Gọi \(\dfrac{1}{a}=x;\dfrac{1}{b}=y\)
Hpt trở thành \(\left\{{}\begin{matrix}x+y=\dfrac{1}{72}\\20x+45y=\dfrac{5}{12}\end{matrix}\right.\)
giải ra =>\(x=\dfrac{1}{120};y=\dfrac{1}{180}\)
Vậy a=120; b=180.
\(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{72}\\\dfrac{20}{a}+\dfrac{45}{b}=\dfrac{5}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{72}\\20\cdot\dfrac{1}{a}+45\cdot\dfrac{1}{b}=\dfrac{5}{12}\end{matrix}\right.\)
đặt : \(\dfrac{1}{a}=x;\dfrac{1}{b}=y\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=\dfrac{1}{72}\left(1\right)\Rightarrow y=\dfrac{1}{72}-x\left(3\right)\\20x+45y=\dfrac{5}{12}\left(2\right)\end{matrix}\right.\)
Thay (3) vào (2)\(\Rightarrow20x+45\cdot\left(\dfrac{1}{72}-x\right)=\dfrac{5}{12}\)
\(\Leftrightarrow20x+\dfrac{5}{8}-45x=\dfrac{5}{12}\)
\(\Leftrightarrow-25x=\dfrac{-5}{24}\Leftrightarrow x=\dfrac{1}{120}\)
Thay \(x=\dfrac{1}{120}vào\left(3\right)\)\(\Rightarrow y=\dfrac{1}{72}-\dfrac{1}{120}=\dfrac{1}{180}\)
Vs \(x=\dfrac{1}{120}\Rightarrow\dfrac{1}{a}=\dfrac{1}{120}\Rightarrow a=120\)
Vs \(y=\dfrac{1}{180}\Rightarrow\dfrac{1}{b}=\dfrac{1}{180}\Rightarrow b=180\)
Nghiệm của hệ (120;180)
CHÚC BẠN HỌC TỐT
\(\left\{{}\begin{matrix}\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{72}\\\dfrac{20}{a}+\dfrac{45}{b}=\dfrac{5}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{20}{a}+\dfrac{20}{b}=\dfrac{20}{72}\\\dfrac{20}{a}+\dfrac{45}{b}=\dfrac{5}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-25}{b}=\dfrac{-5}{36}\\\dfrac{20}{a}+\dfrac{45}{b}=\dfrac{5}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=180\\\dfrac{20}{a}+\dfrac{45}{180}=\dfrac{5}{12}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=180\\\dfrac{20}{a}=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=180\\a=120\end{matrix}\right.\)
