C1: Giải hệ phương trình
a) $\frac{1}{x}$ +$\frac{1}{y}$ =$\frac{5}{24}$
$\frac{3}{x}$ +$\frac{4}{y}$ =$\frac{3}{4}$
b) $\frac{1}{x}$ +$\frac{1}{y}$ =$\frac{5}{36}$
$\frac{6}{x}$ +$\frac{3}{y}$ =$\frac{2}{3}$
C1: Giải hệ phương trình
a) $\frac{1}{x}$ +$\frac{1}{y}$ =$\frac{5}{24}$
$\frac{3}{x}$ +$\frac{4}{y}$ =$\frac{3}{4}$
b) $\frac{1}{x}$ +$\frac{1}{y}$ =$\frac{5}{36}$
$\frac{6}{x}$ +$\frac{3}{y}$ =$\frac{2}{3}$
Đáp án:
$a,\\-\text{Đặt} \dfrac{1}{x}=a;\dfrac{1}{y}=b\\-\text{Ta có}:\\\left \{\matrix {{a+b=\dfrac{5}{24}} \hfill\cr {3a+4b=\dfrac{3}{4}}} \right.\\⇔\left \{\matrix {{3a+3b=\dfrac{5}{8}} \hfill\cr {3a+4b=\dfrac{3}{4}}} \right.\\⇔\left \{\matrix {{b=\dfrac{1}{8}} \hfill\cr {a+\dfrac{1}{8}=\dfrac{5}{24}}} \right.\\⇔\left \{\matrix {{b=\dfrac{1}{8}} \hfill\cr {a=\dfrac{5}{24}-\dfrac{1}{8}}} \right.\\⇔ \left \{\matrix {{b=\dfrac{1}{8}} \hfill\cr {a=\dfrac{1}{12}}} \right.\\-Ta có:\\\dfrac{1}{x}=a⇔\dfrac{1}{x}=\dfrac{1}{12}⇒x=12\\\dfrac{1}{y}=b⇔\dfrac{1}{y}=\dfrac{1}{8}⇒y=8\\\text{Vậy HPT có 1 nghiệm duy nhất là} (x;y)=(12;8)\\b,-\text{Đặt} \dfrac{1}{x}=a;\dfrac{1}{y}=b\\-Ta có:\\\left \{\matrix {{a+b=\dfrac{5}{36}} \hfill\cr {6a+3b=\dfrac{2}{3}}} \right.\\⇔\left \{\matrix {{3a+3b=\dfrac{5}{12}} \hfill\cr {6a+3b=\dfrac{2}{3}}} \right.\\⇔\left \{\matrix {{3a=\dfrac{1}{4}} \hfill\cr {a+b=\dfrac{5}{36}}} \right.\\⇔\left \{\matrix {{a=\dfrac{1}{12}}\hfill\cr {\dfrac{1}{12}+b=\dfrac{5}{36}}} \right.\\⇔ \left \{\matrix {{a=\dfrac{1}{12}} \hfill\cr {b=\dfrac{5}{36}-\dfrac{1}{12}}} \right.\\⇔\left \{\matrix {{a=\dfrac{1}{12}} \hfill\cr {b=\dfrac{1}{18}}} \right.\\ -Ta có:\\\dfrac{1}{x}=a⇔\dfrac{1}{x}=\dfrac{1}{12}⇒x=12\\\dfrac{1}{y}=b⇔\dfrac{1}{y}=\dfrac{1}{17}⇒y=18\\\text{Vậy HPT có 1 nghiệm duy nhất là} (x;y)=(12;18)$
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