x+1/x-3=-x-6x+1/(x-3)(x-1) eching

$$\frac { x + 1 } { x - 3 } = - \frac { x - 6 x + 1 } { ( x - 3 ) ( x - 1 ) }$$

x uchun yechish

$$x=5$$

$$x=0$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 1,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-3\right)\left(x-1\right) ga, x-3,\left(x-3\right)\left(x-1\right) ning eng kichik karralisiga ko‘paytiring.

$$\left(x-1\right)\left(x+1\right)=-\left(x-6x+1\right)$$

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.

$$x^{2}-1=-\left(x-6x+1\right)$$

-5x ni olish uchun x va -6x ni birlashtirish.

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

-5x+1 teskarisini topish uchun har birining teskarisini toping.

$$x^{2}-1=5x-1$$

Ikkala tarafdan 5x ni ayirish.

$$x^{2}-1-5x=-1$$

1 ni ikki tarafga qo’shing.

$$x^{2}-1-5x+1=0$$

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

$$x^{2}-5x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, -5 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}}}{2}$$

\left(-5\right)^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-5\right)±5}{2}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±5}{2}$$

x=\frac{5±5}{2} tenglamasini yeching, bunda ± musbat. 5 ni 5 ga qo'shish.

$$x=\frac{10}{2}$$

10 ni 2 ga bo'lish.

$$x=5$$

x=\frac{5±5}{2} tenglamasini yeching, bunda ± manfiy. 5 dan 5 ni ayirish.

$$x=\frac{0}{2}$$

0 ni 2 ga bo'lish.

$$x=0$$

Tenglama yechildi.

$$x=5 x=0$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 1,3 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-3\right)\left(x-1\right) ga, x-3,\left(x-3\right)\left(x-1\right) ning eng kichik karralisiga ko‘paytiring.

$$\left(x-1\right)\left(x+1\right)=-\left(x-6x+1\right)$$

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.

$$x^{2}-1=-\left(x-6x+1\right)$$

-5x ni olish uchun x va -6x ni birlashtirish.

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

-5x+1 teskarisini topish uchun har birining teskarisini toping.

$$x^{2}-1=5x-1$$

Ikkala tarafdan 5x ni ayirish.

$$x^{2}-1-5x=-1$$

1 ni ikki tarafga qo’shing.

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

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

$$x^{2}-5x=0$$

-5 ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{5}{2} olish uchun. Keyin, -\frac{5}{2} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{5}{2} kvadratini chiqarish.

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

x^{2}-5x+\frac{25}{4} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

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

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

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

Qisqartirish.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

\frac{5}{2} ni tenglamaning ikkala tarafiga qo'shish.

$$x=5 x=0$$