không biết

Ss

Ta xét:

\(\left(x-5\right)\left(y-5\right)=25\Leftrightarrow x-5=y-5=5\Leftrightarrow x=y=10\)

\(x-5=y-5=-5\Leftrightarrow x=y=0\)

\(\left\{{}\begin{matrix}x-5=1\\y-5=25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-5=-1\\y-5=-25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=4\\y=-20\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-5=25\\y-5=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=30\\y=6\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-5=-25\\y-5=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-20\\y=4\end{matrix}\right.\)

Vậy \(\left(x,y\right)\in\left\{\left(10;10\right),\left(0;0\right),\left(6;30\right),\left(4;-20\right),\left(30;6\right),\left(-20;4\right)\right\}\)

Xd

Là 8 bn ạ 

Hai phân số đó có dạng: 

$\frac{1}{a};\frac{1}{b}$(a, b là số tự nhiên, a,b khác 0)

Ta có: $\frac{1}{a}+\frac{1}{b}+\frac{1}{ab}=\frac{1}{2}\iff \frac{a+b+1}{ab}=\frac{1}{2}\iff 2a+2b+2=ab$

<=> $2a+2b+2-ab=0\iff a\left(2-b\right)-2\left(2-b\right)+6=0\iff \left(a-2\right)\left(2-b\right)=-6$

Em lập bảng làm tiếp nhé nhớ đk a, b để loại TH

a)

b) \(\dfrac{x^2}{6}=\dfrac{24}{25}\)

\(\Leftrightarrow\left(5x\right)^2=144\)

\(\Leftrightarrow\left(5x\right)^2=12^2\)

\(\Leftrightarrow\left[{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)

c) \(\dfrac{x-2}{x-1}=\dfrac{x+4}{x+7}\)

\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x-1\right)\left(x+4\right)\)

\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)