Đề : Tìm x???

\(\frac{2}{3}x+\frac{1}{4}x=\frac{-22}{27}\)

\(x\left(\frac{2}{3}+\frac{1}{4}\right)=\frac{-22}{27}\)

\(x\left(\frac{8}{12}+\frac{3}{12}\right)=\frac{-22}{27}\)

\(x.\frac{11}{12}=\frac{-22}{27}\)

\(x=\frac{-22}{27}\div\frac{11}{12}\)

\(x=\frac{-22}{27}.\frac{12}{11}\)

\(x=\frac{-8}{9}\)

Vậy \(x=-\frac{8}{9}\).

S=2+4+6+...+98+100

S=\(\frac{\left[\left(\frac{100-2}{2}+1\right).\left(100+2\right)\right]}{2}=2550\)

S=1+2+3+4+...+2016+2017

S=\(\frac{\left(2017-1+1\right).\left(2017+1\right)}{2}=2035153\)

1.Số lượng số của S= (2017-1)+1=2017 số

tổng=(2016+1).(2016:2)+2017=2 035 153

2.Số lượng số của S=(100-2):2+1=50 số

tổng=(100+2).(50:2)=2 550

\(S=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2016}\right)\)

\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right).....\left(\frac{2016}{2016}-\frac{1}{2016}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2015}{2016}=\frac{1}{2016}\)

\(S=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2015}{2016}\)

\(S=\frac{1\cdot2\cdot3\cdot...\cdot2015}{2\cdot3\cdot4\cdot...\cdot2016}\)

\(S=\frac{1}{2016}\)

S=\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x..x\left(1-\frac{1}{2016}\right)\)

S=\(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2015}{2016}\)

S=1-\(\frac{2015}{2016}=\frac{1}{2016}\)

\(S=1^2+2^2+3^2+...+n^2\)

\(=1.2-1+2.3-2+3.4-3+...+n\left(n+1\right)-n\)

\(=\left[1.2+2.3+3.4+...+n\left(n+1\right)\right]-\left(1+2+3+...+n\right)\)

Theo dạng tổng quát: \(1.2+2.3+3.4+...+n\left(n+1\right)=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

\(\Rightarrow S=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{2n\left(n+1\right)\left(n+2\right)}{6}-\frac{3n\left(n+1\right)}{6}\)

\(=\frac{2n\left(n+1\right)\left(n+2\right)-3n\left(n+1\right)}{6}\)

\(=\frac{n\left(n+1\right).\left[2\left(n+2\right)-3\right]}{6}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Vậy \(S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

Ta có : \(S=1^2+2^2+3^2+...+\)\(n^2\)

\(\Rightarrow S=\frac{n.\left(n+1\right)\left(n+2\right)}{2}\)

Xin lỗi mình nhớ nhầm công thức : \(S=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

uses crt;

var n,i:longint;

s:real;

{------------ham-tinh-giai-thua---------------------}

function gthua(x:longint):real;

var i:longint;

gt:real;

begin

gt:=1;

for i:=1 to x do

gt:=gt*i;

gthua:=gt;

end;

{------------chuong-trinh-chinh------------------}

begin

clrscr;

write('Nhap n='); readln(n);

s:=0;

for i:=1 to n do 

  s:=s+gthua(i);

writeln(s:0:0);

readln;

end.

:) 

35353

565

s=s:)

\(S=1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot3+3\cdot3\cdot4+...+3\cdot99\cdot100\\ 3S=1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\\ 3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+....+99\cdot100\cdot101-98\cdot99\cdot100\\ 3S=99\cdot100\cdot101\\ S=\dfrac{99\cdot100\cdot101}{3}=33\cdot100\cdot101=3300\cdot101=333300\)