m = 5 

n = -1

mình nhầm câu trên

\(a,x< 50\Leftrightarrow\sqrt{x}-1< 5\sqrt{2}-1\\ M=\dfrac{\sqrt{x}-1}{2}\in Z\\ \Leftrightarrow\sqrt{x}-1\in B\left(2\right)=\left\{0;2;4;6\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{1;3;5;7\right\}\\ \Leftrightarrow x\in\left\{1;9;25;49\right\}\\ b,\Leftrightarrow\sqrt{x}-5\inƯ\left(9\right)=\left\{-3;-1;1;3;9\right\}\left(\sqrt{x}-5>-5\right)\\ \Leftrightarrow\sqrt{x}\in\left\{2;4;6;8;14\right\}\\ \Leftrightarrow x\in\left\{4;16;36;64;196\right\}\)

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{2y}{6}\)

\(\dfrac{1}{x}=\dfrac{1+2y}{6}\)

\(6=x\left(1+2y\right)\)

Tự làm típ

\(x\left(x+y\right)=\dfrac{1}{48};y\left(x+y\right)=\dfrac{1}{24}\)

\(x^2+xy=\dfrac{1}{48};xy+y^2=\dfrac{1}{24}\)

\(\Rightarrow x^2+xy-y^2-xy=\dfrac{1}{48}-\dfrac{1}{24}\)

\(x^2-y^2=\dfrac{-1}{24}\)

\(\left(x+y\right)\left(x-y\right)=\dfrac{-1}{24}\)(HĐT số 3)

Làm tips

Ta có :

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+.................+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+............+\dfrac{2}{n\left(n+1\right)}=\dfrac{2003}{2004}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+............+\dfrac{2}{2\left(n+1\right)}=\dfrac{2003}{2004}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{n+1}\right)=\dfrac{2003}{2004}\)

\(\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{2003}{4008}\)

\(\dfrac{1}{n+1}=\dfrac{1}{4008}\)

\(\Rightarrow n+1=4008\)

\(\Rightarrow n=4007\) (Thỏa mãn \(n\in N\))

Vậy \(n=4007\) là giá trị cần tìm

~~Chúc bn học tốt~~

hình như sai đề phải bạn ạ

sai đề r

Áp dụng BĐT Cauchy cho 2 số dương ta có:
\(x^2+\dfrac{1}{x^2}\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)
Tương tự: \(y^2+\dfrac{1}{y^2}\ge2\)
\(z^2+\dfrac{1}{z^2}\ge2\)
Cộng vế theo vế 3 BĐT cùng chiều trên ta được:
\(x^2+y^2+z^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\ge6\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x^2=\dfrac{1}{x^2}\\y^2=\dfrac{1}{y^2}\\z^2=\dfrac{1}{z^2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\) ( Vì x,y,z nguyên dương )
Vậy các số x,y,z thỏa mãn đề bài là (x;y;z)= ( 1;1;1)

Cách khác: Không sử dụng BĐT Cauchy
Pt \(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}\right)+\left(y^2+\dfrac{1}{y^2}\right)+\left(z^2+\dfrac{1}{z^2}\right)=6\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+2+\left(y-\dfrac{1}{y}\right)^2+2+\left(z-\dfrac{1}{z}\right)^2+2=6\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+\left(z-\dfrac{1}{z}\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\\z-\dfrac{1}{z}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\y=\dfrac{1}{y}\\z=\dfrac{1}{z}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\\z=1\end{matrix}\right.\)( Vì x,y,z nguyên dương )

Chậc đây á hả ~~

a,\(\dfrac{x}{3}-\dfrac{1}{y}=\dfrac{1}{2}\)

=> \(\dfrac{1}{y}=\dfrac{x}{3}-\dfrac{1}{2}=>\dfrac{1}{y}=\dfrac{2x-3}{6}\)

=> y(2x-3)=6.1=6

=> y và 2x-3 là Ư (6)= {+-1,+-2,+-3,+-6}

vậy (x;y)= .......................

b,c làm tương tự

chúc bn học tốt