5nx^2-5x-nx-1/x^2-1=2n+1 eching

$$\frac { 5 n x ^ { 2 } - 5 x - n x - 1 } { x ^ { 2 } - 1 } = 2 n + 1$$

n uchun yechish (complex solution)

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

$$x\neq \frac{1+\sqrt{23}i}{6}\text{ and }x\neq \frac{-\sqrt{23}i+1}{6}\text{ and }x\neq -1\text{ and }x\neq 1$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini \left(x-1\right)\left(x+1\right) ga ko'paytirish.

$$5nx^{2}-5x-nx-1=2n\left(x-1\right)\left(x+1\right)+\left(x-1\right)\left(x+1\right)$$

2n ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5nx^{2}-5x-nx-1=\left(2nx-2n\right)\left(x+1\right)+\left(x-1\right)\left(x+1\right)$$

2nx-2n ga x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5nx^{2}-5x-nx-1=2nx^{2}-2n+\left(x-1\right)\left(x+1\right)$$

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$5nx^{2}-5x-nx-1=2nx^{2}-2n+x^{2}-1$$

Ikkala tarafdan 2nx^{2} ni ayirish.

$$5nx^{2}-5x-nx-1-2nx^{2}=-2n+x^{2}-1$$

3nx^{2} ni olish uchun 5nx^{2} va -2nx^{2} ni birlashtirish.

$$3nx^{2}-5x-nx-1=-2n+x^{2}-1$$

2n ni ikki tarafga qo’shing.

$$3nx^{2}-5x-nx-1+2n=x^{2}-1$$

5x ni ikki tarafga qo’shing.

$$3nx^{2}-nx-1+2n=x^{2}-1+5x$$

1 ni ikki tarafga qo’shing.

$$3nx^{2}-nx+2n=x^{2}-1+5x+1$$

0 olish uchun -1 va 1'ni qo'shing.

$$3nx^{2}-nx+2n=x^{2}+5x$$

n'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(3x^{2}-x+2\right)n=x^{2}+5x$$

Ikki tarafini 3x^{2}-x+2 ga bo‘ling.

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

3x^{2}-x+2 ga bo'lish 3x^{2}-x+2 ga ko'paytirishni bekor qiladi.

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

n uchun yechish

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

$$|x|\neq 1$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

Tenglamaning ikkala tarafini \left(x-1\right)\left(x+1\right) ga ko'paytirish.

$$5nx^{2}-5x-nx-1=2n\left(x-1\right)\left(x+1\right)+\left(x-1\right)\left(x+1\right)$$

2n ga x-1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5nx^{2}-5x-nx-1=\left(2nx-2n\right)\left(x+1\right)+\left(x-1\right)\left(x+1\right)$$

2nx-2n ga x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$5nx^{2}-5x-nx-1=2nx^{2}-2n+\left(x-1\right)\left(x+1\right)$$

Hisoblang: \left(x-1\right)\left(x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

$$5nx^{2}-5x-nx-1=2nx^{2}-2n+x^{2}-1$$

Ikkala tarafdan 2nx^{2} ni ayirish.

$$5nx^{2}-5x-nx-1-2nx^{2}=-2n+x^{2}-1$$

3nx^{2} ni olish uchun 5nx^{2} va -2nx^{2} ni birlashtirish.

$$3nx^{2}-5x-nx-1=-2n+x^{2}-1$$

2n ni ikki tarafga qo’shing.

$$3nx^{2}-5x-nx-1+2n=x^{2}-1$$

5x ni ikki tarafga qo’shing.

$$3nx^{2}-nx-1+2n=x^{2}-1+5x$$

1 ni ikki tarafga qo’shing.

$$3nx^{2}-nx+2n=x^{2}-1+5x+1$$

0 olish uchun -1 va 1'ni qo'shing.

$$3nx^{2}-nx+2n=x^{2}+5x$$

n'ga ega bo'lgan barcha shartlarni birlashtirish.

$$\left(3x^{2}-x+2\right)n=x^{2}+5x$$

Ikki tarafini 3x^{2}-x+2 ga bo‘ling.

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

3x^{2}-x+2 ga bo'lish 3x^{2}-x+2 ga ko'paytirishni bekor qiladi.

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

x uchun yechish (complex solution)

$$\left\{\begin{matrix}x=\frac{-\sqrt{25+18n-23n^{2}}+n+5}{2\left(3n-1\right)}\text{, }&n\neq \frac{1}{3}\\x=\frac{\sqrt{25+18n-23n^{2}}+n+5}{2\left(3n-1\right)}\text{, }&n\neq -\frac{2}{3}\text{ and }n\neq \frac{1}{3}\text{ and }n\neq \frac{3}{2}\\x=\frac{1}{8}\text{, }&n=\frac{1}{3}\end{matrix}\right,$$

x uchun yechish

$$\left\{\begin{matrix}x=\frac{-\sqrt{25+18n-23n^{2}}+n+5}{2\left(3n-1\right)}\text{, }&n\neq \frac{1}{3}\text{ and }n\geq \frac{9-4\sqrt{41}}{23}\text{ and }n\leq \frac{4\sqrt{41}+9}{23}\\x=\frac{\sqrt{25+18n-23n^{2}}+n+5}{2\left(3n-1\right)}\text{, }&n\neq \frac{3}{2}\text{ and }n\neq \frac{1}{3}\text{ and }n\geq \frac{9-4\sqrt{41}}{23}\text{ and }n\leq \frac{4\sqrt{41}+9}{23}\text{ and }n\neq -\frac{2}{3}\\x=\frac{1}{8}\text{, }&n=\frac{1}{3}\end{matrix}\right,$$