ដោះស្រាយ x^2+x/x^2-25+x^2-1/x^2+11x+30

$$\frac { x ^ { 2 } + x } { x ^ { 2 } - 25 } + \frac { x ^ { 2 } - 1 } { x ^ { 2 } + 11 x + 30 }$$

វាយតម្លៃ

$$\frac{\left(x+1\right)\left(2x^{2}+5\right)}{\left(x+6\right)\left(x^{2}-25\right)}$$

ជំហានចម្លើយខ្លីៗ

ដាក់ជាកត្តា x^{2}-25។ ដាក់ជាកត្តា x^{2}+11x+30។

$$\frac{x^{2}+x}{\left(x-5\right)\left(x+5\right)}+\frac{x^{2}-1}{\left(x+5\right)\left(x+6\right)}$$

ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ \left(x-5\right)\left(x+5\right) និង \left(x+5\right)\left(x+6\right) គឺ \left(x-5\right)\left(x+5\right)\left(x+6\right)។ គុណ \frac{x^{2}+x}{\left(x-5\right)\left(x+5\right)} ដង \frac{x+6}{x+6}។ គុណ \frac{x^{2}-1}{\left(x+5\right)\left(x+6\right)} ដង \frac{x-5}{x-5}។

$$\frac{\left(x^{2}+x\right)\left(x+6\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}+\frac{\left(x^{2}-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

ដោយសារ \frac{\left(x^{2}+x\right)\left(x+6\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)} និង \frac{\left(x^{2}-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។

$$\frac{\left(x^{2}+x\right)\left(x+6\right)+\left(x^{2}-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

ធ្វើផល​គុណនៅក្នុង \left(x^{2}+x\right)\left(x+6\right)+\left(x^{2}-1\right)\left(x-5\right)។

$$\frac{x^{3}+6x^{2}+x^{2}+6x+x^{3}-5x^{2}-x+5}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

បន្សំដូចជាតួនៅក្នុង x^{3}+6x^{2}+x^{2}+6x+x^{3}-5x^{2}-x+5។

$$\frac{2x^{3}+2x^{2}+5x+5}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

ពន្លាត \left(x-5\right)\left(x+5\right)\left(x+6\right)។

$$\frac{2x^{3}+2x^{2}+5x+5}{x^{3}+6x^{2}-25x-150}$$

ពន្លាត

$$\frac{2x^{3}+2x^{2}+5x+5}{\left(x+6\right)\left(x^{2}-25\right)}$$

ជំហានចម្លើយខ្លីៗ

ដាក់ជាកត្តា x^{2}-25។ ដាក់ជាកត្តា x^{2}+11x+30។

$$\frac{x^{2}+x}{\left(x-5\right)\left(x+5\right)}+\frac{x^{2}-1}{\left(x+5\right)\left(x+6\right)}$$

ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ \left(x-5\right)\left(x+5\right) និង \left(x+5\right)\left(x+6\right) គឺ \left(x-5\right)\left(x+5\right)\left(x+6\right)។ គុណ \frac{x^{2}+x}{\left(x-5\right)\left(x+5\right)} ដង \frac{x+6}{x+6}។ គុណ \frac{x^{2}-1}{\left(x+5\right)\left(x+6\right)} ដង \frac{x-5}{x-5}។

$$\frac{\left(x^{2}+x\right)\left(x+6\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}+\frac{\left(x^{2}-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

ដោយសារ \frac{\left(x^{2}+x\right)\left(x+6\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)} និង \frac{\left(x^{2}-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។

$$\frac{\left(x^{2}+x\right)\left(x+6\right)+\left(x^{2}-1\right)\left(x-5\right)}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

ធ្វើផល​គុណនៅក្នុង \left(x^{2}+x\right)\left(x+6\right)+\left(x^{2}-1\right)\left(x-5\right)។

$$\frac{x^{3}+6x^{2}+x^{2}+6x+x^{3}-5x^{2}-x+5}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

បន្សំដូចជាតួនៅក្នុង x^{3}+6x^{2}+x^{2}+6x+x^{3}-5x^{2}-x+5។

$$\frac{2x^{3}+2x^{2}+5x+5}{\left(x-5\right)\left(x+5\right)\left(x+6\right)}$$

ពន្លាត \left(x-5\right)\left(x+5\right)\left(x+6\right)។

$$\frac{2x^{3}+2x^{2}+5x+5}{x^{3}+6x^{2}-25x-150}$$