جد مجموعة الحل لكل معادلة من المعادلات التالية وتحقق من صحة الحل:

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

نضرب طرفي المعادلة في $\left(4y^2\right)$

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

التحقق:

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\frac{\mathrm{y}+1}{{\mathrm{y}}^2}=\frac{\frac{-2}{6}+1}{{\left(\frac{-2}{3}\right)}^2}=\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{9}{12}=\frac{3}{4}=\mathrm{R}.\mathrm{S} \left(\mathrm{y}=\frac{-2}{3}\right) \\ \mathrm{L}.\mathrm{S}=\frac{\mathrm{y}+1}{{\mathrm{y}}^2}=\frac{2+1}{2^2}=\frac{3}{4}=\mathrm{R}.\mathrm{S} \left(\mathrm{y}=2\right) \end{gathered}$$

تأكد من فهمك

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

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

نضرب طرفي المعادلة في $\left(4{\mathrm{x}}^2\right)L\mathrm{CM}$

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

التحقق:

$$\begin{gathered} L.S=\frac{1}{x}+\frac{1}{2}=1+\frac{1}{2}=\frac{3}{2} \left(x=1\right) \\ R\cdot S=\frac{6}{4x^2}=\frac{6}{4\left(1\right)}=\frac{3}{2}=L\cdot S \\ L.S=\frac{1}{-3}+\frac{1}{2}=\frac{-2+3}{6}=\frac{1}{6} \left(x=-3\right) \\ R\cdot S=\frac{6}{2(-3{)}^2}=\frac{6}{36}=\frac{1}{6}=L\cdot S \end{gathered}$$

$\frac{y}{2}-\frac{7}{5}=\frac{3}{10y}$

نضرب طرفي المعادلة في $\left(10y\right)LCM$

$\begin{array}{rl}& \frac{y}{2}\cdot \frac{7}{5}=\frac{3}{10y}\\ & \Rightarrow 5y^2-14y=3\Rightarrow 5y^2-14y-3=0\Rightarrow (5y+1)(y-3)=0\\ & \Rightarrow \{\begin{array}{l}5y+1=0\Rightarrow y=-\frac{1}{5}\\ or y-3=0\Rightarrow y=3\Rightarrow s=\{-\frac{1}{5},3\}\end{array}\end{array}$

التحقق:

$$\begin{gathered} \begin{array}{rl}& \text{L. S}=\frac{y}{2}-\frac{7}{5}=\frac{\frac{1}{5}}{2}-\frac{7}{5}=\frac{1}{10}-\frac{7}{5}=\frac{-3}{2} \left(y=-\frac{1}{5}\right)\\ & R.S=\frac{3}{10y}=\frac{3}{10-\frac{1}{5}}=\frac{-3}{2}=L.S\end{array} \\ \begin{array}{rl}& \text{L. S}=\frac{y}{2}-\frac{7}{5}=\frac{3}{2}-\frac{7}{5}=\frac{15-14}{10}=\frac{1}{10} \left(y=3\right)\\ & \text{R.}S=\frac{3}{10\left(3\right)}=\frac{1}{10}=L\cdot S\end{array} \end{gathered}$$

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

نضرب طرفي المعادلة في $\left(2x\right)$

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

التحقق:

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

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

نضرب طرفي المعادلة في $\left(4y^2\right)$

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

التحقق:

$$\begin{gathered} \mathrm{L}.\mathrm{S}=\frac{\mathrm{y}+1}{{\mathrm{y}}^2}=\frac{\frac{-2}{6}+1}{{\left(\frac{-2}{3}\right)}^2}=\frac{\frac{1}{3}}{\frac{4}{9}}=\frac{9}{12}=\frac{3}{4}=\mathrm{R}.\mathrm{S} \left(\mathrm{y}=\frac{-2}{3}\right) \\ \mathrm{L}.\mathrm{S}=\frac{\mathrm{y}+1}{{\mathrm{y}}^2}=\frac{2+1}{2^2}=\frac{3}{4}=\mathrm{R}.\mathrm{S} \left(\mathrm{y}=2\right) \end{gathered}$$

$\frac{9x-14}{x-5}=\frac{x^2}{x-5}$

نضرب طرفي المعادلة في $(x-5)$

$\begin{array}{rl}& \frac{9x-14}{x-5}=\frac{x^2}{x-5}\\ & 9x-14=x^2\Rightarrow x^2-9x+14=0\Rightarrow (x-2)(x-7)=0\\ & \begin{array}{rl}& \Rightarrow \{x-2=0\Rightarrow x=2\\ & \text{or }x-7=0\Rightarrow x=-7\Rightarrow s=\{2,7\}\end{array}\end{array}$

التحقق:

$$\begin{gathered} \text{L.}S-\frac{9x-14}{x-5}-\frac{18-14}{2-5}-\frac{4}{-3} \left(x=2\right) \\ \text{R. S}=\frac{2x}{x-5}=\frac{4}{-3}=L.S \\ \mathrm{L}.\mathrm{S}=\frac{9x-14}{x-5}=\frac{63-14}{7-5}=\frac{49}{2} \left(x=7\right) \\ R\cdot S=\frac{x^2}{x-5}=\frac{49}{2}=L.S \end{gathered}$$

$\frac{1}{y^2-6}=\frac{2}{y+3}$

نضرب طرفي المعادلة في $(y^2-6)(y+3)$

$\begin{array}{rl}& \frac{1}{y^2-6}=\frac{2}{y+3}\\ & y+3=2(y^2-6)\Rightarrow 2y^2-12-y-3=0\Rightarrow 2y^2-y-15=0\Rightarrow (2y+5)(y-3)=0\\ & \begin{array}{rl}& \{2y+5=0\Rightarrow y=\frac{-5}{2}\\ & \text{or }y-3=0\Rightarrow y=3\Rightarrow s=\{\frac{-5}{2},3\}\end{array}\end{array}$

التحقق:

$$\begin{gathered} \text{L.}S=\frac{1}{y^2-6}=\frac{1}{\frac{25}{4}-6}=\frac{1}{\frac{1}{4}}=4 \left(y=\frac{-5}{2}\right) \\ R\cdot S=\frac{2}{y+3}=\frac{2}{\frac{5}{2}+3}=\frac{1}{\frac{1}{2}}=4=L\cdot S \\ L.S=\frac{1}{y^2-6}=\frac{1}{9-6}=\frac{1}{3} \left(y=3\right) \\ \text{R.}\mathrm{S}=\frac{2}{\mathrm{y}+3}=\frac{2}{3+3}=\frac{1}{6}=\frac{1}{3}=\mathrm{L}\cdot \mathrm{S} \end{gathered}$$

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

$\frac{y-4}{y+2}-\frac{2}{y-2}=\frac{17}{y^2-4}$

نضرب طرفي المعادلة في $(y+2)(y-1)$

$$\begin{gathered} \begin{array}{rl}& \frac{y-4}{y+2}-\frac{2}{y-2}=\frac{17}{y^2-4}\\ & \Rightarrow \frac{y-4}{y+2}=\frac{2}{y-2}=\frac{17}{(y+2)(y-2)}\Rightarrow (y-4)(y-2)-2(y+2)=17\end{array} \\ \begin{array}{c}\Rightarrow y^2-6y+8-2y-4-17=0\Rightarrow y^2-8y-13=0,a=1,b=-8,c=-13\\ \Rightarrow y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\Rightarrow y=\frac{8\pm \sqrt{64-4\left(1\right)(-13)}}{2}\\ \Rightarrow y=\frac{8\pm \sqrt{29}}{2}\Rightarrow y=4\pm \sqrt{29}\\ \Rightarrow \{y=4+\sqrt{29}\\ \text{or }y=4-\sqrt{29}\Rightarrow s=\{4+\sqrt{29},4-\sqrt{29}\}\end{array} \end{gathered}$$

$\frac{9}{x^2-x-6}-\frac{5}{x-3}=1$

نضرب طرفي المعادلة في $(x-3)(x+2)$

$\begin{array}{rl}& \frac{9}{x^2-x-6}-\frac{5}{x-3}=1\Rightarrow \frac{9}{(x-3)(x+2)}-\frac{5}{x-3}=1\\ & \Rightarrow 9-5(x+2)=(x-3)(x+2)\\ & \Rightarrow 9-5x-10=x^2-x-6\Rightarrow x^2+4x-5=0\Rightarrow (x+5)(x-1)=0\\ & \begin{array}{rl}& \Rightarrow \{x+5=0\Rightarrow x=-5\\ & \text{or }x-1=0\Rightarrow x=1\Rightarrow s=\{-5,1\}\end{array}\end{array}$

$\frac{12}{y^2-16}+\frac{6}{y+4}=2$

نضرب طرفي المعادلة في $(y-4)(y+4)$

$\begin{array}{rl}& \frac{12}{y^2-16}+\frac{6}{y+4}=2\\ & \Rightarrow \frac{12}{(y-4)(y+4)}+\frac{6}{y+4}=2\\ & \Rightarrow 12+6(y-4)=2(y-4)(y+4)\Rightarrow 12+6y-24=2y^2-32\\ & \Rightarrow 2y^2-6y-20=0\Rightarrow (2y+4)(y-5)=0\\ & \begin{array}{rl}& \Rightarrow \{2y+4=0\Rightarrow y=-2\\ & \text{or }y-5=0\Rightarrow y=5\Rightarrow s=\{-2,5\}\end{array}\end{array}$

$\frac{2x}{x+1}+\frac{3x}{x-1}=\frac{8+7x+3x^2}{x^2-1}$

نضرب طرفي المعادلة في $(x+1)(x-1)$

$$\begin{gathered} \begin{array}{rl}& \frac{2x}{x+1}+\frac{3x}{x-1}=\frac{8+7x+3x^2}{(x+1)(x-1)}\Rightarrow 2x(x-1)+3x(x+1)=8+7x+3x^2\\ & \Rightarrow 2x^2-2x+3x^2+3x=8+7x+3x^2\\ & \Rightarrow 2x^2-6x-8=0\Rightarrow x^2-3x-4=0\Rightarrow (x-4)(x+1)=0\end{array} \\ \begin{array}{rl}& \Rightarrow \{\begin{array}{c}x-4=0\Rightarrow x=4\\ \text{or }x+1=0\Rightarrow x=-1\end{array}\\ & \Rightarrow \mathrm{s}=\left\{4\right\}\end{array} \end{gathered}$$

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