تدرب وحل التمرينات

اجمع الحدود الجبرية المتشابهة الآتية:

$\begin{array}{rl}& -5{\mathrm{x}}^2,2{\mathrm{x}}^2,\frac{1}{5}{\mathrm{x}}^2\end{array}$

$\begin{array}{rl}& -5x^2+2x^2+\frac{1}{5}{x}^2\\ & =(-5+2+\frac{1}{5})x^2\\ & =\frac{-25+10+1}{5}{x}^2=\frac{-14}{5}{x}^2\end{array}$

$\begin{array}{rl}& \sqrt{169}{\mathrm{x}}^4\mathrm{y},13{\mathrm{x}}^4\mathrm{y},{\mathrm{x}}^4\mathrm{y}\end{array}$

$\begin{array}{rl}& \sqrt{169}{x}^{4}y+13x^{4}y+x^{4}y\\ =& (\sqrt{169}+13+1)x^{4}y\\ =& (13+13+1)x^{4}y\\ =& 27x^{4}y\end{array}$

$\begin{array}{rl}& \frac{1}{6}\mathrm{xyz},10\mathrm{xyz},20\mathrm{xyz}\end{array}$

$\begin{array}{rl}& \frac{1}{6}xyz+20xyz=(\frac{1}{6}+20)xyz\\ & =20\frac{1}{6}xyz\end{array}$

$\begin{array}{rl}& |-9|{\mathrm{r}}^2\mathrm{v},|-8|{\mathrm{r}}^2\mathrm{v},\left|3\right|{\mathrm{r}}^2\mathrm{v}\end{array}$

$\begin{array}{rl}& |-9|r^{2}v+|-8|r^{2}v+\left|3\right|r^{2}v\\ & =\left(\right|-9|+|-8|+|3\left|\right)r^{2}v\\ & =(9+8+3)r^{2}v=20r^{2}v\end{array}$

$\begin{array}{rl}& \frac{3}{2}\mathrm{gh},\frac{6}{5}\mathrm{gh},\frac{1}{4}\mathrm{gh}\end{array}$

$\begin{array}{rl}& \frac{3}{2}gh+\frac{6}{5}gh+\frac{1}{4}gh\\ & =(\frac{3}{2}+\frac{6}{5}+\frac{1}{4})gh\\ & =\frac{30+24+5}{20}gh=\frac{59}{20}gh\end{array}$

$\begin{array}{rl}& \frac{3}{10}{\mathrm{hk}}^2,\frac{1}{2}h{\mathrm{k}}^2,\frac{1}{5}{\mathrm{hk}}^2\end{array}$

$\begin{array}{rl}& \frac{3}{10}H{K}^2+\frac{1}{2}H{K}^2+\frac{1}{5}H{K}^2\\ & =(\frac{3}{10}+\frac{1}{2}+\frac{1}{5})H{K}^2\\ & =\frac{3+5+2}{10}H{K}^2=\frac{10}{10}H{K}^2=H{K}^2\end{array}$

جد ناتج طرح الحد الجبري الأول من الحد الجبري الثاني في كل مما يأتي:

$\begin{array}{rl}& 24{\mathrm{z}}^2\mathrm{w},48{\mathrm{z}}^2\mathrm{w}\end{array}$

$\begin{array}{rl}& 48z^{2}w-24z^{2}w\\ =& (48-24)z^{2}w=24z^{2}w\end{array}$

$\begin{array}{rl}& 20{\mathrm{r}}^2\mathrm{v},40{\mathrm{r}}^2\mathrm{v}\end{array}$

$\begin{array}{rl}& 40R^{2}V-20R^{2}V\\ =& (40-20)R^{2}V=20R^{2}V\end{array}$

$\begin{array}{rl}& 4\mathrm{xyz},16\mathrm{xyz}\end{array}$

$\begin{array}{rl}& 16xyz-4xyz\\ =& (16-4)xyz=12xyz\end{array}$

$\begin{array}{rl}& |-9|\mathrm{ba},|-20|\mathrm{ab}\end{array}$

$\begin{array}{rl}& |-20|AB-|-9|BA\\ & =\left(\right|-20|-|-9\left|\right)BA\\ & =(20-9)BA=11BA\end{array}$

$\begin{array}{rl}& \frac{1}{5}{\mathrm{h}}^2\mathrm{k},\frac{1}{25}{\mathrm{kh}}^2\end{array}$

$(\frac{1}{25}-\frac{1}{5})H^{2}K=\frac{1-5}{25}{H}^{2}K=\frac{-4}{25}{H}^{2}K$

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

$(12-\frac{1}{3})A{B}^2=\frac{36-1}{3}A{B}^2=\frac{35}{3}A{B}^2$