ແກ້ x^2/x-1leqx+1

$$\frac { x ^ { 2 } } { x - 1 } \leq x + 1$$

ແກ້ສຳລັບ x

$$x<1$$

ຂັ້ນຕອນການແກ້

ລົບ x ອອກຈາກທັງສອງຂ້າງ.

$$\frac{x^{2}}{x-1}-x\leq 1$$

ເພື່ອເພີ່ມ ຫຼື ຫານນິພົດ, ໃຫ້ຂະຫຍາຍພວກມັນເພື່ອໃຫ້ຕົວຄູນມີຈຳນວນດຽວກັນ. ຄູນ x ໃຫ້ກັບ \frac{x-1}{x-1}.

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

ເນື່ອງຈາກ \frac{x^{2}}{x-1} ແລະ \frac{x\left(x-1\right)}{x-1} ມີຕົວຫານດຽວກັນ, ໃຫ້ຫານພວກມັນໂດຍການຫານຈຳນວນທີ່ເປັນເສດໃນເລກເສດສ່ວນຂອງພວກມັນ.

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

ຄູນໃນເສດສ່ວນ x^{2}-x\left(x-1\right).

$$\frac{x^{2}-x^{2}+x}{x-1}\leq 1$$

ຮວມຂໍ້ກຳນົດໃນ x^{2}-x^{2}+x.

$$\frac{x}{x-1}\leq 1$$

Denominator x-1 cannot be zero since division by zero is not defined. There are two cases.

$$x-1>0 x-1<0$$

Consider the case when x-1 is positive. Move -1 to the right hand side.

$$x>1$$

The initial inequality does not change the direction when multiplied by x-1 for x-1>0.

$$x\leq x-1$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$x-x\leq -1$$

ຮວມຄຳສັບ.

$$0\leq -1$$

Consider condition x>1 specified above.

$$x\in \emptyset$$

Now consider the case when x-1 is negative. Move -1 to the right hand side.

$$x<1$$

The initial inequality changes the direction when multiplied by x-1 for x-1<0.

$$x\geq x-1$$

Move the terms containing x to the left hand side and all other terms to the right hand side.

$$x-x\geq -1$$

ຮວມຄຳສັບ.

$$0\geq -1$$

Consider condition x<1 specified above.

$$x<1$$

ວິທີແກ້ສຸດທ້າຍແມ່ນເປັນການຮວມວິທີການທີ່ຊອກມາໄດ້.

$$x<1$$