B1 a, Rút gọn : M=x+2/x+3 -5/x^2 +x -6 + 1/ 2-x
b,Tìm x để M có gtri nguyên
B2 Rút gọn bt: (1-2xy). x+2/2xy-1 + x^2- 3/x+2
B1 a, Rút gọn : M=x+2/x+3 -5/x^2 +x -6 + 1/ 2-x
b,Tìm x để M có gtri nguyên
B2 Rút gọn bt: (1-2xy). x+2/2xy-1 + x^2- 3/x+2
Đáp án:
B1:
a) \(\dfrac{{x – 4}}{{x – 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)DK:x \ne \left\{ { – 3;2} \right\}\\
M = \dfrac{{x + 2}}{{x + 3}} – \dfrac{5}{{{x^2} + x – 6}} + \dfrac{1}{{2 – x}}\\
= \dfrac{{\left( {x + 2} \right)\left( {x – 2} \right) – 5 – x – 3}}{{\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} – 4 – x – 8}}{{\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{{x^2} – x – 12}}{{\left( {x – 2} \right)\left( {x + 3} \right)}} = \dfrac{{\left( {x – 4} \right)\left( {x + 3} \right)}}{{\left( {x – 2} \right)\left( {x + 3} \right)}}\\
= \dfrac{{x – 4}}{{x – 2}}\\
b)M = \dfrac{{x – 2 – 2}}{{x – 2}} = 1 – \dfrac{2}{{x – 2}}\\
M \in Z \to \dfrac{2}{{x – 2}} \in Z\\
\to x – 2 \in U\left( 2 \right)\\
\to \left[ \begin{array}{l}
x – 2 = 2\\
x – 2 = – 2\\
x – 2 = 1\\
x – 2 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 0\\
x = 3\\
x = 1
\end{array} \right.\\
B2:\\
\left( {1 – 2xy} \right).\dfrac{{x + 2}}{{2xy – 1}} + \dfrac{{{x^2} – 3}}{{x + 2}}\\
= – x – 2 + \dfrac{{{x^2} – 3}}{{x + 2}}\\
= \dfrac{{ – {x^2} – 2x – 2x – 4 + {x^2} – 3}}{{x + 2}}\\
= \dfrac{{ – 4x – 7}}{{x + 2}}
\end{array}\)
Mình chỉ làm dc b1 thôi sr nhiều
Chúc bạn học tốt nha~
~Cheese~