Câu \(1\)
- Theo bài ta có : \(\dfrac{G}{A}=\dfrac{2}{3}\left(1\right)\)
- Mà \(A=A_1+A_2\) và \(G_1=\dfrac{1}{2}T_2\) hơn hết \(G_2=X_2=A_1\)
\(\rightarrow G_1+G_2=\dfrac{1}{2}T_2+A_1=1,5A_1=G\)
- Do đó ta thay vào \((1)\) được : \(\dfrac{1,5A_1}{A_1+A_2}=\dfrac{2}{3}\rightarrow2A_1+2A_2=4,5A_1\)
\(\rightarrow2A_2=2,5A_1\) mà \(A_2=T_1=500\left(nu\right)\)
\(\rightarrow A_1=T_2=400\left(nu\right)\)
\(\rightarrow\left\{{}\begin{matrix}A=T=A_1+A_2=900\left(nu\right)\\G=X=\dfrac{2.A}{3}=600\left(nu\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}G_1=X_2=\dfrac{1}{2}T_2=200\left(nu\right)\\G_2=X_1=500\left(nu\right)\end{matrix}\right.\)
\(\rightarrow L=3,4.\dfrac{2A+2G}{2}=5100\left(\overset{o}{A}\right)\)
