اختبار الفصل

(1)- جد ناتج ضرب مقدار جبري في مقدار جبري كل منهما من حدين:

$(x+5{)}^2$

$(x+5{)}^2=x^2+10x+25$

$(v-\sqrt{2})(v+\sqrt{2})$

$(v-\sqrt{2})(v+\sqrt{2})=v^2-2$

$(2-x)(5-x)$

$(2-x)(5-x)=10-7x+x^2$

$(2y-3)(y+9)$

$(2y-3)(y+9)=2y^2+15y-27$

(2)- جد ناتج ضرب مقدار جبري من حدين في مقدار جبري من ثلاثة حدود:

$(x+11)(x^2-11x+121)$

$(x+11)(x^2-11x+121)=x^3+{11}^3=x^3+1331$

$(\frac{1}{3}-y)(\frac{1}{9}+\frac{1}{3}y+y^2)$

$(\frac{1}{3}-y)(\frac{1}{9}+\frac{1}{3}y+y^2)=\frac{1}{27}-y^3$

$(y-1{)}^3$

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

${(z+\frac{1}{4})}^3$

${(z+\frac{1}{4})}^3=z^3+\frac{3}{4}{z}^2+\frac{3}{16}z+\frac{1}{64}$

(3)- حلل المقدار باستعمال العامل المشترك الأكبر (GCF) وتحقق من صحة الحل:

$8x^2-12x$

$$\begin{gathered} 8x^2-12x=4x(2x-3) \\ 4x(2x-3)=8x^2-12x \mathrm{التحقق} \end{gathered}$$

$7y^3+14y^2-21y$

$$\begin{gathered} 7y^3+14y^2-21y=7y(y^2+2y-3)=7y(y+3)(y-1) \\ 7y(y+3)(y-1)=7y(y^2+2y-3)7y^3+14y^2-21y \mathrm{التحقق} \end{gathered}$$

$\sqrt{18}{\mathrm{z}}^3\mathrm{r}+\sqrt{2}({\mathrm{zr}}^2-\mathrm{zr})$

$\begin{array}{rl}& \sqrt{18}{\mathrm{z}}^3\mathrm{r}+\sqrt{2}({\mathrm{zr}}^2-\mathrm{zr})\\ =& 3\sqrt{2}{\mathrm{z}}^3\mathrm{r}+\sqrt{2}{\mathrm{zr}}^2-\sqrt{2}\mathrm{zr}\\ =& \sqrt{2}\mathrm{zr}(3{\mathrm{z}}^2+\mathrm{r}-1)\\ & \sqrt{2}\mathrm{zr}(3{\mathrm{z}}^2+\mathrm{r}-1)=3\sqrt{2}{\mathrm{z}}^3\mathrm{r}+\sqrt{2}{\mathrm{zr}}^2-\sqrt{2}\mathrm{zr} \mathrm{التحقق}\end{array}$

(4)- حلل المقدار باستعمال ثنائية الحد كعامل مشترك أكبر:

$\frac{2}{3}(y+5)+\frac{1}{3}y(y+5)$

$\begin{array}{rl}& \frac{2}{3}(y+5)+\frac{1}{3}y(y+5)\\ & =(y+5)(\frac{2}{3}+\frac{1}{3}y)\end{array}$

$\sqrt{5}z(z^2-1)-\sqrt{2}{z}^2(z^2-1)$

$\begin{array}{rl}& \sqrt{5}z(z^2-1)-\sqrt{2}{z}^2(z^2-1)\\ & =z(z^2-1)(\sqrt{5}-\sqrt{2}z)=(z-1)(z+1)(\sqrt{5}-\sqrt{2}z)\end{array}$

(5)- حلل المقدار باستعمال خاصية التجميع:

$6x^4-18x^3+10x-30$

$\begin{array}{rl}6x^4-18x^3+10x-30& =6x^3(x-3)+10(x-3)\\ & =(x-3)(6x^3+10)\end{array}$

$56-8y+14y^2-2y^3$

$\begin{array}{rl}& 56-8y+14y^2-2y^3=8(7-y)+2y^2(7-y)\\ & =(7-y)(8+2y^2)\end{array}$

(6)- حلل المقدار بالتجميع مع المعكوس:

$9x^3-6x^2+8-12x$

$\begin{array}{rl}& 9x^3-6x^2+8-12x=3x^2(3x-2)-4(3x-2)\\ & =(3x-2)(3x^2-4)\end{array}$

$\sqrt{11}{z}^3-\sqrt{44}{z}^2+5(2-z)$

$\begin{array}{rl}& \sqrt{11}{z}^3-\sqrt{44}{z}^2+5(2-z)=\sqrt{11}{z}^2(z-2)-5(z-2)\\ & =(z-2)(\sqrt{11}{z}^2-5)\end{array}$

(7)- حلل كل مقدار جبري من المقادير الآتية:

$16-x^2$

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

$\frac{1}{3}{z}^2-\frac{1}{27}$

$\frac{1}{3}{z}^2-\frac{1}{27}=\frac{1}{27}(9z^2-1)=\frac{1}{27}(3z+1)(3z-1)$

$\frac{1}{16}v-\frac{1}{2}{v}^4$

$\begin{array}{rl}\frac{1}{16}v-\frac{1}{2}{v}^4& =\frac{1}{16}v(1-8v^3)\\ & =\frac{1}{16}v(1-2v)(1+2v+4v^2)\end{array}$

$8x^3-\frac{1}{125}$

$8x^3-\frac{1}{125}=(2x+\frac{1}{5})(4x^2+\frac{2}{5}x+\frac{1}{25})$

$81-18y+y^2$

$81-18y+y^2=(9-y{)}^2$

$7z^2-36z+5$

$7z^2-36z+5=(7z-1)(z-5)$

(8)- حدد أي من المقادير الجبرية التالية يمثل مربعاً كاملاً وحلله:

$25x^2+30x+9$

$$\begin{gathered} 25x^2+30x+9=(5x+3{)}^2 \\ 2(5x\left)\right(3)=30x \end{gathered}$$

يمثل مربعاً كاملاً.

$49-14y+y^2$

$$\begin{gathered} 49-14y+y^2=(7-y{)}^2 \\ 2(7\left)\right(y)=14y \end{gathered}$$

يمثل مربعاً كاملاً.

$4{\mathrm{v}}^2+4\sqrt{5}\mathrm{v}+5$

$$\begin{gathered} 4v^2+4\sqrt{5}v+5=(2v+\sqrt{5}{)}^2 \\ 2(2v\left)\right(\sqrt{5})=4\sqrt{5}v \end{gathered}$$

يمثل مربعاً كاملاً

(9)- اكتب الحد المفقود في المقدار الجبري ${\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$ ليصبح مربعاً كاملاً وحلله:

$x^2+\dots .+81$

$$\begin{gathered} x^2+\dots +81\Rightarrow x^2+18x+81=(x+9{)}^2 \\ 2(x\left)\right(9)=18x \end{gathered}$$

الحد الوسط = $18x=2\sqrt{81x^2}$

$36-12y+\dots$

$$\begin{gathered} 36-12y+\dots \Rightarrow 36-12y+y^2=(6-y{)}^2 \\ 2(6\left)\right(y)=12y \end{gathered}$$

الحد الأخير = $y^2={\left(\frac{12y}{2\left(6\right)}\right)}^2$

$7-\dots +4z^2$

$$\begin{gathered} 7-\dots +4z^2\Rightarrow 7-4\sqrt{7}z+4z^2=(\sqrt{7}-2z{)}^2 \\ 2(\sqrt{7}\left)\right(2z)=4\sqrt{7}z \end{gathered}$$

الحد الوسط = $4\sqrt{7}z=2\sqrt{7\left(4z^2\right)}$

(10)- حلل كل مقدار من المقادير الجبرية الآتية:

$x^2+7x+10$

$x^2+7x+10=(x+5)(x+2)$

$x^2-5\sqrt{3}x+18$

$\begin{array}{rl}{x}^2-5\sqrt{3}x+18& =x^2-5\sqrt{3}x+\left(2\sqrt{3}\right)\left(3\sqrt{3}\right)\\ & =(x-2\sqrt{3})(x-3\sqrt{3})\end{array}$

$2v^2+9v+7$

$2v^2+9v+7=(2v+7)(v+1)$

$32-16x+2x^2$

$\begin{array}{rl}32-16x+2x^2& =2(16-8x+x^2)\\ & =2(4-x{)}^2\end{array}$

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

$\begin{array}{rl}\frac{1}{4}{y}^2-2y+3& =\frac{1}{4}(y^2-8y+12)\\ & =\frac{1}{4}(y-6)(y-2)\end{array}$

$12-7\sqrt{2}v+2v^2$

$\begin{array}{rl}& 12-7\sqrt{2}v+2v^2\\ & =(4-\sqrt{2}v)(3-\sqrt{2}v)\end{array}$

$8+27x^3$

$8+27x^3=(2+3x)(4-6x+9x^2)$

$125{\mathrm{y}}^3-1$

$125y^3-1=(5y-1)(25y^2+5y+1)$

$\frac{1}{v^3}-\frac{8}{27}$

$\frac{1}{v^3}-\frac{8}{27}=(\frac{1}{v}-\frac{2}{3})(\frac{1}{v^2}+\frac{2}{3v}+\frac{4}{9})$

$1+0.125y^3$

$1+0.125y^3=(1+0.5y)(1-0.5y+0.25y^2)$

$z^3-0.027$

$z^3-0.027=(z-0.3)(z^2+0.3z+0.09)$

$3-\frac{1}{9}{v}^3$

$3-\frac{1}{9}{v}^3=\frac{1}{9}(27-v^3)=\frac{1}{9}(3-v)(9+3v+v^2)$

(11)- اكتب كل مقدار من المقادير التالية على أبسط صورة:

$\frac{27-8z^3}{4z^2-9}÷\frac{9+6z+4z^2}{9+6z}$

$\begin{array}{rl}& \frac{27-8z^3}{-(9-4z^2)}÷\frac{9+6z+4z^2}{9+6z}\\ & =\frac{(3-2z)(9+6z+4z^2)}{-(3-2z)(3+2z)}\times \frac{3(3+2z)}{9+6z+4z^2}=-3\end{array}$

$\frac{7}{x^2-25}-\frac{6}{x^2+10x+25}$

$\begin{array}{rl}& \frac{7}{x^2-25}-\frac{6}{x^2+10x+25}\\ & =\frac{7}{(x-5)(x+5)}-\frac{6}{(x+5{)}^2}=\frac{7(x+5)-6(x-5)}{(x+5{)}^2(x-5)}\\ & =\frac{7x+35-6x+30}{(x+5{)}^2(x-5)}=\frac{x+65}{(x+5{)}^2(x-5)}\end{array}$

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

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

$\frac{z+3}{z+5}-\frac{z-5}{z-3}+\frac{1}{z^2+2z-15}$

$\begin{array}{rl}& \frac{z+3}{z+5}-\frac{z-5}{z-3}+\frac{1}{z^2+2z-15}\\ & =\frac{z+3}{z+5}-\frac{z-5}{z-3}+\frac{1}{(z+5)(z-3)}\\ & =\frac{(z+3)(z-3)-(z-5)(z+5)+1}{(z+5)(z-3)}\\ & =\frac{z^2-9-z^2+25+1}{(z+5)(z-3)}\\ & =\frac{17}{(z+5)(z-3)}\end{array}$