bt m=3x+21/x^2-9+2/x+3-3/x-3 a,tìm x để bt xácho bt m=3x+21/x^2-9+2/x+3-3/x-3c định b,rút gọn m c,tính gtri khi x=1/2
bt m=3x+21/x^2-9+2/x+3-3/x-3 a,tìm x để bt xácho bt m=3x+21/x^2-9+2/x+3-3/x-3c định b,rút gọn m c,tính gtri khi x=1/2
Giải thích các bước giải:
a,
ĐKXĐ: \(\begin{array}{l}
\\
\left\{ \begin{array}{l}
{x^2} – 9 \ne 0\\
x + 3 \ne 0\\
x – 3 \ne 0
\end{array} \right. \Leftrightarrow x \ne \pm 3
\end{array}\)
b,
Ta có:
\(\begin{array}{l}
M = \frac{{3x + 21}}{{{x^2} – 9}} + \frac{2}{{x + 3}} – \frac{3}{{x – 3}}\\
= \frac{{3x + 21}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} + \frac{{2\left( {x – 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} – \frac{{3\left( {x + 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{3x + 21 + 2\left( {x – 3} \right) – 3\left( {x + 3} \right)}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{3x + 21 + 2x – 6 – 3x – 9}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{2x + 6}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{2}{{x – 3}}
\end{array}\)
c,
Thay \(x = \frac{1}{2}\) vào A ta được:
\[A = \frac{2}{{\frac{1}{2} – 3}} = \frac{2}{{ – \frac{5}{2}}} = – \frac{4}{5}\]