جواب کاردرکلاس صفحه 31 درس 2 ریاضی یازدهم تجربی (هندسه)

الف

\( \Rightarrow \frac{a}{b} = \frac{c}{d}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop \Rightarrow \limits_{bd \ne 0}^{ \times bd} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{a}{{\not b}} \times \not bd = \frac{c}{{\not d}} \times b\not d \Rightarrow ad = bc\)

ب

\( \Rightarrow ad = bc{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop \Rightarrow \limits_{bd \ne 0}^{ \times \frac{1}{{bd}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} a\not d \times \frac{1}{{b\not d}} = \not bc \times \frac{1}{{\not bd}} \Rightarrow \frac{a}{b} = \frac{c}{d}\)

پ

\( \Rightarrow \frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop \Rightarrow \limits_{ac \ne 0}^{ \times \frac{1}{{ac}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \not ad \times \frac{1}{{\not ac}} = b\not c \times \frac{1}{{a\not c}} \Rightarrow \frac{b}{a} = \frac{d}{c}\)

ت

\( \Rightarrow \frac{a}{b} = \frac{c}{d} \Rightarrow ad = bc\left\{ {\begin{array}{*{20}{l}}{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop \Rightarrow \limits_{ab \ne 0}^{ \times \frac{1}{{ab}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{\not ad}}{{\not ab}} = \frac{{\not bc}}{{a\not b}} \Rightarrow \frac{c}{a} = \frac{d}{b}}\\{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \mathop \Rightarrow \limits_{cd \ne 0}^{ \times \frac{1}{{cd}}} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \frac{{a\not d}}{{c\not d}} = \frac{{\not bc}}{{c\not b}} \Rightarrow \frac{a}{c} = \frac{b}{d}}\end{array}} \right.\)

ث

\( \Rightarrow \frac{a}{b} = \frac{c}{d} \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}\frac{a}{b} + 1 = \frac{c}{d} + 1 \Rightarrow \frac{{a + b}}{b} = \frac{{c + d}}{d}\\\end{array}\\{\frac{b}{a} = \frac{d}{c} \Rightarrow \frac{b}{a} + 1 = \frac{d}{c} + 1 \Rightarrow \frac{{a + b}}{a} = \frac{{c + d}}{c} \Rightarrow \frac{a}{{a + b}} = \frac{c}{{c + d}}}\end{array}} \right.\)

ج

\( \Rightarrow \frac{a}{b} = \frac{c}{d} \Rightarrow \left\{ {\begin{array}{*{20}{l}}\begin{array}{l}\frac{a}{b} - 1 = \frac{c}{d} - 1 \Rightarrow \frac{{a - b}}{b} = \frac{{c - d}}{d}\\\end{array}\\{\frac{b}{a} = \frac{d}{c} \Rightarrow \frac{b}{a} - 1 = \frac{d}{c} - 1 \Rightarrow \frac{{b - a}}{a} = \frac{{d - c}}{c} \Rightarrow \frac{a}{{b - a}} = \frac{c}{{d - c}}}\end{array}} \right.\)

الف

\(\frac{5}{{12}} = \frac{{15}}{{42}}\;\; \Rightarrow \;\;5 \times \;42\; = 15 \times \;14\)

ب

\(3 \times 40 = 12 \times 10\;\; \Rightarrow \;\;\frac{3}{{\;10\;\;}} = \frac{{12}}{{\;40\;\;}}\)

پ

\(\frac{7}{{10}} = \frac{{21}}{{30}}\;\; \Rightarrow \;\;\frac{{10}}{7} = \frac{{30}}{{21}}\)

ت

\(\frac{6}{{11}} = \frac{{18}}{{33}}\;\; \Rightarrow \;\;\frac{6}{{18}} = \frac{{11}}{{33}}\;\;,\;\;\frac{{33}}{{11}} = \frac{{18}}{6}\)

ث

\(\frac{4}{{14}} = \frac{{10}}{{35}}\;\; \Rightarrow \;\;\frac{{18}}{{14}} = \frac{{45}}{{35}}\;\;,\;\;\frac{4}{{18}} = \frac{{10}}{{45}}\)

ج

\(\frac{5}{{12}} = \frac{{10}}{{24}}\;\; \Rightarrow \;\;\frac{{ - 7}}{{12}} = \frac{{ - 14}}{{24}}\;\;,\;\;\frac{5}{{ - 7}} = \frac{{10}}{{ - 14}}\)