2^n+1=128
(7n-11)^3=2^5.5^2+200. Nhanh giùm mình nha
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\(\Rightarrow\left(7n-11\right)^3=32\times25+200\)
\(\Rightarrow\left(7n-11\right)^3=1000\)
\(\Rightarrow7n-11=10\)
\(\Rightarrow7n=10+11\)
\(\Rightarrow n=21:7=3\)
\(\left(7n-11\right)^3=2^5.5^2+200\)
\(\left(7n-11\right)^3=32.25+200\)
\(\left(7n-11\right)^3=800+200\)
\(\left(7n-11\right)^3=1000\)
=>TH1:(7n-11)3=103
=>7n-11=10
=>7n=10+11=21
=>n=21:7=3
TH2: (7n-11)3=-103
=>7n-11=-10
=>7n=-10+11
=>7n=1
=>n=1:7=1/7
Mà n thuộc N nên n=3
Kết luận n=3
câu a đề sai
2n-15=17
=> 2n=17+15
=>2n=32
=> n=5
c,\(2.3^n=162=>3^n=162:2=>3^n=81\)
=>\(n=4\)
d, (7n-11)^3=2^5.5^2+200 = (7n-11)^3=32.25+200
= (7n-11)^3=800 +200 = 1000
câu này đè sai
(7x−11)3 = 25 . 52 + 200
(7x−11)3 = 32 . 25 + 200
(7x−11)3 = 1000
(7x−11)3 = 103
7x−11 = 10
7x = 10 + 11
7x = 21
x = 21 : 7
x = 3
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)
\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{255}{128}+\frac{3}{512}\)
\(=\frac{1020}{512}+\frac{3}{512}\)
\(=\frac{1023}{512}\)
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
\(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
\(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)
Ta có \(\left(n^2+7n+9\right)⋮\left(n+3\right)\)
\(\Leftrightarrow\left[\left(n^2+3n\right)+\left(4n+12\right)-3\right]⋮\left(n+3\right)\)
\(\Leftrightarrow\left[n\left(n+3\right)+4\left(n+3\right)-3\right]⋮\left(n+3\right)\)
\(\Rightarrow-3⋮\left(n+3\right)\)Hay \(n+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\)
n + 3 | - 3 | - 1 | 1 | 3 |
n | - 6 | - 4 | - 2 | 0 |
Vậy \(n\in\left\{-6;-4;-2;0\right\}\)
Ta có: \(\frac{n^2+7n+9}{n+3}=\frac{n^2+3n+3n+9}{n+3}+\frac{n}{n+3}\)
= \(\frac{\left(n+3\right)^2}{n+3}+\frac{n+3-3}{n+3}=n+3+1-\frac{3}{n+3}\)=> x + 4 - 3/n+3
Do n thuộc N => n+ 4 thuộc N; Để \(n^2+7n+9⋮n+3=>3⋮n+3\)
Hay n+3 thuộc Ư(3)
=> n+ 3 thuộc { -3;-1;1;3}
=> n thuộc { -6; -4; -2;0}
Mà n thuộc N nên n =0
bạn làm chi tiết hộ mình 3 câu C,D,E. Mình đang gấp! tung nguyen viet
ko hieu bang cach hop la the nao
4C=4^2+4^3+..+4^(n+1)
4C-C=4^(n+1)-4=4(4^n-1)
C=4.(4^n-1)/3
5^2.D=5^2+5^4+..+5^202
5^2D-D=5^202-1
D=(5^202-1)/24
E=(2^202-1)/3