[tex]\begin{aligned}&\textsf{Hasil penyederhanaan}:\\&{\sf1.\ \ }\frac{2\sqrt{50}+4\sqrt{8}}{2\sqrt{32}}=\boxed{\,\bf\frac{9}{4}\,}=\boxed{\,\bf2\frac{1}{4}\,}\\&{\sf2.\ \ }\frac{5\sqrt{3}+3\sqrt{27}}{2\sqrt{2}}=\boxed{\,\bf\frac{7\sqrt{6}}{2}\,}\\&{\sf3.\ \ }\frac{2\sqrt{5}\times5\sqrt{10}}{2\sqrt{2}}=\boxed{\,\bf25\,}\end{aligned}[/tex]
Menyederhanakan Bentuk Akar
Nomor 1
[tex]\begin{aligned}\frac{2\sqrt{50}+4\sqrt{8}}{2\sqrt{32}}&=\frac{2\sqrt{25\cdot2}+4\sqrt{4\cdot2}}{2\sqrt{16\cdot2}}\\&=\frac{2\cdot5\sqrt{2}+4\cdot2\sqrt{2}}{2\cdot4\sqrt{2}}\\&=\frac{10\sqrt{2}+8\sqrt{2}}{8\sqrt{2}}\\&=\frac{10\cancel{\sqrt{2}}}{8\cancel{\sqrt{2}}}+\frac{\cancel{8\sqrt{2}}}{\cancel{8\sqrt{2}}}\\&=\frac{10}{8}+1\\&=\frac{5}{4}+1\\&=\frac{5+4}{4}\\\frac{2\sqrt{50}+4\sqrt{8}}{2\sqrt{32}}&=\boxed{\,\bf\frac{9}{4}\,}=\boxed{\,\bf2\frac{1}{4}\,}\end{aligned}[/tex]
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Nomor 2
[tex]\begin{aligned}\frac{5\sqrt{3}+3\sqrt{27}}{2\sqrt{2}}&=\frac{5\sqrt{3}+3\sqrt{9\cdot3}}{2\sqrt{2}}\\&=\frac{5\sqrt{3}+3\cdot3\sqrt{3}}{2\sqrt{2}}\\&=\frac{5\sqrt{3}+9\sqrt{3}}{2\sqrt{2}}\\&=\frac{(5+9)\sqrt{3}}{2\sqrt{2}}\\&=\frac{14\sqrt{3}}{2\sqrt{2}}\\&=\frac{\cancel{2}\cdot7\sqrt{3}}{\cancel{2}\sqrt{2}}\\&=\frac{7\sqrt{3}}{\sqrt{2}}\\&=\frac{7\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\&=\frac{7\sqrt{3}\sqrt{2}}{2}\end{aligned}[/tex][tex]\begin{aligned}\frac{5\sqrt{3}+3\sqrt{27}}{2\sqrt{2}}&=\boxed{\,\bf\frac{7\sqrt{6}}{2}\,}\end{aligned}[/tex]
Nomor 3
[tex]\begin{aligned}\frac{2\sqrt{5}\times5\sqrt{10}}{2\sqrt{2}}&=\frac{2\sqrt{5}\cdot5\sqrt{2}\sqrt{5}}{2\sqrt{2}}\\&=\frac{(\cancel{2}\cdot5)\sqrt{5}\sqrt{5}\cancel{\sqrt{2}}}{\cancel{2}\cancel{\sqrt{2}}}\\&=5\cdot\sqrt{5}\sqrt{5}\\&=5\cdot5\\\frac{2\sqrt{5}\times5\sqrt{10}}{2\sqrt{2}}&=\boxed{\,\bf25\,}\end{aligned}[/tex]
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Verified answer
[tex]\begin{aligned}&\textsf{Hasil penyederhanaan}:\\&{\sf1.\ \ }\frac{2\sqrt{50}+4\sqrt{8}}{2\sqrt{32}}=\boxed{\,\bf\frac{9}{4}\,}=\boxed{\,\bf2\frac{1}{4}\,}\\&{\sf2.\ \ }\frac{5\sqrt{3}+3\sqrt{27}}{2\sqrt{2}}=\boxed{\,\bf\frac{7\sqrt{6}}{2}\,}\\&{\sf3.\ \ }\frac{2\sqrt{5}\times5\sqrt{10}}{2\sqrt{2}}=\boxed{\,\bf25\,}\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
Menyederhanakan Bentuk Akar
Nomor 1
[tex]\begin{aligned}\frac{2\sqrt{50}+4\sqrt{8}}{2\sqrt{32}}&=\frac{2\sqrt{25\cdot2}+4\sqrt{4\cdot2}}{2\sqrt{16\cdot2}}\\&=\frac{2\cdot5\sqrt{2}+4\cdot2\sqrt{2}}{2\cdot4\sqrt{2}}\\&=\frac{10\sqrt{2}+8\sqrt{2}}{8\sqrt{2}}\\&=\frac{10\cancel{\sqrt{2}}}{8\cancel{\sqrt{2}}}+\frac{\cancel{8\sqrt{2}}}{\cancel{8\sqrt{2}}}\\&=\frac{10}{8}+1\\&=\frac{5}{4}+1\\&=\frac{5+4}{4}\\\frac{2\sqrt{50}+4\sqrt{8}}{2\sqrt{32}}&=\boxed{\,\bf\frac{9}{4}\,}=\boxed{\,\bf2\frac{1}{4}\,}\end{aligned}[/tex]
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Nomor 2
[tex]\begin{aligned}\frac{5\sqrt{3}+3\sqrt{27}}{2\sqrt{2}}&=\frac{5\sqrt{3}+3\sqrt{9\cdot3}}{2\sqrt{2}}\\&=\frac{5\sqrt{3}+3\cdot3\sqrt{3}}{2\sqrt{2}}\\&=\frac{5\sqrt{3}+9\sqrt{3}}{2\sqrt{2}}\\&=\frac{(5+9)\sqrt{3}}{2\sqrt{2}}\\&=\frac{14\sqrt{3}}{2\sqrt{2}}\\&=\frac{\cancel{2}\cdot7\sqrt{3}}{\cancel{2}\sqrt{2}}\\&=\frac{7\sqrt{3}}{\sqrt{2}}\\&=\frac{7\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}\\&=\frac{7\sqrt{3}\sqrt{2}}{2}\end{aligned}[/tex]
[tex]\begin{aligned}\frac{5\sqrt{3}+3\sqrt{27}}{2\sqrt{2}}&=\boxed{\,\bf\frac{7\sqrt{6}}{2}\,}\end{aligned}[/tex]
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Nomor 3
[tex]\begin{aligned}\frac{2\sqrt{5}\times5\sqrt{10}}{2\sqrt{2}}&=\frac{2\sqrt{5}\cdot5\sqrt{2}\sqrt{5}}{2\sqrt{2}}\\&=\frac{(\cancel{2}\cdot5)\sqrt{5}\sqrt{5}\cancel{\sqrt{2}}}{\cancel{2}\cancel{\sqrt{2}}}\\&=5\cdot\sqrt{5}\sqrt{5}\\&=5\cdot5\\\frac{2\sqrt{5}\times5\sqrt{10}}{2\sqrt{2}}&=\boxed{\,\bf25\,}\end{aligned}[/tex]