ដោះស្រាយ 2/x^2+7x+10+3/x^2-25

$$\frac { 2 } { x ^ { 2 } + 7 x + 10 } + \frac { 3 } { x ^ { 2 } - 25 }$$

វាយតម្លៃ

$$\frac{5x-4}{\left(x+2\right)\left(x^{2}-25\right)}$$

ជំហានចម្លើយខ្លីៗ

ដាក់ជាកត្តា x^{2}+7x+10។ ដាក់ជាកត្តា x^{2}-25។

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

ដើម្បីបូក ឬដក​កន្សោម ពន្លាតពួកវាដើម្បីធ្វើឲ្យភាគបែងរបស់ពួកវាដូចគ្នា។ ពហុគុណរួមតូចបំផុតនៃ \left(x+2\right)\left(x+5\right) និង \left(x-5\right)\left(x+5\right) គឺ \left(x-5\right)\left(x+2\right)\left(x+5\right)។ គុណ \frac{2}{\left(x+2\right)\left(x+5\right)} ដង \frac{x-5}{x-5}។ គុណ \frac{3}{\left(x-5\right)\left(x+5\right)} ដង \frac{x+2}{x+2}។

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

ដោយសារ \frac{2\left(x-5\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)} និង \frac{3\left(x+2\right)}{\left(x-5\right)\left(x+2\right)\left(x+5\right)} មានភាគបែងដូចគ្នា សូមបូកពួកវាដោយការបូកភាគយករបស់ពួកវា។

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

ធ្វើផល​គុណនៅក្នុង 2\left(x-5\right)+3\left(x+2\right)។

$$\frac{2x-10+3x+6}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}$$

បន្សំដូចជាតួនៅក្នុង 2x-10+3x+6។

$$\frac{5x-4}{\left(x-5\right)\left(x+2\right)\left(x+5\right)}$$

ពន្លាត \left(x-5\right)\left(x+2\right)\left(x+5\right)។

$$\frac{5x-4}{x^{3}+2x^{2}-25x-50}$$

ធ្វើឌីផេរ៉ងស្យែល w.r.t. x

$$\frac{2\left(-5x^{3}+x^{2}+8x-175\right)}{\left(\left(x+2\right)\left(x^{2}-25\right)\right)^{2}}$$