x^2/x-1leqx+1 eching

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

x uchun yechish

$$x<1$$

Yechim qadamlari

Ikkala tarafdan x ni ayirish.

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

Ifodalarni qo‘shish yoki ayirish uchun ularni yoyib, maxrajlarini bir xil qiling. x ni \frac{x-1}{x-1} marotabaga ko'paytirish.

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

\frac{x^{2}}{x-1} va \frac{x\left(x-1\right)}{x-1} da bir xil maxraji bor, ularning suratini ayirish orqali ayiring.

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

x^{2}-x\left(x-1\right) ichidagi ko‘paytirishlarni bajaring.

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

x^{2}-x^{2}+x kabi iboralarga o‘xshab birlashtiring.

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

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.

$$x-x\leq -1$$

O'xshash hadlarni birlashtirish.

$$0\leq -1$$

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.

$$x-x\geq -1$$

O'xshash hadlarni birlashtirish.

$$0\geq -1$$

Consider condition x<1 specified above.

$$x<1$$

Oxirgi yechim olingan yechimlarning birlashmasidir.

$$x<1$$