cho bieu thuc M=(2x/x+3 + x/x-3 – 3xbinh phuong + 3/x binh phog -9
a,rut gon bieu thuc M
b,tinh gia tri cua bieu thuc M tai|x-1|=3
c,tim x de M <0
cho bieu thuc M=(2x/x+3 + x/x-3 – 3xbinh phuong + 3/x binh phog -9
a,rut gon bieu thuc M
b,tinh gia tri cua bieu thuc M tai|x-1|=3
c,tim x de M <0
Đáp án:
$\begin{array}{l}
a)M = \left( {\frac{{2x}}{{x + 3}} + \frac{x}{{x – 3}} – \frac{{3{x^2} + 3}}{{{x^2} – 9}}} \right)\left( {x \ne 3;x \ne – 3} \right)\\
M = \frac{{2x}}{{x + 3}} + \frac{x}{{x – 3}} – \frac{{3{x^2} + 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{2x\left( {x – 3} \right) + x\left( {x + 3} \right) – 3{x^2} – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{2{x^2} – 6x + {x^2} + 3x – 3{x^2} – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \frac{{ – 3x – 3}}{{\left( {x – 3} \right)\left( {x + 3} \right)}} = \frac{{ – 3\left( {x + 1} \right)}}{{{x^2} – 9}}\\
b)\left| {x – 1} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
x – 1 = 3\\
x – 1 = – 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 4\\
x = – 2
\end{array} \right.\\
+ Khi\,x = 4 \Rightarrow M = \frac{{ – 3.5}}{{{4^2} – 9}} = – \frac{{15}}{7}\\
+ Khi\,x = – 2 \Rightarrow M = \frac{{ – 3.\left( { – 1} \right)}}{{4 – 9}} = \frac{3}{5}\\
c)M < 0\\
\Rightarrow \frac{{ – 3\left( {x + 1} \right)}}{{{x^2} – 9}} < 0\\
\Rightarrow \frac{{x + 1}}{{\left( {x + 3} \right)\left( {x – 3} \right)}} > 0\\
\Rightarrow \left[ \begin{array}{l}
– 3 < x < – 1\\
x > 3
\end{array} \right.
\end{array}$