Đáp án:
Câu 13:
$T=11$
Câu 14:
$T=4$
Giải thích các bước giải:
Câu 13:
$\begin{cases}\dfrac{6}{x}+\dfrac{5}{y}=3\\\dfrac{9}{x}-\dfrac{10}{y}=1\end{cases}\,\,(x\ne 0,\,y\ne 0)$
Đặt $a=\dfrac{1}{x},\,b=\dfrac{1}{y}$
$\to \begin{cases}6a+5b=3\\9a-10b=1\end{cases}$
$\to \begin{cases}12a+10b=6\\9a-10b=1\end{cases}$
$\to \begin{cases}21a=7\\9a-10b=1\end{cases}$
$\to \begin{cases}a=\dfrac{1}{3}\\9.\dfrac{1}{3}-10b=1\end{cases}$
$\to \begin{cases}a=\dfrac{1}{3}\\b=\dfrac{1}{5}\end{cases}$
$\to \begin{cases}\dfrac{1}{x}=\dfrac{1}{3}\\\dfrac{1}{y}=\dfrac{1}{5}\end{cases}$
$\to \begin{cases}x=3\\y=5\end{cases}$ (thỏa mãn)
$\to (x_0,\,y_0)=(3,\,5)$
$\to T=2x_0+y_0=2.3+5=11$
Câu 14:
$\begin{cases}x+y+xy=\dfrac{7}{2}\\x^2y+xy^2=\dfrac{5}{2}\end{cases}$
$\to \begin{cases}x+y=\dfrac{7}{2}-xy\\xy(x+y)=\dfrac{5}{2}\end{cases}$
$\to \begin{cases}x+y=\dfrac{7}{2}-xy\\xy\Bigg(\dfrac{7}{2}-xy\Bigg)=\dfrac{5}{2}\end{cases}$
$\to \begin{cases}x+y=\dfrac{7}{2}-xy\\(xy)^2-\dfrac{7}{2}xy+\dfrac{5}{2}=0\,\,(1)\end{cases}$
$(1)↔(xy)^2-\dfrac{7}{2}xy+\dfrac{5}{2}=0$
Đặt $A=xy$
$\to A^2-\dfrac{7}{2}A+\dfrac{5}{2}=0$
$\to A^2-A-\dfrac{5}{2}A+\dfrac{5}{2}=0$
$\to A(A-1)-\dfrac{5}{2}(A-1)=0$
$\to (A-1)\Bigg(A-\dfrac{5}{2}\Bigg)=0$
$\to \left[\begin{array}{l}A=1\\A=\dfrac{5}{2}\end{array}\right.$
$\to \left[\begin{array}{l}xy=1\\xy=\dfrac{5}{2}\end{array}\right.$
$\to \left[\begin{array}{l}\begin{cases}x+y=\dfrac{5}{2}\\xy=1\end{cases}\\\begin{cases}x+y=1\\xy=\dfrac{5}{2}\end{cases}\end{array}\right.$
$\to \left[\begin{array}{l}\begin{cases}y=\dfrac{5}{2}-x\\\Bigg(\dfrac{5}{2}-x\Bigg)x=1\end{cases}\\\begin{cases}y=1-x\\(1-x)x=\dfrac{5}{2}\end{cases}\end{array}\right.$
$\to \left[\begin{array}{l}\begin{cases}y=\dfrac{5}{2}-x\\x^2-\dfrac{5}{2}x+1=0\end{cases}\\\begin{cases}y=1-x\\x^2-x+\dfrac{5}{2}=0 \,\,(\text{vô nghiệm})\end{cases}\end{array}\right.$
$\to \begin{cases}y=\dfrac{5}{2}-x\\\left[\begin{array}{l}x=2\\x=\dfrac{1}{2}\end{array}\right.\end{cases}$
$\to \left[\begin{array}{l}\begin{cases}x=2\\y=\dfrac{1}{2}\end{cases}\\\begin{cases}x=\dfrac{1}{2}\\y=2\end{cases}\end{array}\right.$
$\to \begin{cases}(x_1;\,y_1)=\Bigg(2;\,\dfrac{1}{2}\Bigg)\\(x_2;\,y_2)=\Bigg(\dfrac{1}{2};\,2\Bigg)\end{cases}$
$\to T=x_1+y_1-x_2+y_2=2+\dfrac{1}{2}-\dfrac{1}{2}+2=4$
