1,Q= 2x/ x+3 Tính Q khi x= -1/3 2,P= x+1/x-3+ 11x-3/x²-9 M= Q+P Rút gọn M 3, Tìm x thuộc Z để M thuộc Z

Đáp án:

 2) \(\dfrac{{3x}}{{x – 3}}\)

Giải thích các bước giải:

\(\begin{array}{l}
1)Thay:x =  – \dfrac{1}{3}\\
 \to Q = \dfrac{{2.\left( { – \dfrac{1}{3}} \right)}}{{ – \dfrac{1}{3} + 3}} =  – \dfrac{1}{4}\\
2)P = \dfrac{{x + 1}}{{x – 3}} + \dfrac{{11x – 3}}{{{x^2} – 9}} = \dfrac{{\left( {x + 1} \right)\left( {x + 3} \right) + 11x – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{{x^2} + 4x + 3 + 11x – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{{x^2} + 15x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
M = Q + P = \dfrac{{{x^2} + 15x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \dfrac{{2x}}{{x + 3}}\\
 = \dfrac{{{x^2} + 15x + 2x\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} = \dfrac{{{x^2} + 15x + 2{x^2} – 6x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
 = \dfrac{{3{x^2} + 9x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} = \dfrac{{3x}}{{x – 3}}\\
3)M = \dfrac{{3x}}{{x – 3}} = \dfrac{{3\left( {x – 3} \right) + 9}}{{x – 3}} = 3 + \dfrac{9}{{x – 3}}\\
M \in Z\\
 \Leftrightarrow \dfrac{9}{{x – 3}} \in Z\\
 \Leftrightarrow x – 3 \in U\left( 9 \right)\\
 \to \left[ \begin{array}{l}
x – 3 = 9\\
x – 3 =  – 9\\
x – 3 = 3\\
x – 3 =  – 3\\
x – 3 = 1\\
x – 3 =  – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x =  – 6\\
x = 6\\
x = 0\\
x = 4\\
x = 2
\end{array} \right.
\end{array}\)