Giải pt sau:
a) (x + 3)$^{2}$ – (x – 3)$^{2}$ = 6(x + 3)
b) $\frac{3}{2x+10}$ + $\frac{2x}{25-x^{2}}$ + $\frac{3}{x-5}$ = 0
c) $\frac{12x^2+30-21}{16x^2}$ – $\frac{3x-7}{3-4x}$ = $\frac{6x+5}{4x+3}$
d) $\frac{x+5}{x-1}$ – $\frac{x+1}{x-3}$ = $\frac{8}{x^2-4x+3}$
Đáp án:
d) x=11
Giải thích các bước giải:
\(\begin{array}{l}
a){x^2} + 6x + 9 – {x^2} + 6x – 9 = 6x + 18\\
\to 6x = 18\\
\to x = 3\\
b)DK:x \ne \pm 5\\
\dfrac{3}{{2\left( {x + 5} \right)}} – \dfrac{{2x}}{{\left( {x – 5} \right)\left( {x + 5} \right)}} + \dfrac{3}{{x – 5}} = 0\\
\to \dfrac{{3\left( {x – 5} \right) – 4x + 3\left( {x + 5} \right)}}{{2\left( {x – 5} \right)\left( {x + 5} \right)}} = 0\\
\to 3x – 15 – 4x + 3x + 15 = 0\\
\to 2x = 0 \to x = 0\\
d)DK:x \ne \left\{ {1;3} \right\}\\
\dfrac{{x + 5}}{{x – 1}} – \dfrac{{x + 1}}{{x – 3}} = \dfrac{8}{{{x^2} – 4x + 3}}\\
\to \dfrac{{\left( {x + 5} \right)\left( {x – 3} \right) – \left( {x + 1} \right)\left( {x – 1} \right) – 8}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to \dfrac{{{x^2} + 2x – 15 – {x^2} + 1 – 8}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to 2x – 22 = 0\\
\to x = 11
\end{array}\)
( bạn xem lại đề câu c nha )