Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
N=\(\dfrac{2^{10}.13+2^9+130}{2^8.104}\)
N=\(\dfrac{13312+642}{26624}\)
N=\(\dfrac{3954}{26624}\)=\(\dfrac{6977}{13312}\)
\(B=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^9.33+3^9.15}{3^9.2^4}\)
\(=\frac{3^9\left(33+15\right)}{3^9.2^4}=\frac{3^9.48}{3^9.16}\)
\(=\frac{48}{16}=3\)
\(B=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{3^{10}.\left(11+5\right)}{3^9.8}\)
\(=\frac{3^{10}.16}{3^9.8}\)
\(=\frac{3.2}{1}\)
\(=6\)
Lời giải:
a.
\(\frac{n+1}{n+2}=\frac{n+1}{n+2}+1-1=\frac{2n+3}{n+2}-1\)
\(> \frac{2n+3}{n+3}-1=\frac{(n+3)+n}{n+3}-1=\frac{n}{n+3}\)
b.
\(10A=\frac{10^{12}-10}{10^{12}-1}=\frac{(10^{12}-1)-9}{10^{12}-1}=1-\frac{9}{10^{12}-1}<1\)
\(10B=\frac{10^{11}+10}{10^{11}+1}=\frac{(10^{11}+1)+9}{10^{11}+1}=1+\frac{9}{10^{11}+1}>1\)
$\Rightarrow 10A< 10B\Rightarrow A< B$
Lời giải:
$S=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+...+\frac{69}{7^{70}}$
$7S=\frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+...+\frac{69}{7^{69}}$
$6S=7S-S=\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}}$
$42S=1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}}$
$\Rightarrow 42S-6S=(1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{68}}-\frac{69}{7^{69}})-(\frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+....+\frac{1}{7^{69}}-\frac{69}{7^{70}})$
$\Rightarrow 36S=1-\frac{69}{7^{69}}-\frac{1}{7^{69}}+\frac{69}{7^{70}}$
Hay $36S=1-\frac{69.7-7-69}{7^{70}}=1-\frac{407}{7^{70}}$
$\Rightarrow S=\frac{1}{36}(1-\frac{407}{7^{70}})$
N=1/2+1/22+...+1/210
2N=1+1/2+...+1/29
2N-N=1-1/210=1-1/1024=1023/1024
Giải:
N=1/2+1/22+1/23+...+1/29+1/210
2N=1+1/2+1/22+...+1/28+1/29
2N-N=(1+1/2+1/22+...+1/28+1/29)-(1/2+1/22+1/23+...+1/29+1/210)
N=1-1/210=1023/1024
Chúc bạn học tốt!
\(A=\frac{3^{10}.11+3^{10}.5}{3^9.2^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}=\frac{3^{10}.16}{3^9.16}=3\)
= (1+99) + (3 + 97) + ( 5 + 95 ) + ....... + ( 49 + 51 )
có tất cả 25 cặp
= 100 x 25 = 2500
Nhớ k mình nhé! Thanks bạn
(1+99)+(3+97)+....+(49+51)=100+100+...+100=5000
Luu y:co 50 so 100
\(\Rightarrow-5\left(n+3\right)+42⋮n+3\\ \Rightarrow n+3\inƯ\left(42\right)=\left\{-42;-21;-14;-7;-6;-3;-2;-1;1;2;3;6;7;14;21;42\right\}\\ \Rightarrow n\in\left\{-45;-24;-17;-10;-9;-6;-5;-4;-2;-1;0;3;4;11;17;39\right\}\)
\(N=\dfrac{3^{10}.11+3^9.15}{3^9.2^4}=\dfrac{3^9.33+3^9.15}{3^9.2^4}\)
\(=\dfrac{3^9\left(33+15\right)}{3^9.16}\)
\(=\dfrac{48}{16}=3\)