Giải phương trình chứa ẩn ở mẫu
a) 1 + $\frac{2x-5}{x-2}$ – $\frac{3x-5}{x-1}$ = 0
b) $\frac{x-3}{x-2}$ – $\frac{x-2}{x-4}$ = -1
c) $\frac{3x+2}{x-7}$ = $\frac{6x+1}{2x-3}$
d) $\frac{x+1}{x-2}$ – $\frac{x-1}{x+2}$ – $\frac{2(x ²+2}{x ² -4}$
Đáp án:
c) \(x = \dfrac{1}{4}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne \left\{ {1;2} \right\}\\
1 + \dfrac{{2x – 5}}{{x – 2}} – \dfrac{{3x – 5}}{{x – 1}} = 0\\
\to \dfrac{{\left( {x – 1} \right)\left( {x – 2} \right) + \left( {2x – 5} \right)\left( {x – 1} \right) – \left( {3x – 5} \right)\left( {x – 2} \right)}}{{\left( {x – 1} \right)\left( {x – 2} \right)}} = 0\\
\to {x^2} – 3x + 2 + 2{x^2} – 7x + 5 – 3{x^2} + 11x – 10 = 0\\
\to x = 3\\
b)DK:x \ne \left\{ {2;4} \right\}\\
\dfrac{{\left( {x – 3} \right)\left( {x – 4} \right) – {{\left( {x – 2} \right)}^2} + \left( {x – 2} \right)\left( {x – 4} \right)}}{{\left( {x – 2} \right)\left( {x – 4} \right)}} = 0\\
\to {x^2} – 7x + 12 – {x^2} + 4x – 4 + {x^2} – 6x + 8 = 0\\
\to {x^2} – 9x + 16 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{9 + \sqrt {17} }}{2}\\
x = \dfrac{{9 – \sqrt {17} }}{2}
\end{array} \right.\\
c)DK:x \ne \left\{ {\dfrac{2}{3};7} \right\}\\
\dfrac{{3x + 2}}{{x – 7}} = \dfrac{{6x + 1}}{{2x – 3}}\\
\to \dfrac{{\left( {3x + 2} \right)\left( {2x – 3} \right) – \left( {6x + 1} \right)\left( {x – 7} \right)}}{{\left( {x – 7} \right)\left( {2x – 3} \right)}} = 0\\
\to 6{x^2} – 5x – 6 – 6{x^2} + x + 7 = 0\\
\to – 4x + 1 = 0\\
\to x = \dfrac{1}{4}\\
d)DK:x \ne \pm 2\\
\dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) – \left( {x – 1} \right)\left( {x – 2} \right) – 2{x^2} – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{{x^2} + 3x + 2 – {x^2} + 3x – 2 – 2{x^2} – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
= \dfrac{{ – 2{x^2} + 6x – 4}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}
\end{array}\)