CMR
D=\(\frac{2!}{3!}+\frac{2!}{4!}+\frac{2!}{5!}+.....+\frac{2!}{n!}
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~
b)
program hotrotinhoc;
var s: real;
i,n: byte;
function t(x: byte): longint;
var j: byte;
t1: longint;
begin
t1:=1;
for j:=1 to x do
t1:=t1*j;
t1:=t;
end;
begin
readln(n);
s:=0;
for i:=1 to n do
s:=s+1/t(i);
write(s:1:2);
readln
end.
c) Đề em ghi sai rồi thế này với đúng :
\(T=1+\frac{2}{2^2}+\frac{3}{3^2}+\frac{4}{4^2}+...+\frac{n}{n^2}\)
program hotrotinhoc;
var t: real;
n,i: byte;
begin
readln(n);
t:=0;
for i:=1 to n do
t:=t+i/(i*i);
write(t:1:2);
readln
end.
Bài 3:
\(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
Vậy P = \(\frac{-272}{45}\)
\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)
\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)
\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)
\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}=1\)
Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !
c, \(\frac{-32}{-2^n}=4\)
\(\Rightarrow-2^n=-32:4\)
\(\Rightarrow-2^n=-8\)
\(\Rightarrow-2^n=-2^3\Rightarrow n=3\)
d, \(\frac{8}{2^n}=2\)
\(\Rightarrow2^n=8:2\)
\(\Rightarrow2^n=4\)
\(\Rightarrow2^n=2^2\Rightarrow n=2\)
e, \(\frac{25^3}{5^n}=25\)
\(\Rightarrow5^n=25^3:25\)
\(\Rightarrow5^n=25^2\)
\(\Rightarrow5^n=5^4\Rightarrow n=4\)
i , \(8^{10}:2^n=4^5\)
\(\Rightarrow2^n=8^{10}:4^5\)
\(\Rightarrow2^n=\left(2^3\right)^{10}:\left(2^2\right)^5\)
\(\Rightarrow2^n=2^{30}:2^{10}\)
\(\Rightarrow2^n=2^{20}\Rightarrow n=20\)
k, \(2^n.81^4=27^{10}\)
\(\Rightarrow2^n=27^{10}:81^4\)
\(\Rightarrow2^n=\left(3^3\right)^{10}:\left(3^4\right)^4\)
\(\Rightarrow2^n=3^{30}:3^{16}\)
\(\Rightarrow2^n=3^{14}\)
\(\Rightarrow2^n=4782969\)Không chia hết cho 2 nên ko có Gt n thỏa mãn