ແກ້ u/1+u+u/1-u^2

$$\frac { u } { 1 + u } + \frac { u } { 1 - u ^ { 2 } }$$

ປະເມີນ

$$\frac{u\left(2-u\right)}{1-u^{2}}$$

ຂັ້ນຕອນທາງອອກສັ້ນ

ຕົວປະກອບ 1-u^{2}.

$$\frac{u}{1+u}+\frac{u}{\left(u-1\right)\left(-u-1\right)}$$

ເພື່ອເພີ່ມ ຫຼື ຫານນິພົດ, ໃຫ້ຂະຫຍາຍພວກມັນເພື່ອໃຫ້ຕົວຄູນມີຈຳນວນດຽວກັນ. ຈຳນວນຄູນທີ່ນິຍົມໜ້ອຍທີ່ສຸດຂອງ 1+u ກັບ \left(u-1\right)\left(-u-1\right) ແມ່ນ \left(u-1\right)\left(u+1\right). ຄູນ \frac{u}{1+u} ໃຫ້ກັບ \frac{u-1}{u-1}. ຄູນ \frac{u}{\left(u-1\right)\left(-u-1\right)} ໃຫ້ກັບ \frac{-1}{-1}.

$$\frac{u\left(u-1\right)}{\left(u-1\right)\left(u+1\right)}+\frac{-u}{\left(u-1\right)\left(u+1\right)}$$

ເນື່ອງຈາກ \frac{u\left(u-1\right)}{\left(u-1\right)\left(u+1\right)} ແລະ \frac{-u}{\left(u-1\right)\left(u+1\right)} ມີຕົວຫານດຽວກັນ, ໃຫ້ເພີ່ມພວກມັນໂດຍການເພີ່ມຈຳນວນທີ່ເປັນເສດໃນເລກເສດສ່ວນຂອງພວກມັນ.

$$\frac{u\left(u-1\right)-u}{\left(u-1\right)\left(u+1\right)}$$

ຄູນໃນເສດສ່ວນ u\left(u-1\right)-u.

$$\frac{u^{2}-u-u}{\left(u-1\right)\left(u+1\right)}$$

ຮວມຂໍ້ກຳນົດໃນ u^{2}-u-u.

$$\frac{u^{2}-2u}{\left(u-1\right)\left(u+1\right)}$$

ຂະຫຍາຍ \left(u-1\right)\left(u+1\right).

$$\frac{u^{2}-2u}{u^{2}-1}$$

ຕົວປະກອບ

$$\frac{u\left(2-u\right)}{1-u^{2}}$$