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(1)- حل المعادلات التالية باستعمال الفرق بين مربعين وتحقق من صحة الحل:

$x^2-16=0$

$$\begin{gathered} \begin{array}{rl}{x}^2-16=0\Rightarrow (x-4)(x+4)=0& \Rightarrow x-4=0 \text{or }x+4=0\\ & \Rightarrow x=4 \text{or }x-4\Rightarrow S=\{4,-4\}\end{array} \\ \mathrm{L}.\mathrm{S}=16-16=0=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=4\right) \\ \mathrm{L}.\mathrm{S}=(-4{)}^2-16=16-16=0=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=-4\right) \end{gathered}$$

$81-y^2=0$

$$\begin{gathered} \begin{array}{rl}81-y^2-0\Rightarrow (9-y)(9+y)=0& \Rightarrow 9-y=0 \text{or }9+y=0\\ & \Rightarrow y=9 \text{or }y=-9\Rightarrow S=\{9,-9\}\end{array} \\ \begin{array}{rl}& L.S=81-81=0=R.S \left(y=9\right)\\ & L.S=81-(-9{)}^2=81-81=0=R.S \left(y=-9\right)\end{array} \end{gathered}$$

$2{\mathrm{z}}^2-8=0$

$$\begin{gathered} \begin{array}{rl}2z^2-8=0\Rightarrow 2(z^2-4)& =0\Rightarrow z^2-4=0\Rightarrow (z-2)(z+2)=0\\ \Rightarrow z-2& =0 \text{or }z+2=0\Rightarrow z=2 \text{or }z=-2\Rightarrow S=\{2,-2\}\end{array} \\ L.S=2\left(4\right)-8=8-8=0=\mathrm{R}.S \left(z=2\right) \\ L.S=2(-2{)}^2-8=8-8=0=\text{R.S }\left(y=-9\right) \end{gathered}$$

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

$4x^2-9=0$

$\begin{array}{c}4x^2-9=0\Rightarrow (2x-3)(2x+3)=0\Rightarrow 2x-3=0 \text{or }2x+3=0\\ \Rightarrow x=\frac{3}{2} \text{or }x=\frac{-3}{2}\end{array}$

مجموعة الحل $S=\{\frac{3}{2},\frac{-3}{2}\}$

$5y^2-20=0$

$\begin{array}{rl}5y^2-20=0\Rightarrow & 5(y^2-4)=0\Rightarrow y^2-4=0\Rightarrow (y-2)(y+2)=0\\ & \Rightarrow y-2=0 \text{or }y+2=0\Rightarrow y=2 \text{or }y=-2\end{array}$

مجموعة الحل $S=\{2,-2\}$

$(y+2{)}^2-49=0$

$\begin{array}{rl}(y+2{)}^2-49=0& \Rightarrow \left(\right(y+2)-7)(y+2)+7)=0\\ & \Rightarrow \left(\right(y+2)-7)=0 \text{or }\left(\right(y+2)+7)=0\Rightarrow y= 5 \text{or }y=-9\end{array}$

مجموعة الحل $\mathrm{S}=\{5,-9\}$

$(3-z{)}^2-1=0$

$\begin{array}{rl}(3-z{)}^2-i-0& \Rightarrow \left(\right(3-z)-1)\left(\right(3-z)+1)=0\\ & \Rightarrow \left(\right(3-z)-1)=0 \text{or }\left(\right(3-z)+1)=0\\ & \Rightarrow 2-z=0 \text{or }4-z=0\Rightarrow z=2 \text{or }z=4\end{array}$

مجموعة الحل $S=\{2,4\}$

$x^2-3=0$

$\begin{array}{rl}& x^2-3=0\Rightarrow (x-\sqrt{3})(x+\sqrt{3})=0\Rightarrow x-\sqrt{3}=0 \text{or }x+\sqrt{3}=0\\ & \Rightarrow x=\sqrt{3} \text{or }x=-\sqrt{3}\end{array}$

مجموعة الحل $S=\{\sqrt{3},-\sqrt{3}\}$

$y^2-\frac{1}{9}=0$

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

مجموعة الحل $\mathrm{S}=\{\frac{1}{3},-\frac{1}{3}\}$

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

$x^2=64$

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

$Z^2=7$

$z^2=7\Rightarrow z=\pm \sqrt{7}\Rightarrow S=\{\sqrt{7},-\sqrt{7}\}$

$2y^2=\frac{49}{8}$

$2y^2=\frac{49}{8}\Rightarrow y^2=\frac{49}{16}\Rightarrow y=\pm \sqrt{\frac{49}{16}}\Rightarrow y=\pm \frac{7}{4}\Rightarrow S=\{\frac{7}{4},\frac{-7}{4}\}$

$6z^2-5=0$

$6z^2-5=0\Rightarrow z^2=\frac{5}{6}\Rightarrow z=\pm \sqrt{\frac{5}{6}}\Rightarrow S=\{\sqrt{\frac{5}{6}},\sqrt{\frac{5}{6}}\}$

$4(x^2-12)=33$

$\begin{array}{rl}4(x^2-12)=33& \Rightarrow x^2-12=\frac{33}{4}=x^2=\frac{33}{4}+12\\ & \Rightarrow x^2=\frac{81}{4}\Rightarrow x=\pm \frac{9}{2}\Rightarrow S=\{\frac{9}{2},\frac{-9}{2}\end{array}$

$z^2+\frac{2}{3}=\frac{5}{6}$

${\mathrm{z}}^2+\frac{2}{3}=\frac{5}{6}\Rightarrow {\mathrm{z}}^2=\frac{5}{6}-\frac{2}{3}\Rightarrow {\mathrm{z}}^2=\frac{1}{6}\Rightarrow \mathrm{z}=\pm \sqrt{\frac{1}{6}}\Rightarrow \mathrm{S}=\{\frac{1}{\sqrt{6}},\frac{-1}{\sqrt{6}}\}$

(4)- حل المعادلات التالية:

$3\sqrt{x}=15$

$3\sqrt{x}=15\Rightarrow \sqrt{x}=5\Rightarrow (\sqrt{x}{)}^2=5^2\Rightarrow x=25\Rightarrow S=\{25\}$

$\sqrt{y-5}=2$

$\sqrt{y-5}=2\Rightarrow y-5=4\Rightarrow y=9\Rightarrow S=\left\{9\right\}$

$\sqrt{2z}=6$

$\sqrt{2z}=6\Rightarrow 2z=36\Rightarrow z=18\Rightarrow S\left\{18\right\}$