10 câu Trắc nghiệm Toán 9 Bài 3: Giải hệ phương trình bằng phương pháp thế có đáp án (Thông hiểu)

Đáp án B

Ta có

$$\begin{gathered} \left\{\begin{array}{l}x-y=5\\ 3x+2y=18\end{array}\right.\iff \left\{\begin{array}{l}x=y+5\\ 3.\left(y+5\right)+2y=18\end{array}\right. \\ \iff \left\{\begin{array}{l}x=y+5\\ 3y+15+2y=18\end{array}\right.\iff \left\{\begin{array}{l}x=y+5\\ 5y=3\end{array}\right. \\ \iff \left\{\begin{array}{l}y=\frac{3}{5}\\ x=5+\frac{3}{5}\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{28}{5}\\ y=\frac{3}{5}\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có nghiệm duy nhất $\left(x{}_0;{\text{y}}_0\right)=\left(\frac{28}{5};\frac{3}{5}\right)\Rightarrow x_0.{\text{y}}_0\text{=}\frac{84}{25}$

Đáp án D

Ta có

$$\begin{gathered} \left\{\begin{array}{l}x-y=3\\ 3x-4y=2\end{array}\right.\iff \left\{\begin{array}{l}x=y+3\\ 3\left(y+3\right)-4y=2\end{array}\right. \\ \iff \left\{\begin{array}{l}x=y+3\\ y=7\end{array}\right.\iff \left\{\begin{array}{l}x=10\\ y=7\end{array}\right. \end{gathered}$$

Vậy hệ  phương trình có nghiệm duy nhất (x; y) = (10; 7)

Do đó: $x^{2}y={10}^2.7=700$

Đáp án D

Ta có

$$\begin{gathered} \left\{\begin{array}{l}2x-7y=8\\ 1-x+3y=21\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{8+7y}{2}\\ 10.\left(\frac{8+7y}{2}\right)+3y=21\end{array}\right. \\ \iff \left\{\begin{array}{l}x=\frac{8+7y}{2}\\ 40+35y+3y=21\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{{\displaystyle 8+7y}}{{\displaystyle 2}}\\ 38y=-19\end{array}\right. \\ \iff \left\{\begin{array}{l}x=\frac{{\displaystyle 8+7y}}{{\displaystyle 2}}\\ y=-\frac{{\displaystyle 1}}{{\displaystyle 2}}\end{array}\right.\iff \left\{\begin{array}{l}y=-\frac{{\displaystyle 1}}{{\displaystyle 2}}\\ x=\frac{{\displaystyle 9}}{{\displaystyle 4}}\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có nghiệm duy nhất $(x; y)=\left(\frac{9}{4};-\frac{1}{2}\right)\Rightarrow x+y=\frac{7}{4}$

Đáp án C

Ta có

$$\begin{gathered} \left\{\begin{array}{l}7x-3y=5\\ 4x+y=2\end{array}\right.\iff \left\{\begin{array}{l}7x-3\left(2-4x\right)=5\\ y=2-4x\end{array}\right. \\ \iff \left\{\begin{array}{l}x=\frac{11}{19}\\ y=2-4.\frac{11}{19}\end{array}\right.\iff \left\{\begin{array}{l}x=\frac{11}{19}\\ y=-\frac{6}{19}\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có nghiệm duy nhất $(x; y)=\left(\frac{11}{19};-\frac{6}{19}\right)\Rightarrow x+y=\frac{5}{19}$

Đáp án D

Ta có $\left\{\begin{array}{l}-x-\sqrt{2}y=\sqrt{3}\\ \sqrt{2}x+2y=-\sqrt{6}\end{array}\right.$

$$\begin{gathered} \iff \left\{\begin{array}{l}x=-\sqrt{2}y-\sqrt{3}\\ \sqrt{2}\left(-\sqrt{2}y-\sqrt{3}\right)+2y=-\sqrt{6}\end{array}\right. \\ \iff \left\{\begin{array}{l}x=-\sqrt{2}y-\sqrt{3}\\ -2y-\sqrt{6}+2y=-\sqrt{6}\end{array}\right. \\ \iff \left\{\begin{array}{l}x=-\sqrt{2}y-\sqrt{3}\\ -\sqrt{6}=-\sqrt{6}\end{array}\right.\iff \left\{\begin{array}{l}y\in ℝ\\ x=-\sqrt{2}y-\sqrt{3}\end{array}\right. \end{gathered}$$

Vậy hệ phương trình có vô số nghiệm

Đáp án A

Ta có

$$\begin{gathered} \left\{\begin{array}{l}x\sqrt{2}-y\sqrt{3}=1\\ x+y\sqrt{3}=\sqrt{2}\end{array}\right.\iff \left\{\begin{array}{l}\left(\sqrt{2}-y\sqrt{3}\right)\sqrt{2}-y\sqrt{3}=1\\ \end{array}\right. \\ \iff \left\{\begin{array}{l}2-y\left(\sqrt{6}+\sqrt{3}\right)=1\\ x=\sqrt{2}-y\sqrt{3}\end{array}\right.\iff \left\{\begin{array}{l}y\left(\sqrt{6}+\sqrt{3}\right)=1\\ x=\sqrt{2}-y\sqrt{3}\end{array}\right. \\ \iff \left\{\begin{array}{l}y=\frac{{\displaystyle \sqrt{6}-\sqrt{3}}}{{\displaystyle 3}}\\ x=\sqrt{2}-y\sqrt{3}\end{array}\right.\iff \left\{\begin{array}{l}y=\frac{{\displaystyle \sqrt{6}-\sqrt{3}}}{{\displaystyle 3}}\\ x=1\end{array}\right. \end{gathered}$$

Vậy hệ phương trình đã cho có nghiệm duy nhất $(x; y)=\left(1;\frac{\sqrt{6}-\sqrt{3}}{3}\right)$