Question

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Answer 1

Answer 2

Answer 3

Answer 4

Câu 1 : B

Câu 2 : C

Câu 3 : C

Câu 4 : D

Câu 5 : A

Câu 6 : D

Answer 5

Câu 9 :

Ta có : \(\left\{{}\begin{matrix}\dfrac{2x+1}{x-2}-\dfrac{3y+4}{y+3}=9\\\dfrac{3x-4}{x-2}+\dfrac{y-1}{y+3}=2\end{matrix}\right.\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\y\ne-3\end{matrix}\right.\)

HPT \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2x-4+5}{x-2}-\dfrac{3y+9-5}{y+3}=9\\\dfrac{3x-6+2}{x-2}+\dfrac{y+3-4}{y+3}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-2}+2-3+\dfrac{5}{y+3}=9\\\dfrac{2}{x-2}+3+1+\dfrac{-4}{y+3}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-2}+\dfrac{5}{y+3}=10\\\dfrac{2}{x-2}+\dfrac{-4}{y+3}=-2\end{matrix}\right.\)

- Đặt \(\dfrac{1}{x-2}=a;\dfrac{1}{y+3}=b\) ta được : \(\left\{{}\begin{matrix}5a+5b=10\\2a-4b=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

- Thay lại ta được : \(\left\{{}\begin{matrix}x-2=1\\y+3=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-2\end{matrix}\right.\)

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