S=1+4+7+..+n

Tổng S có số số hạng là \(\frac{\left(n-1\right)}{3}+1=\frac{n+2}{3}\)

Tổng S có giá trị là

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

\(\left[{}\begin{matrix}\dfrac{x}{3}-\dfrac{12}{x}=0\\5:x-1=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{x^2-36}{3x}=0\\\dfrac{5}{x}=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}=>x=6và\left(-6\right)\\=>x=5\end{matrix}\right.\)

do \(\left(x-2\right)^2\ge o\forall x\)

\(\Rightarrow\left(x-2\right)^2+5\ge5\)

\(\Rightarrow\frac{6}{\left(x-2\right)^2+5}\ge\frac{6}{5}\)

Suy ra \(\frac{6}{\left(x-2\right)^2+5}\)đạt giá trị nhỏ nhất là \(\frac{6}{5}\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Vậy

cảm ơn bạn nhiu

a) \(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)

\(\Rightarrow\left[\begin{array}{nghiempt}x+\frac{1}{2}=2x+3\\x+\frac{1}{2}=-\left(2x+3\right)\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x-x=\frac{1}{2}-3\\x+\frac{1}{2}=-2x-3\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x+2x=-3-\frac{1}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\3x=\frac{-7}{2}\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-5}{2}\\x=\frac{-7}{6}\end{array}\right.\)

Vậy \(x\in\left\{\frac{-5}{2};\frac{-7}{6}\right\}\)

\(\left|x+\frac{1}{2}\right|=\left|2x+3\right|\)

\(Ta\) \(có\): \(x+\frac{1}{2}=2x+3\)

\(x+\frac{1}{2}=x+x+3\\\)

\(x+\frac{1}{2}=x+\left(x+3\right)\)

\(\Rightarrow\frac{1}{2}=x+3\)

\(\Rightarrow x=\frac{1}{2}-3\)

\(\Rightarrow x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}\)

b, \(\left|x+\frac{1}{5}\right|+\left|x+\frac{2}{5}\right|+\left|x+1\frac{2}{5}\right|=4x\)

\(Ta\) \(có\)

\(x+\frac{1}{5}+x+\frac{2}{5}+x+1\frac{2}{5}\)\(=4x\)

\(3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)

\(3x+2=4x\)

\(3x+2=3x+x\)

\(\Rightarrow x=2\)

Vậy \(x=2\)

b) Vì \(\left|x+\frac{1}{5}\right|\ge0;\left|x+\frac{2}{5}\right|\ge0;\left|x+1\frac{2}{5}\right|\ge0\forall x\)

\(\Rightarrow4x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\) ta có:

\(\left(x+\frac{1}{5}\right)+\left(x+\frac{2}{5}\right)+\left(x+1\frac{2}{5}\right)=4x\)

\(\Rightarrow3x+\left(\frac{1}{5}+\frac{2}{5}+1\frac{2}{5}\right)=4x\)

\(\Rightarrow2=4x-3x=x\)

Vậy x = 2

program dgv;

uses crt;

var x,y,z,n,i: integer;

p,S,A: real;

begin

clrscr;

writeln('Nhap x: '); read(x);

writeln('Nhap y: '); read(y);

writeln('Nhap z: '); read(z);

writeln('Nhap n: '); read(n);

if x mod 2 = 0 then write (x,' la so chan') else writeln(x,'la so le');

if y mod 2 = 0 then write (y,' la so chan') else writeln(y,'la so le');

if z mod 2 = 0 then write (z,' la so chan') else writeln(z,'la so le');

if n mod 2 = 0 then write (n,' la so chan') else writeln(n,'la so le');

if n mod 3 <> 1 then writeln('Tong khong co quy luat') else

begin

S:=0;

for i:= 0 to (n-1)/3 do

S:= S+(3i+1);

writeln('S = ',S);

A:=0;

p:=1;

for i:=1 to n do

begin

p:= p*(x/(x+y));

A:=A+p;

end;

writeln('A = ',A);

readln

end.