Biết nghiệm của hệ phương trình

A

- Hướng dẫn giải

Đáp án A

Điều kiện: x $\ne$ 0; y $\ne$ 0

Ta có

$\left\{\begin{array}{l}\frac{1}{3x}+\frac{1}{3y}=\frac{1}{4}\\ \frac{5}{6x}+\frac{1}{y}=\frac{2}{3}\end{array}\right.\iff \left\{\begin{array}{l}\frac{1}{3}.\frac{1}{x}+\frac{1}{3}.\frac{1}{y}=\frac{1}{4}\\ \frac{5}{6}.\frac{1}{x}+\frac{1}{y}=\frac{2}{3}\end{array}\right.$

Đặt $\frac{1}{x}=a;\frac{1}{y}=b$ khi đó ta có hệ phương trình

$$\begin{gathered} \left\{\begin{array}{l}\frac{1}{3}.a+\frac{1}{3}.b=\frac{1}{4}\\ \frac{5}{6}.a+b=\frac{2}{3}\end{array}\right.\iff \left\{\begin{array}{l}\frac{1}{3}.a+\frac{1}{3}.b=\frac{1}{4}\\ b=\frac{2}{3}-\frac{5}{6}a\end{array}\right. \\ \iff \left\{\begin{array}{l}b=\frac{2}{3}-\frac{5}{6}a\\ \frac{1}{3}a+\frac{1}{3}b=\frac{1}{4}\end{array}\right.\iff \left\{\begin{array}{l}b=\frac{2}{3}-\frac{5}{6}a\\ \frac{1}{3}a+\frac{1}{3}\left(\frac{2}{3}-\frac{5}{6}a\right)=\frac{1}{4}\end{array}\right. \\ \iff \left\{\begin{array}{l}b=\frac{{\displaystyle 2}}{{\displaystyle 3}}-\frac{{\displaystyle 5}}{{\displaystyle 6}}a\\ \frac{{\displaystyle 1}}{{\displaystyle 3}}a+\frac{{\displaystyle 2}}{{\displaystyle 9}}-\frac{{\displaystyle 5}}{{\displaystyle 18}}a=\frac{{\displaystyle 1}}{{\displaystyle 4}}\end{array}\right.\iff \left\{\begin{array}{l}b=\frac{{\displaystyle 2}}{{\displaystyle 3}}-\frac{{\displaystyle 5}}{{\displaystyle 6}}a\\ \frac{{\displaystyle 1}}{{\displaystyle 18}}b=\frac{{\displaystyle 1}}{{\displaystyle 36}}\end{array}\right. \\ \iff \left\{\begin{array}{l}b=\frac{{\displaystyle 1}}{{\displaystyle 2}}\\ a=\frac{{\displaystyle 1}}{{\displaystyle 4}}\end{array}\right. \end{gathered}$$

Thay lại cách đặt ta được

$\left\{\begin{array}{l}\frac{1}{x}=\frac{1}{4}\\ \frac{1}{y}=\frac{1}{2}\end{array}\right.\iff \left\{\begin{array}{l}x=4\\ y=2\end{array}\right.$ (Thỏa mãn điều kiện)

Khi đó x – 3y = 4 – 3.2 = −2