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Câu 1:
-Đánh răng rửa mặt mỗi ngày
-Đi học
-Đi ngủ
Câu 2:
a) Sai
Sửa lại: for i:=10 downto 1 do write('a');
b) Đúng
c) Đúng
Câu 3:
Lặp n-1+1=n(lần)
Câu 4:
S=15
Câu 5:
uses crt;
var s,i:integer;
begin
clrscr;
s:=0;
for i:=1 to 10 do
s:=s+i;
writeln('Tong cua 10 so tu nhien dau tien la: ',s);
readln;
end.
a) Sai
Sửa lại: for i:=10 downto 1 do write('a');
b) Đúng
c) Đúng
?????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????/??bố thằng nào mà biết được
a) \(5+3^{x+1}=86\)
\(=>3^{x+1}=86-5\)
\(=>3^{x+1}=81=3^4\)
\(=>x+1=4\) ( cùng cơ số )
\(=>x=4-1\)
\(=>x=3\)
b) \(15:\left(x+2\right)=\left(3^3+3\right):10\)
\(=>15:\left(x+2\right)=\left(27+3\right):10\)
\(=>15:\left(x+2\right)=30:10=3\)
\(=>x+2=15:3\)
\(=>x+2=5\)
\(=>x=5-2\)
\(=>x=3\)
c) \(\left(9x+2\right).4=80\)
\(=>9x+2=80:4\)
\(=>9x+2=20\)
\(=>9x=20-2\)
\(=>9x=18\)
\(=>x=18:9\)
\(=>x=2\)
d) \(\left(245-x\right)+7^2=14\)
\(=>\left(245-x\right)+14=14\)
\(=>245-x=14-14\)
\(=>245-x=0\)
\(=>x=245-0\)
\(=>x=245\)
đơn giản
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x.\left(x+1\right)}=\frac{39}{40}\)
\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{39}{40}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}:2\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{39}{80}\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}\)
\(\frac{1}{x+1}=\frac{1}{80}\)
\(\Rightarrow x+1=80\)
\(\Rightarrow x=80-1=79\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{x\left(x+1\right)}=\frac{39}{40}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{39}{40}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{39}{40}\div2=\frac{39}{80}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{39}{80}=\frac{1}{80}\)
\(\Leftrightarrow x+1=80\Rightarrow x=80-1=79\)
Vậy \(x=79\)
Bài 3:
\(a,\dfrac{x-1}{10}+\dfrac{x-1}{11}=\dfrac{x-1}{12}+\dfrac{x-1}{13}\)
\(\Rightarrow\dfrac{x-1}{10}+\dfrac{x-1}{11}-\dfrac{x-1}{12}-\dfrac{x-1}{13}=0\)
\(\Rightarrow\left(x-1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)=0\)
Mà \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\ne0\)
\(\Rightarrow x-1=0\Rightarrow x=1\)
Vậy x = 1
b, \(\dfrac{x-2000}{10}+\dfrac{x-1999}{9}=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}\)
\(\Rightarrow\dfrac{x-2000}{10}+1+\dfrac{x-1999}{9}+1=\dfrac{x-1998}{8}+\dfrac{x-1997}{7}+1\)
\(\Rightarrow\dfrac{x-1990}{10}+\dfrac{x-1990}{9}-\dfrac{x-1990}{8}-\dfrac{x-1990}{7}=0\)
\(\Rightarrow\left(x-1990\right)\left(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\right)=0\)
Mà \(\dfrac{1}{10}+\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{7}\ne0\)
\(\Rightarrow x-1990=0\Rightarrow x=1990\)
2, 3, và 49 nha
1 + 1 = 2
1 + 2 = 3
39 + 10 = 49