(\ge\)5^n\(\ge\)5 . 25
5) (n^54)^2 = n
6) 243\(\ge\)3^n\(\ge\)9.27
7) 2^n+3 . 2^n = 144
8)3<3^n\(\le\)234
9) 8. 16\(\ge\)2^n\(\ge\)4
10) 4^15. 9^15<2^n.3^n< 18^16. 2^16
11) 4^11. 25^11\(\le\)2^n. 5^n\(\le\)20^12. 5^12
12)\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}\).\(\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}\)= 2^n
13) 9. 27^n= 3^5
14) (2^3 : 4) . 2^n= 4
15) 3^-2 . 3^4. 3^n = 3^7
16)2^-1. 2^n +4.2^n=9.2^5
1) lim \(\frac{3n^2+5n+4}{2-n^2}\)
2) lim \(\frac{2n^3-4n^2+3n+7}{n^3-7n+5}\)
3) lim \(\left(\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1}\right)\)
4) lim \(\frac{1+3^n}{4+3^n}\)
5) lim \(\frac{4.3^n+7^{n+1}}{2.5^n+7^n}\)
1.
\(\lim \frac{3n^2+5n+4}{2-n^2}=\lim \frac{\frac{3n^2+5n+4}{n^2}}{\frac{2-n^2}{n^2}}=\lim \frac{3+\frac{5}{n}+\frac{4}{n^2}}{\frac{2}{n^2}-1}=\frac{3}{-1}=-3\)
2.
\(\lim \frac{2n^3-4n^2+3n+7}{n^3-7n+5}=\lim \frac{\frac{2n^3-4n^2+3n+7}{n^3}}{\frac{n^3-7n+5}{n^3}}=\lim \frac{2-\frac{4}{n}+\frac{3}{n^2}+\frac{7}{n^3}}{1-\frac{7}{n^2}+\frac{5}{n^3}}=\frac{2}{1}=2\)
3.
\(\lim (\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1})=\lim (n-\frac{3n}{2n^2+3}+\frac{1}{5}-n-\frac{1}{5n+1})\)
\(=\frac{1}{5}-\lim (\frac{3n}{2n^2+3}+\frac{1}{5n+1})=\frac{1}{5}-\lim (\frac{3}{2n+\frac{3}{n}}+\frac{1}{5n+1})=\frac{1}{5}-0=\frac{1}{5}\)
4.
\(\lim \frac{1+3^n}{4+3^n}=\lim (1-\frac{3}{4+3^n})=1-\lim \frac{3}{4+3^n}=1-0=1\)
5.
\(\lim \frac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim \frac{\frac{4.3^n+7^{n+1}}{7^n}}{\frac{2.5^n+7^n}{7^n}}\)
\(=\lim \frac{4.(\frac{3}{7})^n+7}{2.(\frac{5}{7})^n+1}=\frac{7}{1}=7\)