រំលងទៅមាតិកាមេ
វាយតម្លៃ
Tick mark Image
ពន្លាត
Tick mark Image
ក្រាហ្វ

បញ្ហាស្រដៀងគ្នាពី Web Search

ចែករំលែក

\frac{\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}+\frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ x+1 និង x-2 គឺ \left(x-2\right)\left(x+1\right)។ គុណ \frac{x-2}{x+1} ដង \frac{x-2}{x-2}។ គុណ \frac{5-x}{x-2} ដង \frac{x+1}{x+1}។
\frac{\frac{\left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដោយសារ \frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)} និង \frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។
\frac{\frac{x^{2}-2x-2x+4+5x+5-x^{2}-x}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ធ្វើផល​គុណនៅក្នុង \left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right)។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
បន្សំដូចជាតួនៅក្នុង x^{2}-2x-2x+4+5x+5-x^{2}-x។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដាក់ជាកត្តា x^{2}-x-2។ ដាក់ជាកត្តា x^{2}+3x+2។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ \left(x-2\right)\left(x+1\right) និង \left(x+1\right)\left(x+2\right) គឺ \left(x-2\right)\left(x+1\right)\left(x+2\right)។ គុណ \frac{1}{\left(x-2\right)\left(x+1\right)} ដង \frac{x+2}{x+2}។ គុណ \frac{1}{\left(x+1\right)\left(x+2\right)} ដង \frac{x-2}{x-2}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដោយសារ \frac{x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} និង \frac{x-2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} មានភាគបែងដូចគ្នា សូមដកពួកវាដោយការដកភាគយករបស់ពួកវា។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ធ្វើផល​គុណនៅក្នុង x+2-\left(x-2\right)។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
បន្សំដូចជាតួនៅក្នុង x+2-x+2។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}
ដាក់ជាកត្តា x^{2}+x។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}
ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ x និង x\left(x+1\right) គឺ x\left(x+1\right)។ គុណ \frac{x+1}{x} ដង \frac{x+1}{x+1}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{\left(x+1\right)\left(x+1\right)+3-x^{2}}{x\left(x+1\right)}}
ដោយសារ \frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)} និង \frac{3-x^{2}}{x\left(x+1\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{x^{2}+x+1+x+3-x^{2}}{x\left(x+1\right)}}
ធ្វើផល​គុណនៅក្នុង \left(x+1\right)\left(x+1\right)+3-x^{2}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{2x+4}{x\left(x+1\right)}}
បន្សំដូចជាតួនៅក្នុង x^{2}+x+1+x+3-x^{2}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}}
គុណ \frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} ដង \frac{2x+4}{x\left(x+1\right)} ដោយការគុណភាគយក​ចំនួនដងនៃភាគយក និងភាគបែងចំនួនដងនៃភាគបែង។
\frac{9\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}{\left(x-2\right)\left(x+1\right)\times 4\left(2x+4\right)}
ចែក \frac{9}{\left(x-2\right)\left(x+1\right)} នឹង \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)} ដោយការគុណ \frac{9}{\left(x-2\right)\left(x+1\right)} នឹងប្រភាគផ្ទុយគ្នានៃ \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}.
\frac{9x\left(x+1\right)\left(x+2\right)}{4\left(2x+4\right)}
សម្រួល \left(x-2\right)\left(x+1\right) ទាំងនៅក្នុងភាគយក និងភាគបែង។
\frac{9x\left(x+1\right)\left(x+2\right)}{2\times 4\left(x+2\right)}
ដាក់ជាកត្តានូវកន្សោមមិនទាន់បានលើកជាកត្តារួច។
\frac{9x\left(x+1\right)}{2\times 4}
សម្រួល x+2 ទាំងនៅក្នុងភាគយក និងភាគបែង។
\frac{9x^{2}+9x}{8}
ពង្រីកកន្សោម។
\frac{\frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}+\frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ x+1 និង x-2 គឺ \left(x-2\right)\left(x+1\right)។ គុណ \frac{x-2}{x+1} ដង \frac{x-2}{x-2}។ គុណ \frac{5-x}{x-2} ដង \frac{x+1}{x+1}។
\frac{\frac{\left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដោយសារ \frac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+1\right)} និង \frac{\left(5-x\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។
\frac{\frac{x^{2}-2x-2x+4+5x+5-x^{2}-x}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ធ្វើផល​គុណនៅក្នុង \left(x-2\right)\left(x-2\right)+\left(5-x\right)\left(x+1\right)។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{x^{2}-x-2}-\frac{1}{x^{2}+3x+2}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
បន្សំដូចជាតួនៅក្នុង x^{2}-2x-2x+4+5x+5-x^{2}-x។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{1}{\left(x-2\right)\left(x+1\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដាក់ជាកត្តា x^{2}-x-2។ ដាក់ជាកត្តា x^{2}+3x+2។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\left(\frac{x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\right)\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ \left(x-2\right)\left(x+1\right) និង \left(x+1\right)\left(x+2\right) គឺ \left(x-2\right)\left(x+1\right)\left(x+2\right)។ គុណ \frac{1}{\left(x-2\right)\left(x+1\right)} ដង \frac{x+2}{x+2}។ គុណ \frac{1}{\left(x+1\right)\left(x+2\right)} ដង \frac{x-2}{x-2}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ដោយសារ \frac{x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} និង \frac{x-2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} មានភាគបែងដូចគ្នា សូមដកពួកវាដោយការដកភាគយករបស់ពួកវា។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{x+2-x+2}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
ធ្វើផល​គុណនៅក្នុង x+2-\left(x-2\right)។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x^{2}+x}\right)}
បន្សំដូចជាតួនៅក្នុង x+2-x+2។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{x+1}{x}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}
ដាក់ជាកត្តា x^{2}+x។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\left(\frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}+\frac{3-x^{2}}{x\left(x+1\right)}\right)}
ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ x និង x\left(x+1\right) គឺ x\left(x+1\right)។ គុណ \frac{x+1}{x} ដង \frac{x+1}{x+1}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{\left(x+1\right)\left(x+1\right)+3-x^{2}}{x\left(x+1\right)}}
ដោយសារ \frac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)} និង \frac{3-x^{2}}{x\left(x+1\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{x^{2}+x+1+x+3-x^{2}}{x\left(x+1\right)}}
ធ្វើផល​គុណនៅក្នុង \left(x+1\right)\left(x+1\right)+3-x^{2}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)}\times \frac{2x+4}{x\left(x+1\right)}}
បន្សំដូចជាតួនៅក្នុង x^{2}+x+1+x+3-x^{2}។
\frac{\frac{9}{\left(x-2\right)\left(x+1\right)}}{\frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}}
គុណ \frac{4}{\left(x-2\right)\left(x+1\right)\left(x+2\right)} ដង \frac{2x+4}{x\left(x+1\right)} ដោយការគុណភាគយក​ចំនួនដងនៃភាគយក និងភាគបែងចំនួនដងនៃភាគបែង។
\frac{9\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}{\left(x-2\right)\left(x+1\right)\times 4\left(2x+4\right)}
ចែក \frac{9}{\left(x-2\right)\left(x+1\right)} នឹង \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)} ដោយការគុណ \frac{9}{\left(x-2\right)\left(x+1\right)} នឹងប្រភាគផ្ទុយគ្នានៃ \frac{4\left(2x+4\right)}{\left(x-2\right)\left(x+1\right)\left(x+2\right)x\left(x+1\right)}.
\frac{9x\left(x+1\right)\left(x+2\right)}{4\left(2x+4\right)}
សម្រួល \left(x-2\right)\left(x+1\right) ទាំងនៅក្នុងភាគយក និងភាគបែង។
\frac{9x\left(x+1\right)\left(x+2\right)}{2\times 4\left(x+2\right)}
ដាក់ជាកត្តានូវកន្សោមមិនទាន់បានលើកជាកត្តារួច។
\frac{9x\left(x+1\right)}{2\times 4}
សម្រួល x+2 ទាំងនៅក្នុងភាគយក និងភាគបែង។
\frac{9x^{2}+9x}{8}
ពង្រីកកន្សោម។