تدرب وحل التمرينات

(1)- حلل كل مقدار من المقادير الآتية إلى أبسط صورة:

$25-4x^2$

$25-4x^2=(5+2x)(5-2x)$

$y^2-121$

$y^2-121=(y-11)(y+11)$

$x^2-16z^2$

$x^2-16z^2=(x+4z)(x-4z)$

$12-3{\mathrm{t}}^2$

$12-3t^2=3(4-t^2)=3(2+t)(2-t)$

$8y^{3}x-2x^{3}y$

$\begin{array}{rl}8y^{3}x-2x^{3}y& =2xy(4y^2-x^2)\\ & =2xy(2y+x)(2y-x)\end{array}$

$\frac{1}{4}{y}^2-\frac{1}{8}$

$\begin{array}{rl}\frac{1}{4}{y}^2-\frac{1}{8}& =\frac{1}{8}(2y^2-1)\\ & =\frac{1}{8}(\sqrt{2}y+1)(\sqrt{2}y-1)\end{array}$

$\frac{1}{3}{z}^5-\frac{1}{12}z$

$\begin{array}{rl}\frac{1}{3}{z}^3-\frac{1}{12}z& =\frac{1}{12}z(4z^4-1)\\ & =\frac{1}{12}z(2z^2+1)(2z^2-1)\\ & =\frac{1}{12}z(2z^2+1)(\sqrt{2}z+1)(\sqrt{2}z-1)\end{array}$

$4x^2+20x+25$

$4x^2+20x+25=(2x+5{)}^2$

$3z^2-6z+3$

$3z^2-6z+3=3(z^2-2z+1)=3(z-1{)}^2$

$16n^2+8\sqrt{3}n+3$

$16n^2+8\sqrt{3}n+3=(4n+\sqrt{3}{)}^2$

$4t^3-12t^2+9t$

$\begin{array}{rl}4t^3-12t^2+9t& =t(4t^2-12t+9)\\ & =t(2t-3{)}^2\end{array}$

$1-4m+4m^2$

$1-4m+4m^2=(1-2m{)}^2$

(2)- حدد أي مقدار من المقادير التالية يمثل مربعاً كاملاً وحلله:

$4x^2+18x+16$

$$\begin{gathered} 4x^2-18x+16= \\ 2\left(2x\right)\left(4\right)=16x\ne 18x \end{gathered}$$

لا يمثل مربعاً كاملاً.

$y^2+10y+25$

$$\begin{gathered} {\mathrm{y}}^2-10\mathrm{y}+25=(\mathrm{y}+5{)}^2 \\ 2(y\left)\right(5)=10y \end{gathered}$$

يمثل مربعاً كاملاً.

$49-7v+v^2$

$2\left(7\right)\left(v\right)=14v\ne 7v$

لا يمثل مربعاً كاملاً.

$2h^2-12h-18$

لا يمثل مربعاً كاملاً لأن الحد الأخير سالب (18-)

$4v^2+4v+4$

$$\begin{gathered} 4v^2+4v+4 \\ 2\left(2v\right)\left(2\right)=8v\ne 4v \end{gathered}$$

لا يمثل مربعاً كاملاً.

$3-2\sqrt{3}z+z^2$

$$\begin{gathered} 3-2\sqrt{3}+z^2=(\sqrt{3}-z{)}^2 \\ 2(\sqrt{3}\left)\right(z)=2\sqrt{3}z \end{gathered}$$

يمثل مربعاً كاملاً.

(3)- اكتب الحد المفقود في المقدار الجبري ax2 + bx + c ليصبح مربعاً كاملاً وحلله:

$y^2+\dots +36$

$\begin{array}{rl}{y}^2+\dots \dots ..+36& =y^2+2\left(y\right)\left(6\right)+6^2\\ & =y^2+12y+36\\ & =(y+6{)}^2\end{array}$

الحد الوسط = $12y=\left(2\right)\left(y\right)\left(6\right)$

$25-20x+\dots$

$\begin{array}{rl}25-20x+\dots \dots .& =5^2-2\left(5\right)\left(2x\right)+2^2{x}^2\\ & =25-20x+4x^2\\ & =(5-2x{)}^2\end{array}$

الحد الأخير = $4x^2={\left(\frac{20x}{\left(2\right)\left(5\right)}\right)}^2$

$25-4x^{2}4v^2+8v+...$

$\begin{array}{rl}& 4{\mathrm{v}}^2+8\mathrm{v}+\dots \dots =2^2{\mathrm{v}}^2+2\left(2\mathrm{v}\right)\left(2\right)+2^2\\ & =4{\mathrm{v}}^2+8\mathrm{v}+4=(2\mathrm{v}+2{)}^2\end{array}$

الحد الأخير = $4={\left(\frac{8v}{\left(2\right)\left(2v\right)}\right)}^2$

$5-\dots +16x^2$

$$\begin{gathered} 5-\dots \dots ..+16x^2=(\sqrt{5}{)}^2-2(\sqrt{5}\left)\right(4x)+4^2{x}^2=5-8\sqrt{5}x+16x^2 \\ =(\sqrt{5}-4x{)}^2 \end{gathered}$$

الحد الوسط = $8\sqrt{5}x=\left(2\right)\left(\sqrt{5}\right)\left(4x\right)=$

$81+18z+\dots$

$81+18z+\dots \dots {.9}^2+2\left(9\right)\left(z\right)+z^2=81+18z+z^2=(9+z{)}^2$

الحد الأخير = ${\mathrm{z}}^2={\left(\frac{18\mathrm{z}}{\left(2\right)\left(9\right)}\right)}^2$

$9h^2+6\sqrt{2}h+\dots$

$$\begin{gathered} 9h^2+6\sqrt{2}h+\dots \dots =3^2{h}^2+2\left(3h\right)\left(\sqrt{2}\right)+(\sqrt{2}{)}^2=9h^2+6\sqrt{2}h+2 \\ =(3h+\sqrt{2}{)}^2 \end{gathered}$$

الحد الأخير = $2={\left(\frac{6\sqrt{2}h}{\left(2\right)\left(3\right)h}\right)}^2$