تدرب وحل التمرينات

(1)- حل المعادلات التالية وتحقق من صحة الحل:

$x^2=49$

$$\begin{gathered} x^2=49\Rightarrow x=\pm 7\Rightarrow S=\{7,-7\} \\ \mathrm{L}.\mathrm{S}=7^2=49=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=7\right) \\ \text{L.S}=(-7{)}^2=49=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=-7\right) \end{gathered}$$

$5y^2-10=0$

$$\begin{gathered} \begin{array}{rl}5y^2-10=0& \Rightarrow 5(y^2-2)=0\Rightarrow y^2-2=0\Rightarrow (y-\sqrt{2})(y+\sqrt{2})=0\\ & \Rightarrow y-\sqrt{2}=0 \text{or }\Rightarrow y+\sqrt{2}=0\\ & \Rightarrow y=\sqrt{2} \text{or }y=-\sqrt{2}\Rightarrow S=\{\sqrt{2},-\sqrt{2}\}\end{array} \\ \text{L.S}=5(\sqrt{2}{)}^2-10=10-10=0=\mathrm{R}.S \left(y=\sqrt{2}\right) \\ \text{L.S}=5(-\sqrt{2}{)}^2-10=5\left(2\right)-10=10-10=0=\text{R.S }\left(y=-\sqrt{2}\right) \end{gathered}$$

$3z^2-27=0$

$$\begin{gathered} \begin{array}{rl}3z^2-27=0& \Rightarrow 3(z^2-9)=0\Rightarrow z^2-9=0\Rightarrow (z-3)(z+3)=0\\ & \Rightarrow z-3=0 \text{or }\Rightarrow z+3=0\\ & \Rightarrow z=3 \text{or }z=-3\Rightarrow S=\{3,-3\}\end{array} \\ \text{L.S}=3\left(9\right)-27=0=27-27=0=\text{R.S }\left(z=3\right) \\ \mathrm{L}.S=3(-3{)}^2-27=3(9)-27=27-27=0=\mathrm{R}.\mathrm{S} \left(\mathrm{z}=-3\right) \end{gathered}$$

(2)- حل المعادلات التالية باستعمال الفرق بين مربعين:

$9x^2-36=0$

$\begin{array}{rl}9x^2-36=0& \Rightarrow 9(x^2-4)=0\Rightarrow x^2-4=0\Rightarrow (x-2)(x+2)=0\\ & \Rightarrow x-2=0 \text{or }x+2=0\\ & \Rightarrow x=2 \text{or }x=-2\Rightarrow S=\{2,-2\}\end{array}$

$7y^2-28=0$

$\begin{array}{rl}7y^2-28=0& \Rightarrow 7(y^2-4)=0\Rightarrow y^2-4=0\Rightarrow (y-2)(y+2)=0\\ & \Rightarrow y-2=0 \text{or }y+2=0\\ & \Rightarrow y=2 \text{or }y=-2\Rightarrow S=\{2,-2\}\end{array}$

$9(x^2-1)-7=0$

$\begin{array}{rl}9(x^2-1)-7& =0\Rightarrow 9x^2-9-7=0\Rightarrow 9x^2-16=0\Rightarrow (3x-4)(3x+4)=0\\ & \Rightarrow 3x-4=0 \text{or }\Rightarrow 3x+4=0\\ & \Rightarrow x=\frac{4}{3} \text{or }x=\frac{-4}{3}\Rightarrow S=\{\frac{4}{3},\frac{-4}{3}\}\end{array}$

$(y+5{)}^2-64=0$

$\begin{array}{rl}(y+5{)}^2-64=0& \Rightarrow \left(\right(y+5)-8)\left(\right(y+5)+8)=0\\ & \Rightarrow y-3=0 \text{or }\Rightarrow y+13=0\\ & \Rightarrow y=3 \text{or }y=-13\Rightarrow S=\{3,-13\}\end{array}$

$x^2-2=0$

$\begin{array}{rl}{x}^2-2=0\Rightarrow (x-\sqrt{2})(x+\sqrt{2})& =0\\ \Rightarrow x-\sqrt{2}=0& \text{ or }x+\sqrt{2}=0\\ \Rightarrow x=\sqrt{2}& \text{ or }x=-\sqrt{2}\Rightarrow S=\{\sqrt{2},-\sqrt{2}\}\end{array}$

$y^2-\frac{1}{36}=0$

$\begin{array}{rl}{y}^2-\frac{1}{36}=0& \Rightarrow (y-\frac{1}{6})(y+\frac{1}{6})=0\Rightarrow \\ & \Rightarrow y-\frac{1}{6}=0 \text{or }y+\frac{1}{6}=0\\ & \Rightarrow y=\frac{1}{6} \text{or }y=\frac{-1}{6}\Rightarrow S=\{\frac{1}{6},\frac{-1}{6}\}\end{array}$

(3)- حل المعادلات التالية باستعمال قاعدة الجذر التربيعي:

$x^2=121$

$x^2=121\Rightarrow x=\pm 11\Rightarrow S=\{11,-11\}$

$50-2y^2=0$

$\begin{array}{rl}50-2y^2=0& \Rightarrow 2(25-y^2)=0\Rightarrow 25-y^2=0\\ & \Rightarrow y^2=25\Rightarrow y=\pm 5\Rightarrow S=\{5,-5\}\end{array}$

$x^2=\frac{1}{64}$

$x^2=\frac{1}{64}\Rightarrow x=\pm \frac{1}{8}\Rightarrow S=\{\frac{1}{8},\frac{-1}{8}\}$

$3y^2=\frac{25}{3}$

$3y^2=\frac{25}{3}\Rightarrow y^2=\frac{25}{9}\Rightarrow y=\pm \frac{5}{3}\Rightarrow S=\{\frac{5}{3},\frac{-5}{3}\}$

$7(x^2-2)=50$

$\begin{array}{rl}7(x^2-2)=50& \Rightarrow 7x^2-14=50\Rightarrow 7x^2=64\\ & \Rightarrow x^2=\frac{64}{7}\Rightarrow x=\sqrt{\frac{8}{\sqrt{7}}}\Rightarrow S=\sqrt{\frac{8}{\sqrt{7}}}\sqrt{\frac{-8}{\sqrt{7}}\}}\end{array}$

$\frac{1}{5}{y}^2=\frac{1}{3}$

$\frac{1}{5}{y}^2=\frac{1}{3}\Rightarrow y^2=\frac{5}{3}\Rightarrow y=\pm \frac{5}{3}\Rightarrow S=\{\frac{5}{3},-\frac{5}{3}\}$

(4)- حل المعادلات التالية:

$6\sqrt{x}=30$

$6\sqrt{x}=30\Rightarrow \sqrt{x}=5\Rightarrow x=25\Rightarrow S=\left\{25\right\}$

$\sqrt{y-9}=4$

$\sqrt{y-9}=4\Rightarrow y-9=16\Rightarrow y=25\Rightarrow S=\left\{25\right\}$

$\sqrt{4z}=8$

$\sqrt{4z}=8\Rightarrow 4z=64\Rightarrow z=16\Rightarrow S=\left\{16\right\}$