Toán Lớp 8: Bài 2. Cho biểu thức A = 2x/x+3 + x+1/x-3 + 3-11x/9-x^2 và B = X-3 /x+1 với 0
Toán Lớp 8: Bài 2. Cho biểu thức A = 2x/x+3 + x+1/x-3 + 3-11x/9-x^2 và B = X-3 /x+1 với 0
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a)A = \dfrac{{3x}}{{x – 3}}\\
b)x = – \dfrac{{15}}{7}\\
c)x > – 1;x \ne 3\\
d)\left[ \begin{array}{l}
x = 2\\
x = – 4\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)
a)DK:x \ne \left\{ { – 3; – 1;3} \right\}\\
A = \dfrac{{2x}}{{x + 3}} + \dfrac{{x + 1}}{{x – 3}} + \dfrac{{3 – 11x}}{{9 – {x^2}}}\\
= \dfrac{{2x\left( {x – 3} \right) + \left( {x + 1} \right)\left( {x + 3} \right) – 3 + 11x}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{2{x^2} – 6x + {x^2} + 4x + 3 – 3 + 11x}}{{\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}}\\
b)B = \dfrac{9}{2}\\
\to \dfrac{{x – 3}}{{x + 1}} = \dfrac{9}{2}\\
\to 2x – 6 = 9x + 9\\
\to 7x = – 15\\
\to x = – \dfrac{{15}}{7}\\
c)B < 1\\
\to \dfrac{{x – 3}}{{x + 1}} < 1\\
\to \dfrac{{x – 3 – x – 1}}{{x + 1}} < 0\\
\to \dfrac{{ – 4}}{{x + 1}} < 0\\
\to x + 1 > 0\\
\to x > – 1;x \ne 3\\
d)P = A.B = \dfrac{{3x}}{{x – 3}}.\dfrac{{x – 3}}{{x + 1}}\\
= \dfrac{{3x}}{{x + 1}} = \dfrac{{3\left( {x + 1} \right) – 3}}{{x + 1}}\\
= 3 – \dfrac{3}{{x + 1}}\\
P \in Z \to \dfrac{3}{{x + 1}} \in Z\\
\to x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = – 3\\
x + 1 = 1\\
x + 1 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = – 4\\
x = 0\\
x = – 2
\end{array} \right.
\end{array}\)