[tex]\begin{aligned}\sf a.\ &\frac{2a}{3}\times\frac{4a}{7}=\boxed{\,\frac{8a^2}{21}\,}\\\sf b.\ &\frac{6a+1}{3x}\times\frac{bx}{2}=\boxed{\,\frac{6ab+b}{6}\,}\\\sf c.\ &\frac{2a-1}{3b}\times\frac{bc}{2}=\boxed{\,\frac{2ac-c}{6}\,}\\\sf d.\ &\frac{a+3}{2b}\,:\,\frac{ab}{3b}=\boxed{\,\frac{3a+9}{2ab}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\\\sf e.\ &\frac{2a}{3}\,:\,\frac{4a}{7}=\boxed{\,\frac{7}{6}\,}\\\sf f.\ &\frac{2a+b}{2a}\,:\,\frac{3}{b}=\boxed{\,\frac{2ab+b^2}{6a}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\end{aligned}[/tex]
Penyederhanaan Bentuk Aljabar
Untuk semua soal yang diberikan, kita perlu mengingat beberapa aturan operasi aritmatika perkalian dan pembagian pada pecahan.
Penyelesaian
Soal a.
[tex]\begin{aligned}\frac{2a}{3}\times\frac{4a}{7}&=\frac{2a\times4a}{3\times7}\\&=\boxed{\,\frac{8a^2}{21}\,}\end{aligned}[/tex]
Soal b.
[tex]\begin{aligned}\frac{6a+1}{3x}\times\frac{bx}{2}&=\frac{(6a+1)\times b\cancel{x}}{2\times3\cancel{x}}\\&=\frac{(6a+1)b}{6}\\&=\boxed{\,\frac{6ab+b}{6}\,}\\&=\frac{\cancel{6}ab}{\cancel{6}}+\frac{b}{6}\\&=\boxed{\,ab+\frac{b}{6}\,}\end{aligned}[/tex]
Soal c.
[tex]\begin{aligned}\frac{2a-1}{3b}\times\frac{bc}{2}&=\frac{(2a-1)\times \cancel{b}c}{3\cancel{b}\times2}\\&=\frac{(2a-1)c}{3\times2}\\&=\boxed{\,\frac{2ac-c}{6}\,}\end{aligned}[/tex]
Soal d.
[tex]\begin{aligned}\frac{a+3}{2b}\,:\,\frac{ab}{3b}&=\frac{a+3}{2b}\times\frac{3\cancel{b}}{a\cancel{b}}\\&=\frac{3(a+3)}{2ab}\\&=\boxed{\,\frac{3a+9}{2ab}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\end{aligned}[/tex]
Soal e.
[tex]\begin{aligned}\frac{2a}{3}\,:\,\frac{4a}{7}&=\frac{2a}{3}\times\frac{7}{4a}\\&=\frac{\cancel{2a}}{3}\times\frac{7}{2\cdot\cancel{2a}}\\&=\frac{7}{3\times2}\\&=\boxed{\,\frac{7}{6}\,}\end{aligned}[/tex]
Soal f.
[tex]\begin{aligned}\frac{2a+b}{2a}\,:\,\frac{3}{b}&=\frac{2a+b}{2a}\times\frac{b}{3}\\&=\frac{(2a+b)b}{2a\times3}\\&=\boxed{\,\frac{2ab+b^2}{6a}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\end{aligned}[/tex] [tex]\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}[/tex]
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[tex]\begin{aligned}\sf a.\ &\frac{2a}{3}\times\frac{4a}{7}=\boxed{\,\frac{8a^2}{21}\,}\\\sf b.\ &\frac{6a+1}{3x}\times\frac{bx}{2}=\boxed{\,\frac{6ab+b}{6}\,}\\\sf c.\ &\frac{2a-1}{3b}\times\frac{bc}{2}=\boxed{\,\frac{2ac-c}{6}\,}\\\sf d.\ &\frac{a+3}{2b}\,:\,\frac{ab}{3b}=\boxed{\,\frac{3a+9}{2ab}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\\\sf e.\ &\frac{2a}{3}\,:\,\frac{4a}{7}=\boxed{\,\frac{7}{6}\,}\\\sf f.\ &\frac{2a+b}{2a}\,:\,\frac{3}{b}=\boxed{\,\frac{2ab+b^2}{6a}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\end{aligned}[/tex]
Penjelasan
Penyederhanaan Bentuk Aljabar
Untuk semua soal yang diberikan, kita perlu mengingat beberapa aturan operasi aritmatika perkalian dan pembagian pada pecahan.
(a/b) : (c/d) = (a/b) × (d/c) = (ad)/(bc)
atau
(a/b) : (c/d) = (a/c) : (b/d) = (a/c) × (d/b) = (ad)/(bc)
Penyelesaian
Soal a.
[tex]\begin{aligned}\frac{2a}{3}\times\frac{4a}{7}&=\frac{2a\times4a}{3\times7}\\&=\boxed{\,\frac{8a^2}{21}\,}\end{aligned}[/tex]
Soal b.
[tex]\begin{aligned}\frac{6a+1}{3x}\times\frac{bx}{2}&=\frac{(6a+1)\times b\cancel{x}}{2\times3\cancel{x}}\\&=\frac{(6a+1)b}{6}\\&=\boxed{\,\frac{6ab+b}{6}\,}\\&=\frac{\cancel{6}ab}{\cancel{6}}+\frac{b}{6}\\&=\boxed{\,ab+\frac{b}{6}\,}\end{aligned}[/tex]
Soal c.
[tex]\begin{aligned}\frac{2a-1}{3b}\times\frac{bc}{2}&=\frac{(2a-1)\times \cancel{b}c}{3\cancel{b}\times2}\\&=\frac{(2a-1)c}{3\times2}\\&=\boxed{\,\frac{2ac-c}{6}\,}\end{aligned}[/tex]
Soal d.
[tex]\begin{aligned}\frac{a+3}{2b}\,:\,\frac{ab}{3b}&=\frac{a+3}{2b}\times\frac{3\cancel{b}}{a\cancel{b}}\\&=\frac{3(a+3)}{2ab}\\&=\boxed{\,\frac{3a+9}{2ab}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\end{aligned}[/tex]
Soal e.
[tex]\begin{aligned}\frac{2a}{3}\,:\,\frac{4a}{7}&=\frac{2a}{3}\times\frac{7}{4a}\\&=\frac{\cancel{2a}}{3}\times\frac{7}{2\cdot\cancel{2a}}\\&=\frac{7}{3\times2}\\&=\boxed{\,\frac{7}{6}\,}\end{aligned}[/tex]
Soal f.
[tex]\begin{aligned}\frac{2a+b}{2a}\,:\,\frac{3}{b}&=\frac{2a+b}{2a}\times\frac{b}{3}\\&=\frac{(2a+b)b}{2a\times3}\\&=\boxed{\,\frac{2ab+b^2}{6a}\,,\ a\ne0{\sf\ dan\ }b\ne0\,}\end{aligned}[/tex]
[tex]\overline{\begin{array}{l}\small\textsf{Duc In Altum}\\\small\text{bertolaklah\;ke\;tempat}\\\small\text{yang\;lebih\;dalam}\end{array}}[/tex]