Toán Lớp 8: Cho biểu thức A = $\frac{3x+15}{x²+10x+25}$ : ($\frac{x}{x+3}$ – $\frac{2x}{3-x}$ -$\frac{3x²+9}{x²-9}$)
a, Tìm đkxđ và rút gọn biểu thức A
b, Tìm x để A < 1
, Tìm x để A =
$\frac{2x-3}{x+1}$
giúp mình với mình cảm ơn ạ.
Leave a reply
Comments ( 1 )
Giải đáp:
\(\begin{array}{l}
a)\dfrac{{x + 3}}{{x + 5}}\\
b)x > – 5\\
c)x = – 6
\end{array}\)
Lời giải và giải thích chi tiết:
\(\begin{array}{l}
a)DK:x \ne \left\{ { – 5; – 3;3} \right\}\\
A = \dfrac{{3x + 15}}{{{x^2} + 10x + 25}}:\left( {\dfrac{x}{{x + 3}} – \dfrac{{2x}}{{3 – x}} – \dfrac{{3{x^2} + 9}}{{{x^2} – 9}}} \right)\\
= \dfrac{{3\left( {x + 5} \right)}}{{{{\left( {x + 5} \right)}^2}}}:\dfrac{{x\left( {x – 3} \right) + 2x\left( {x + 3} \right) – 3{x^2} – 9}}{{\left( {x – 3} \right)\left( {x + 3} \right)}}\\
= \dfrac{{3\left( {x + 5} \right)}}{{{{\left( {x + 5} \right)}^2}}}.\dfrac{{\left( {x – 3} \right)\left( {x + 3} \right)}}{{{x^2} – 3x + 2{x^2} + 6x – 3{x^2} – 9}}\\
= \dfrac{{3\left( {x + 5} \right)}}{{{{\left( {x + 5} \right)}^2}}}.\dfrac{{\left( {x – 3} \right)\left( {x + 3} \right)}}{{3x – 9}}\\
= \dfrac{{3\left( {x + 5} \right)}}{{{{\left( {x + 5} \right)}^2}}}.\dfrac{{\left( {x – 3} \right)\left( {x + 3} \right)}}{{3\left( {x – 3} \right)}}\\
= \dfrac{{x + 3}}{{x + 5}}\\
b)A < 1\\
\to \dfrac{{x + 3}}{{x + 5}} < 1\\
\to \dfrac{{x + 3 – x – 5}}{{x + 5}} < 0\\
\to \dfrac{{ – 2}}{{x + 5}} < 0\\
\to x + 5 > 0\\
\to x > – 5\\
c)A = \dfrac{{2x – 3}}{{x + 1}}\\
\to \dfrac{{x + 3}}{{x + 5}} = \dfrac{{2x – 3}}{{x + 1}}\\
\to \left( {x + 3} \right)\left( {x + 1} \right) = \left( {2x – 3} \right)\left( {x + 5} \right)\left( {DK:x \ne – 1} \right)\\
\to {x^2} + 4x + 3 = 2{x^2} + 7x – 15\\
\to {x^2} + 3x – 18 = 0\\
\to \left( {x + 6} \right)\left( {x – 3} \right) = 0\\
\to \left[ \begin{array}{l}
x = – 6\\
x = 3\left( l \right)
\end{array} \right.
\end{array}\)