حل كلاً من المتباينتين الآتيتين:

$\frac{x+1}{x-3}-1\le 2$

$\begin{array}{rl}& \frac{x+1}{x-3}\le 2+1\\ & \frac{x+1}{x-3}\le 3\to \therefore x+1\le 3x-9\\ & \therefore 1+9\le 3x-x\\ & \therefore 10\le 2x\to \therefore 5\le x\end{array}$

$\frac{6}{x}+2\ge 0$

$\begin{array}{rl}& \frac{6}{x}\ge -2\therefore 6\ge -2x\\ & \therefore x\ge -3\end{array}$

حل أسئلة مراجعة تراكمية

بسط كل عبارة مما يأتي:

64) $\frac{2r-4}{r-2}$

$\frac{2(r-2)}{r-2}=2$

65) $\frac{r^2-7r-30}{r^2-5r-24}$

$\frac{(r-10)(r+3)}{(r-8)(r+3)}=\frac{r-10}{r-8}$

66) $\frac{y}{4}-\frac{4y}{3x}+\frac{3y}{4x}$

$\frac{3xy-16y+9y}{12x}=\frac{y(3x-16+9)}{12x}=\frac{y(3x-7)}{12x}$

67) $\frac{\frac{1}{a}+\frac{1}{4}}{\frac{1}{a^2}-\frac{1}{16}}$

$$\begin{gathered} \frac{\frac{4-a}{4a}}{\frac{16-a^2}{16a^2}}=\frac{4-a}{4a}÷\frac{(4-a)(4+a)}{16a^2} \\ \begin{array}{rl}& =\frac{4-a^1}{4a_1}\times \frac{16a^{4a}}{(4-a)(4+a)}\\ & =\frac{4a}{4+a}\end{array} \end{gathered}$$

68) $\frac{6x^2-11x+4}{6x^2+x-2}\cdot \frac{12x^2+11x+2}{8x^2+14x+3}$

$\frac{(2x-1)(3x-4)}{(2x-1)(3x+2)}\cdot \frac{(4x+1)(3x+2)}{(4x+1)(2x+3)}=\frac{3x-4}{2x+3}$

حل كلاً من المعادلتين الآتيتين:

69) $\frac{8}{x}=1+\frac{2}{x-2}$

$\begin{array}{rl}& \frac{8}{x}=\frac{x-2+2}{x-2}\\ & \therefore \frac{8}{x}=\frac{x}{x-2}\to \therefore x^2=8x-16\\ & \therefore x^2-8x+16=0\end{array}$

70) $x-\frac{1}{2}=\frac{3}{2}$

$x=\frac{3}{2}+\frac{1}{2}=\frac{4}{2}=2$

حل كلاً من المتباينتين الآتيتين:

71) $\frac{x+1}{x-3}-1\le 2$

$\begin{array}{rl}& \frac{x+1}{x-3}\le 2+1\\ & \frac{x+1}{x-3}\le 3\to \therefore x+1\le 3x-9\\ & \therefore 1+9\le 3x-x\\ & \therefore 10\le 2x\to \therefore 5\le x\end{array}$

72) $\frac{6}{x}+2\ge 0$

$\begin{array}{rl}& \frac{6}{x}\ge -2\therefore 6\ge -2x\\ & \therefore x\ge -3\end{array}$