حل المعادلات التالية باستعمال خاصية الضرب الصفري:

$\begin{array}{rl}& (5t-7)(t+9)=0\end{array}$

• $s=\{\frac{7}{5},\frac{1}{9}\}$
• $s=\{\frac{5}{7},\frac{1}{9}\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\frac{\mathbf{7}}{\mathbf{5}}\mathbf{,}\mathbf{-}\mathbf{9}\mathbf{\}}$
• $s=\{\frac{-7}{5},9\}$

[4-3] حل معادلات من الدرجة الثانية بمتغير واحد في R

اختر الإجابة الصحيحة لكل مما يأتي:

حل المعادلات التالية باستعمال الجذر التربيعي:

$\begin{array}{rl}& 9{\mathrm{y}}^2=1\end{array}$

• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{-}\mathbf{3}}\mathbf{\}}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\frac{1}{9},\frac{1}{-9}\}\end{array}$
• $\begin{array}{rl}& s=\{\frac{1}{-3},\frac{1}{-9}\}\end{array}$
• $\begin{array}{rl}& s=\{\frac{1}{3},\frac{1}{9}\}\end{array}$

$\begin{array}{rl}& 21{\mathrm{x}}^2=3\end{array}$

• $\begin{array}{rl}& \mathrm{s}=\{\frac{1}{7},\frac{1}{-7}\}\end{array}$
• $\begin{array}{rl}& \mathrm{s}=\{\frac{1}{\sqrt{7}},\frac{1}{7}\}\end{array}$
• $\begin{array}{rl}& \mathbf{s}\mathbf{=}\mathbf{\{}\frac{\mathbf{1}}{\sqrt{\mathbf{7}}}\mathbf{,}\frac{\mathbf{1}}{\mathbf{-}\sqrt{\mathbf{7}}}\mathbf{\}}\end{array}$
• $\begin{array}{rl}& s=\{\frac{1}{-7},\frac{1}{\sqrt{7}}\}\end{array}$

$\begin{array}{rl}& \frac{1}{3}{\mathrm{t}}^2=12\end{array}$

• $\mathrm{s}=\{6,0\}$
• $\mathrm{s}=\{-6,0\}$
• $s=\{2,-2\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\mathbf{6}\mathbf{,}\mathbf{-}\mathbf{6}\mathbf{\}}$

$8+h^2=32$

• $s=\{2\sqrt{6},-2\sqrt{3}\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\mathbf{2}\sqrt{\mathbf{6}}\mathbf{,}\mathbf{-}\mathbf{2}\sqrt{\mathbf{6}}\mathbf{\}}$
• $s=\{2\sqrt{3},-2\sqrt{3}\}$
• $s=\{4\sqrt{6},3\sqrt{6}\}$

$n^2-\frac{2}{8}=\frac{3}{8}$

• $s=\{\sqrt{\frac{8}{5}},-\sqrt{\frac{8}{5}}\}$
• $s=\{\sqrt{\frac{1}{8}},-\sqrt{\frac{1}{8}}\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\sqrt{\frac{\mathbf{5}}{\mathbf{8}}}\mathbf{,}\mathbf{-}\sqrt{\frac{\mathbf{5}}{\mathbf{8}}}\mathbf{\}}$
• $s=\{\sqrt{\frac{5}{8}},-\sqrt{\frac{8}{5}}\}$

حل المعادلات التالية باستعمال خاصية الضرب الصفري:

$\begin{array}{rl}& (x-3)(x+1)=0\end{array}$

• $\mathrm{s}=\{-3,-1\}$
• $\mathrm{s}=\{3,1\}$
• $s=\{1,-3\}$
• $\mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{3}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{\}}$

$\begin{array}{rl}& (5t-7)(t+9)=0\end{array}$

• $s=\{\frac{7}{5},\frac{1}{9}\}$
• $s=\{\frac{5}{7},\frac{1}{9}\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\frac{\mathbf{7}}{\mathbf{5}}\mathbf{,}\mathbf{-}\mathbf{9}\mathbf{\}}$
• $s=\{\frac{-7}{5},9\}$

$\begin{array}{rl}& x^2-x=0\end{array}$

• $s=\{0,-1\}$
• $s=\{1,1\}$
• $s=\{1,-1\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{\}}$

$\begin{array}{rl}& 6z^2-36z=0\end{array}$

• $\mathit{s}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\mathbf{6}\mathbf{\}}$
• $s=\{0,-6\}$
• $s=\{6,-6\}$
• $s=\{6,-1\}$

$\begin{array}{rl}& \sqrt{12}{v}^2-2v=0\end{array}$

• $s=\{0,\frac{1}{\sqrt{6}}\}$
• $s=\{0,-\frac{1}{\sqrt{6}}\}$
• $\mathbf{s}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}\mathbf{\}}$
• $\mathrm{s}=\{0,-\frac{1}{\sqrt{3}}\}$

$\begin{array}{rl}& \frac{1}{5}r-\frac{1}{\sqrt{5}}{r}^2=0\end{array}$

• $\mathrm{s}=\{0,\frac{1}{5}\}$
• $s=\{0,\frac{-1}{\sqrt{5}}\}$
• $\mathrm{s}=\{0,\frac{-1}{5}\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\mathbf{0}\mathbf{,}\frac{\mathbf{1}}{\sqrt{\mathbf{5}}}\mathbf{\}}$

$(x+\sqrt{3})(x-\sqrt{2})=0$

• $s=\{\sqrt{3},\sqrt{2}\}$
• $\mathit{s}\mathbf{=}\mathbf{\{}\mathbf{-}\sqrt{\mathbf{3}}\mathbf{,}\sqrt{\mathbf{2}}\mathbf{\}}$
• $s=\{\sqrt{3},-\sqrt{2}\}$
• $s=\{-\sqrt{3},-\sqrt{2}\}$