Giải phương trình : (1/1*51+1/2*52+.....+1/10*60)x=(1/1*11+1/2*12+.....+1/40*50)
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a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
Hình như câu này chỉ sử dụng câu lệnh for... to... do hay sao thoii í cậu. Thầy tớ gợi í thế
a)
uses crt;
var b:real;
i:integer;
begin
clrscr;
i:=10;
b:=1;
while i<=30 do
begin
b:=b*i;
i:=i+1;
end;
writeln('B=',b:0:0);
readln;
end.
b) uses crt;
var c,j:integer;
begin
clrscr;
j:=50;
c:=0;
while j<=100 do
begin
c:=c+j;
j:=j+1;
end;
writeln('C=',c);
readln;
end.
c) uses crt;
var i,d:integer;
begin
clrscr;
i:=-50;
d:=0;
while i<=50 do
begin
d:=d+i;
inc(i);
end;
writeln('D=',d);
readln;
end.
d) uses crt;
var n,i:integer;
e:real;
begin
clrscr;
write('n='); readln(n);
e:=0;
for i:=1 to n do
e:=e+1/(i*(i+2));
writeln('E=',e:4:2);
readln;
end.
a) 50 + 48 + 46 + ... + 4 - 47 - 45 - 43 - ... - 1
= (50 - 45) + (48 - 43) + (46 - 41) + ... + (6 - 1) + (4 - 47)
=72
Cứ gộp nhóm làm sao cho trong ngoặc đó bằng 5
b) 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 + ... + 50 - 51 - 52 + 53 + 54
= (1 + 54) + (2 + 53) - (3 + 52) - (4 + 51) + ... + (25 + 30) + (26 + 29) - (27 + 28)
=55
Cứ gộp nhóm làm sao cho trong ngoặc đó bằng 55. Còn dấu đằng trước nhóm thì theo dấu đề bài cho
~ Học tốt ~
(1 + \(\dfrac{1}{49}\))\(\times\)(1 + \(\dfrac{1}{50}\))\(\times\)(1 + \(\dfrac{1}{51}\))\(\times\)(1 + \(\dfrac{1}{52}\))\(\times\)...\(\times\)(1 + \(\dfrac{1}{60}\))
= \(\dfrac{49+1}{49}\) \(\times\) \(\dfrac{50+1}{50}\)\(\times\) \(\dfrac{51+1}{51}\)\(\times\)\(\dfrac{52+1}{52}\)\(\times\)...\(\times\)\(\dfrac{61}{60}\)
= \(\dfrac{50}{49}\)\(\times\)\(\dfrac{51}{50}\)\(\times\)\(\dfrac{52}{51}\)\(\times\)...\(\times\)\(\dfrac{61}{60}\)
= \(\dfrac{50\times51\times52\times53\times...\times60}{50\times51\times52\times53\times...\times60}\)\(\times\)\(\dfrac{61}{49}\)
= \(\dfrac{61}{49}\)
\(1,\left(dk:x\ne0,-1,4\right)\)
\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)
\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)
\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)
\(\Leftrightarrow-x=-44\)
\(\Leftrightarrow x=44\left(tm\right)\)
\(2,\left(đk:x\ne4\right)\)
\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)
\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)
\(\Leftrightarrow28-12-6x-9+5x-20=0\)
\(\Leftrightarrow-x=13\)
\(\Leftrightarrow x=-13\left(tm\right)\)