giúp mình với
a) $\frac{2x}{x+3}$ – $\frac{x-1}{x-2}$ = $\frac{x-7}{x^2+x-6}$
b) $\frac{x-1}{2}$ – $\frac{2x+3}{3}$ ≤ $\frac{x}{5}$ + 4
giúp mình với
a) $\frac{2x}{x+3}$ – $\frac{x-1}{x-2}$ = $\frac{x-7}{x^2+x-6}$
b) $\frac{x-1}{2}$ – $\frac{2x+3}{3}$ ≤ $\frac{x}{5}$ + 4
Đáp án:
a. x=5
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ne \left\{ { – 3;2} \right\}\\
Pt \to \frac{{2x\left( {x – 2} \right) – \left( {x – 1} \right)\left( {x + 3} \right) – x + 7}}{{\left( {x + 3} \right)\left( {x – 2} \right)}} = 0\\
\to 2{x^2} – 4x – {x^2} – 2x + 3 – x + 7 = 0\\
\to {x^2} – 7x + 10 = 0\\
\to {x^2} – 5x – 2x + 10 = 0\\
\to x\left( {x – 5} \right) – 2\left( {x – 5} \right) = 0\\
\to \left[ \begin{array}{l}
x – 5 = 0\\
x – 2 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 5\\
x = 2\left( l \right)
\end{array} \right.
\end{array}\)
\(\begin{array}{l}
b.\frac{{15x – 15 – 10\left( {2x + 3} \right) – 6x – 4.30}}{{30}} \le 0\\
\to 15x – 15 – 20x – 30 – 6x – 120 \le 0\\
\to – 11x \le 165\\
\to x \ge – 15
\end{array}\)