Question

Giải hệ pt:

\(\left\{{}\begin{matrix}\dfrac{60}{x_1+x_2}=\dfrac{4}{3}\\\dfrac{60}{x_2}_{_{ }}-\dfrac{60}{x_1}=2\end{matrix}\right.\)

Answer 1

Lời giải:

PT (1)\(\rightarrow x_1+x_2=\frac{60.3}{4}=45\)

\(\Rightarrow x_2=45-x_1\)

Thay vào pt (2)

\(\frac{60}{x_2}-\frac{60}{x_1}=2\)

\(\Leftrightarrow \frac{60}{45-x_1}-\frac{60}{x_1}=2\)

\(\Leftrightarrow \frac{1}{45-x_1}-\frac{1}{x_1}=\frac{1}{30}\Leftrightarrow \frac{x_1-(45-x_1)}{x_1(45-x_1)}=\frac{1}{30}\)

\(\Leftrightarrow 30(2x_1-45)=x_1(45-x_1)\)

\(\Leftrightarrow x_1^2+15x_1-1350=0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=30\rightarrow x_2=15\\x_1=-45\rightarrow x_2=90\end{matrix}\right.\)

(đều thỏa mãn)

Vậy \((x_1,x_2)=(30;15);(-45;90)\)