تدرب وحل التمرينات

(1)- جد مجموعة الحل لكل معادلة من المعادلات التالية وتحقق من صحة الحل:

$\frac{4}{6x^2}+\frac{1}{3}=\frac{1}{x}$

نضرب طرفي المعادلة في $6x^2$

$\begin{array}{rl}& \frac{4}{6x^2}+\frac{1}{3}=\frac{1}{x}\\ & \Rightarrow 4+2x^2=6x\Rightarrow 2x^2-6x+4=0\Rightarrow x^2-3x+2=0\\ & \begin{array}{rl}& \Rightarrow (x-2)(x-1)=0\Rightarrow \{x-2=0\Rightarrow x=2\\ & \text{or }\mathrm{x}-1=0\Rightarrow \mathrm{x}=1\Rightarrow \mathrm{s}=\{2,1\}\end{array}\end{array}$

التحقق:

$$\begin{gathered} \begin{array}{rl}& \text{L.}S=\frac{4}{6x^2}+\frac{1}{3}=\frac{1}{6}+\frac{1}{3}=\frac{3}{6}=\frac{1}{2} \left(x=2\right)\\ & \text{R.}S=\frac{1}{x}=\frac{1}{2}=L.S\end{array} \\ \begin{array}{rl}& \mathrm{L}\cdot \mathrm{S}=\frac{4}{6{\mathrm{x}}^2}+\frac{1}{3}=\frac{4}{6}+\frac{1}{3}=\frac{2}{3}+\frac{1}{3}=1 \left(x=1\right)\\ & \text{R.}\mathrm{S}=\frac{1}{\mathrm{x}}=1=\mathrm{L}\cdot \mathrm{S}\end{array} \end{gathered}$$

$\frac{3y}{4}-\frac{6}{12y}+\frac{1}{4}=0$

نضرب طرفي المعادلة في $12y$

$\begin{array}{rl}& \frac{3y}{4}\cdot \frac{6}{12y}+\frac{1}{4}=0\\ & \Rightarrow 9y^2-6+3y=0\Rightarrow 3y^2+y-2=0\Rightarrow (3y-2)(y+1)=0\\ & \begin{array}{rl}\Rightarrow & \{y-2=0\Rightarrow y=\frac{2}{3}\\ & \text{or }y+1=0\Rightarrow y=-1\Rightarrow s=\{\frac{2}{3},-1\}\end{array}\end{array}$

التحقق:

$$\begin{gathered} \text{L. S}=\frac{3y}{4}-\frac{6}{12y}+\frac{1}{4}=\frac{3\left(\frac{2}{3}\right)}{4}-\frac{6}{12\left(\frac{2}{3}\right)}+\frac{1}{4}=\frac{1}{2}-\frac{3}{4}+\frac{1}{4}=0=R\cdot S \left(y=\frac{2}{3}\right) \\ \mathrm{L}.\mathrm{S}=\frac{3\mathrm{y}}{4}-\frac{6}{12\mathrm{y}}+\frac{1}{4}=\frac{3(-1)}{4}-\frac{6}{12(-1)}+\frac{1}{4}=\frac{-3}{4}+\frac{1}{2}+\frac{1}{4}=0=\mathrm{R}.\mathrm{S} \left(\mathrm{y}=-1\right) \end{gathered}$$

$\frac{9x+22}{x^2}=1$

نضرب طرفي المعادلة في ${\mathrm{x}}^2$

$\begin{array}{rl}& \frac{9x+22}{x^2}\\ & 9x+22=x^2\Rightarrow x^2-9x-22=0\Rightarrow (x-11)(x+2)=0\\ & \Rightarrow \{\begin{array}{l}x-11=0\Rightarrow x=11\\ \text{or }x+2=0\Rightarrow x=-2\Rightarrow s=\{11,-2\}\end{array}\end{array}$

التحقق:

$$\begin{gathered} \mathrm{L}\cdot \mathrm{S}=\frac{9\left(11\right)+22}{(11{)}^2}=\frac{99+22}{121}=1=\mathrm{R}\cdot \mathrm{S} \left(\mathrm{x}=11\right) \\ \mathrm{L}\cdot \mathrm{S}=\frac{9(-2)+22}{(-2{)}^2}=\frac{-18+22}{4}=1=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=-2\right) \end{gathered}$$

$\frac{9}{(y+2{)}^2}=\frac{3y}{y+2}$

نضرب طرفي المعادلة في $(y+2{)}^2$

$\begin{array}{rl}& \frac{9}{(y+2{)}^2}=\frac{3y}{y+2}\\ & \Rightarrow 9=3y(y+2)\Rightarrow 3y^2+6y-9=0\\ & \Rightarrow y^2+2y-3=0\Rightarrow (y-1)(y+3)=0\\ & or y+3=0\Rightarrow y=-3\Rightarrow s=\{1,-3\}\end{array}$

التحقق:

$$\begin{gathered} L\cdot S=\frac{9}{(1+2{)}^2}=1,\mathrm{R}\cdot \mathrm{S}=\frac{3\left(1\right)}{1+2}=1=\mathrm{L}\cdot \mathrm{S} \left(\mathrm{y}=1\right) \\ \text{L. S}=\frac{9}{(-3+2{)}^2}=9,\mathrm{R}\cdot \mathrm{S}=\frac{3(-3)}{-3+2}=\frac{-9}{-1}=9=\mathrm{L}.\mathrm{S} \left(\mathrm{y}=-3\right) \end{gathered}$$

(2)- جد مجموعة الحل لكل معادلة من المعادلات التالية:

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

نضرب طرفي المعادلة في $(x-4)(x-3)$

$$\begin{gathered} \begin{array}{rl}& \frac{3}{x-4}-\frac{2}{x-3}=1\\ & \Rightarrow 3(x-3)-2(x-4)=(x-4)(x-3)\Rightarrow 3x-9-2x+8=x^2-7x+12\\ & \Rightarrow x^2-8x+13=0,a=1,b=-8,c=13\end{array} \\ \begin{array}{rl}& x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\Rightarrow x=\frac{8\pm \sqrt{64-4\left(1\right)(13})}{2}\\ & \Rightarrow x=\frac{-8\pm \sqrt{64-52}}{2}\Rightarrow x=\frac{8\pm \sqrt{12}}{2}\Rightarrow x=4\pm \sqrt{3}\\ & \Rightarrow \{\begin{array}{l}x=4+\sqrt{3}\\ or x=4-\sqrt{3}\Rightarrow s=\{4+\sqrt{3},4-\sqrt{3}\}\end{array}\end{array} \end{gathered}$$

$\frac{y-5}{y+5}-\frac{y+5}{y-5}=\frac{4y^2-24}{y^2-25}$

نضرب طرفي المعادلة في $(y-5)(y+5)$

$$\begin{gathered} \begin{array}{rl}& \frac{y-5}{y+5}-\frac{y+5}{y-5}=\frac{4y^2-24}{y^2-25}\\ & \Rightarrow \frac{y-5}{y+5}-\frac{y+5}{y-5}=\frac{4y^2-24}{(y-5)(y+5)}\end{array} \\ \begin{array}{rl}& \Rightarrow (y-5{)}^2-(y+5{)}^2=4y^2-24\Rightarrow y^2-10y+25-y^2-10y-25=4y^2-24\\ & \Rightarrow -20y=4y^2-24\Rightarrow y^2+5y-6=(y+6)(y-1)=0\\ & \begin{array}{rl}& \Rightarrow \{y+6=0\Rightarrow y=-6\\ & \text{or }y-1=0\Rightarrow y=1\Rightarrow s=\{-6,1\}\end{array}\end{array} \end{gathered}$$

$\frac{6-x}{x^2+x-12}-\frac{2}{x+4}=1$

نضرب طرفي المعادلة في $(x-3)(x+4)$

$\begin{array}{rl}& \frac{6-x}{x^2+x-12}-\frac{2}{x+4}=1\Rightarrow \frac{6-x}{(x-3)(x+4)}-\frac{2}{x+4}=1\\ & \Rightarrow 6-x-2(x-3)=(x-3)(x+4)\Rightarrow 6-x-2x+6=x^2+x-12\\ & \Rightarrow x^2+4x-24=0\\ & x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\Rightarrow x=\frac{-4\pm \sqrt{16+4\left(1\right)\left(24\right)}}{2}\Rightarrow x=\frac{-4\pm \sqrt{112}}{2}\\ & \Rightarrow x=\frac{-4\pm 4\sqrt{7}}{2}\Rightarrow x=-2\pm 2\sqrt{7}\\ & \Rightarrow \{\begin{array}{l}x=-2+2\sqrt{7}\\ or x=-2-2\sqrt{7}\Rightarrow s=\{-2+2\sqrt{7},-2-2\sqrt{7}\}\end{array}\end{array}$

$\frac{4+8y}{y^2-9}+\frac{6}{y-3}=3$

نضرب طرفي المعادلة في $(y-3)(y+3)$

$$\begin{gathered} \frac{4+8y}{y^2-9}+\frac{6}{y-3}=3\Rightarrow \frac{4+8y}{(y-3)(y+3)}+\frac{6}{y-3}=1 \\ \begin{array}{rl}& 4+8y+6(y+3)=3(y-3)(y+3)\\ & \Rightarrow 4+8y+6y+18=3y^2-27\Rightarrow 3y^2-14y-49=0\\ & a=3,b=-14,c=-49\end{array} \\ \begin{array}{rl}& y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\Rightarrow y=\frac{14\pm \sqrt{196+4\left(3\right)\left(49\right)}}{6}\\ & \Rightarrow y=\frac{-14\pm \sqrt{78}4}{6}\Rightarrow y=\frac{14\pm 28}{6}\Rightarrow y=\frac{7\pm 14}{3}\\ & \{\begin{array}{l}y=\frac{7+14}{3}=7\\ or y=\frac{7-14}{3}=\frac{-7}{3}\Rightarrow s=\{7,\frac{-7}{3}\}\end{array}\end{array} \end{gathered}$$