5/2+5/x-2=6/x^2-4 eching

$$\frac { 5 } { 2 } + \frac { 5 } { x - 2 } = \frac { 6 } { x ^ { 2 } - 4 }$$

x uchun yechish

$$x=\frac{\sqrt{85}}{5}-1\approx 0,843908891$$

$$x=-\frac{\sqrt{85}}{5}-1\approx -2,843908891$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -2,2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2\left(x-2\right)\left(x+2\right) ga, 2,x-2,x^{2}-4 ning eng kichik karralisiga ko‘paytiring.

$$2\left(x-2\right)\left(x+2\right)\times \frac{5}{2}+\left(2x+4\right)\times 5=2\times 6$$

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

$$\left(2x-4\right)\left(x+2\right)\times \frac{5}{2}+\left(2x+4\right)\times 5=2\times 6$$

2x-4 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(2x^{2}-8\right)\times \frac{5}{2}+\left(2x+4\right)\times 5=2\times 6$$

2x^{2}-8 ga \frac{5}{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}-20+\left(2x+4\right)\times 5=2\times 6$$

2x+4 ga 5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}-20+10x+20=2\times 6$$

0 olish uchun -20 va 20'ni qo'shing.

$$5x^{2}+10x=2\times 6$$

12 hosil qilish uchun 2 va 6 ni ko'paytirish.

$$5x^{2}+10x=12$$

Ikkala tarafdan 12 ni ayirish.

$$5x^{2}+10x-12=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 10 ni b va -12 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\times 5\left(-12\right)}}{2\times 5}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\times 5\left(-12\right)}}{2\times 5}$$

-4 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100-20\left(-12\right)}}{2\times 5}$$

-20 ni -12 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+240}}{2\times 5}$$

100 ni 240 ga qo'shish.

$$x=\frac{-10±\sqrt{340}}{2\times 5}$$

340 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±2\sqrt{85}}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-10±2\sqrt{85}}{10}$$

x=\frac{-10±2\sqrt{85}}{10} tenglamasini yeching, bunda ± musbat. -10 ni 2\sqrt{85} ga qo'shish.

$$x=\frac{2\sqrt{85}-10}{10}$$

-10+2\sqrt{85} ni 10 ga bo'lish.

$$x=\frac{\sqrt{85}}{5}-1$$

x=\frac{-10±2\sqrt{85}}{10} tenglamasini yeching, bunda ± manfiy. -10 dan 2\sqrt{85} ni ayirish.

$$x=\frac{-2\sqrt{85}-10}{10}$$

-10-2\sqrt{85} ni 10 ga bo'lish.

$$x=-\frac{\sqrt{85}}{5}-1$$

Tenglama yechildi.

$$x=\frac{\sqrt{85}}{5}-1 x=-\frac{\sqrt{85}}{5}-1$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -2,2 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 2\left(x-2\right)\left(x+2\right) ga, 2,x-2,x^{2}-4 ning eng kichik karralisiga ko‘paytiring.

$$2\left(x-2\right)\left(x+2\right)\times \frac{5}{2}+\left(2x+4\right)\times 5=2\times 6$$

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

$$\left(2x-4\right)\left(x+2\right)\times \frac{5}{2}+\left(2x+4\right)\times 5=2\times 6$$

2x-4 ga x+2 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$\left(2x^{2}-8\right)\times \frac{5}{2}+\left(2x+4\right)\times 5=2\times 6$$

2x^{2}-8 ga \frac{5}{2} ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}-20+\left(2x+4\right)\times 5=2\times 6$$

2x+4 ga 5 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$5x^{2}-20+10x+20=2\times 6$$

0 olish uchun -20 va 20'ni qo'shing.

$$5x^{2}+10x=2\times 6$$

12 hosil qilish uchun 2 va 6 ni ko'paytirish.

$$5x^{2}+10x=12$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+10x}{5}=\frac{12}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{10}{5}x=\frac{12}{5}$$

10 ni 5 ga bo'lish.

$$x^{2}+2x=\frac{12}{5}$$

2 ni bo‘lish, x shartining koeffitsienti, 2 ga 1 olish uchun. Keyin, 1 ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+2x+1^{2}=\frac{12}{5}+1^{2}$$

1 kvadratini chiqarish.

$$x^{2}+2x+1=\frac{12}{5}+1$$

\frac{12}{5} ni 1 ga qo'shish.

$$x^{2}+2x+1=\frac{17}{5}$$

x^{2}+2x+1 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+1\right)^{2}=\frac{17}{5}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{\frac{17}{5}}$$

Qisqartirish.

$$x+1=\frac{\sqrt{85}}{5} x+1=-\frac{\sqrt{85}}{5}$$

Tenglamaning ikkala tarafidan 1 ni ayirish.

$$x=\frac{\sqrt{85}}{5}-1 x=-\frac{\sqrt{85}}{5}-1$$