a) $\frac{6}{x^2-1}$ +5 =$\frac{8x-1}{4x+4}$ -$\frac{12x-1}{4-4x}$
b) $\frac{x+4}{x^2-3x+2}$ – $\frac{x+1}{x^2-4x+3}$ = $\frac{2x+5}{x^2-4x+3}$
a) $\frac{6}{x^2-1}$ +5 =$\frac{8x-1}{4x+4}$ -$\frac{12x-1}{4-4x}$
b) $\frac{x+4}{x^2-3x+2}$ – $\frac{x+1}{x^2-4x+3}$ = $\frac{2x+5}{x^2-4x+3}$
Đáp án:
a=2
Giải thích các bước giải:
( trog hình^^) vote ctlhn nhé
Đáp án:
a. x=2
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \pm 1\\
\frac{{6 + 5\left( {{x^2} – 1} \right)}}{{\left( {x – 1} \right)\left( {x + 1} \right)}} = \frac{{\left( {8x – 1} \right)\left( {x – 1} \right) + \left( {12x – 1} \right)\left( {x + 1} \right)}}{{4\left( {x – 1} \right)\left( {x + 1} \right)}}\\
\to 24 + 20{x^2} – 20 = 8{x^2} – 9x + 1 + 12{x^2} + 11x – 1\\
\to 2x = 4\\
\to x = 2\left( {TM} \right)
\end{array}\)
\(\begin{array}{l}
b.DK:x \ne \left\{ {1;2;3} \right\}\\
\frac{{x + 4}}{{\left( {x – 2} \right)\left( {x – 1} \right)}} – \frac{{x + 1}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} – \frac{{2x + 5}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to \frac{{x + 4}}{{\left( {x – 2} \right)\left( {x – 1} \right)}} – \frac{{x + 1 – 2x – 5}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to \frac{{x + 4}}{{\left( {x – 2} \right)\left( {x – 1} \right)}} – \frac{{ – x – 4}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to \frac{{x + 4}}{{\left( {x – 2} \right)\left( {x – 1} \right)}} + \frac{{x + 4}}{{\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to \frac{{\left( {x + 4} \right)\left( {x – 3} \right) + \left( {x + 4} \right)\left( {x – 2} \right)}}{{\left( {x – 2} \right)\left( {x – 1} \right)\left( {x – 3} \right)}} = 0\\
\to {x^2} + x – 12 + {x^2} + 2x – 8 = 0\\
\to 2{x^2} + 3x – 20 = 0\\
\to 2{x^2} + 8x – 5x – 20 = 0\\
\to 2x\left( {x + 4} \right) – 5\left( {x + 4} \right) = 0\\
\to \left( {x + 4} \right)\left( {2x – 5} \right) = 0\\
\to \left[ \begin{array}{l}
x = – 4\\
x = \frac{5}{3}
\end{array} \right.
\end{array}\)