Tìm n ,biết : 2^n+3+2^n=36
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 6 :
a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)
b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)
c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)
d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)
Bài 7 :
a) \(3^x+3^{x+2}=9^{17}+27^{12}\)
\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)
\(\Rightarrow10.3^x=3^{34}+3^{36}\)
\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)
\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)
b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)
\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)
\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)
c) Bài C bạn xem lại đề
d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)
\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)
\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)
\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)
\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)
\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)
Xét công thức: 1+2+3+.....+n = n(n+1):2
1+2+.....+n = 36
=> n(n+1) : 2 = 36
=> n(n+1) = 36 x 2 = 72
Mà n(n+1) =8 x (8+1)
Vậy n = 8
\(\frac{1}{21}+\frac{1}{27}+\frac{1}{36}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)
\(\frac{2}{42}+\frac{2}{54}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)
\(\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{n\left(n+1\right)}=\frac{2}{9}\)
\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{n}-\frac{1}{n+1}=\frac{1}{9}\)
\(\Rightarrow\frac{1}{6}-\frac{1}{n+1}=\frac{n+1-6}{6n+6}=\frac{1}{9}\)
\(\frac{n-5}{6n+6}=\frac{1}{9}\)
\(9n-45=6n+6\)
\(9n-6n=6+45=51\)
\(n=51:3=17\)
\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{1}{3.7}+\frac{1}{4.7}+\frac{1}{4.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{2}{2.3.7}+\frac{2}{2.4.7}+\frac{2}{2.4.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{n}.\left(n+1\right)=\frac{2}{9}\)
\(\Leftrightarrow2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n}+1\right)=\frac{2}{9}\)
\(\Leftrightarrow2.\left(\frac{1}{6}-\frac{1}{n}+1\right)=\frac{2}{9}\)
\(\Leftrightarrow\frac{1}{6}-\frac{1}{n}+1=\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{n}+1=\frac{1}{6}-\frac{1}{9}\)
\(\Leftrightarrow\frac{1}{n}+1=\frac{1}{18}\)
\(\Leftrightarrow n+1=18\)
\(\Leftrightarrow n=17\)
Vậy \(n=17\)
Con " Nguyễn Huyền Trang " đéo biết thì trả lời làm cái l*n gì
mình trả lời bài 1 thôi nhé :
Gọi biểu thức trên là A.
Theo bài ra ta có:A=1/1.6+1/6.11+1/11.16+...+1/(5n+1)+1/(5n+6)
=1/5(1-1/6+1/6-1/11+1/11-1/16+...+1/5n+1-1/5n+6)
=1/5(1-1/5n+6)
=1/5( 5n+6/5n+6-1/5n+6)
=1/5(5n+6-1/5n+6)
=1/5.5n+5/5n+6
=n+1/5n+6
=ĐIỀU PHẢI CHỨNG MINH
x- 20/11.13 - 20/13.15 - 20/13.15 - 20/15.17 -...- 20/53.55=3/11
x-10.(2/11.13+2/13.15+2/15.17+...+2/53.55=3/11
x-10.(1/11-1/13+1/13-1/15+1/15-1/17+...+1/53-1/55)=3/11
x-10.(1/11-1/55)=3/11
x-10.4/55=3/11
x-8/11=3/11
x = 3/11+8/11
x=11/11=1
****
(-36)^1000:(-9)^1000=2^n
[(-36):9]^1000=2^n
4^1000=2^n
2^(2.1000)=2^n
2^2000=2^n
vậy n = 2000
\(\left(-36\right)^{1000}:\left(9\right)^{1000}=2^n\)
\(36^{1000}:9^{1000}=2^n\)
\(\left(36\div9\right)^{1000}=2^n\)
\(4^{1000}=2^n\)
\(\left(2^2\right)^{1000}=2^n\)
\(2^{2000}=2^n\)
=> n = 2000
(-36)^1000:(-9)^1000=2^n
[(-36):9]^1000=2^n
4^1000=2^n
2^(2.1000)=2^n
2^2000=2^n
vậy n = 2000
\(2^{n+3}+2^n=36\)
\(\Leftrightarrow2^n.\left(2^3+1\right)=36\Leftrightarrow2^n.9=36\Leftrightarrow2^n=4\Rightarrow n=2\)
Vậy n = 2
Ta có
\(2^{n+3}+2^n=36\)
\(=2^n.2^3+2^n=36\)
\(\Leftrightarrow2^n\left(2^3+1\right)=36\)
\(\Rightarrow2^n.9=36\)
\(\Rightarrow2^n=4\Leftrightarrow2^n=2^2\)
\(\Rightarrow n=2\)