1/x-1+1/x-4=5/4 eching

$$\frac { 1 } { x - 1 } + \frac { 1 } { x - 4 } = \frac { 5 } { 4 }$$

x uchun yechish

$$x=5$$

$$x = \frac{8}{5} = 1\frac{3}{5} = 1,6$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 1,4 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 4\left(x-4\right)\left(x-1\right) ga, x-1,x-4,4 ning eng kichik karralisiga ko‘paytiring.

$$4x-16+4x-4=5\left(x-4\right)\left(x-1\right)$$

8x ni olish uchun 4x va 4x ni birlashtirish.

$$8x-16-4=5\left(x-4\right)\left(x-1\right)$$

-20 olish uchun -16 dan 4 ni ayirish.

$$8x-20=5\left(x-4\right)\left(x-1\right)$$

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

$$8x-20=\left(5x-20\right)\left(x-1\right)$$

5x-20 ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$8x-20=5x^{2}-25x+20$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$8x-20-5x^{2}=-25x+20$$

25x ni ikki tarafga qo’shing.

$$8x-20-5x^{2}+25x=20$$

33x ni olish uchun 8x va 25x ni birlashtirish.

$$33x-20-5x^{2}=20$$

Ikkala tarafdan 20 ni ayirish.

$$33x-20-5x^{2}-20=0$$

-40 olish uchun -20 dan 20 ni ayirish.

$$33x-40-5x^{2}=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$-5x^{2}+33x-40=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -5 ni a, 33 ni b va -40 ni c bilan almashtiring.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-5\right)\left(-40\right)}}{2\left(-5\right)}$$

33 kvadratini chiqarish.

$$x=\frac{-33±\sqrt{1089-4\left(-5\right)\left(-40\right)}}{2\left(-5\right)}$$

-4 ni -5 marotabaga ko'paytirish.

$$x=\frac{-33±\sqrt{1089+20\left(-40\right)}}{2\left(-5\right)}$$

20 ni -40 marotabaga ko'paytirish.

$$x=\frac{-33±\sqrt{1089-800}}{2\left(-5\right)}$$

1089 ni -800 ga qo'shish.

$$x=\frac{-33±\sqrt{289}}{2\left(-5\right)}$$

289 ning kvadrat ildizini chiqarish.

$$x=\frac{-33±17}{2\left(-5\right)}$$

2 ni -5 marotabaga ko'paytirish.

$$x=\frac{-33±17}{-10}$$

x=\frac{-33±17}{-10} tenglamasini yeching, bunda ± musbat. -33 ni 17 ga qo'shish.

$$x=-\frac{16}{-10}$$

\frac{-16}{-10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=\frac{8}{5}$$

x=\frac{-33±17}{-10} tenglamasini yeching, bunda ± manfiy. -33 dan 17 ni ayirish.

$$x=-\frac{50}{-10}$$

-50 ni -10 ga bo'lish.

$$x=5$$

Tenglama yechildi.

$$x=\frac{8}{5} x=5$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 1,4 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 4\left(x-4\right)\left(x-1\right) ga, x-1,x-4,4 ning eng kichik karralisiga ko‘paytiring.

$$4x-16+4x-4=5\left(x-4\right)\left(x-1\right)$$

8x ni olish uchun 4x va 4x ni birlashtirish.

$$8x-16-4=5\left(x-4\right)\left(x-1\right)$$

-20 olish uchun -16 dan 4 ni ayirish.

$$8x-20=5\left(x-4\right)\left(x-1\right)$$

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

$$8x-20=\left(5x-20\right)\left(x-1\right)$$

5x-20 ga x-1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$8x-20=5x^{2}-25x+20$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$8x-20-5x^{2}=-25x+20$$

25x ni ikki tarafga qo’shing.

$$8x-20-5x^{2}+25x=20$$

33x ni olish uchun 8x va 25x ni birlashtirish.

$$33x-20-5x^{2}=20$$

20 ni ikki tarafga qo’shing.

$$33x-5x^{2}=20+20$$

40 olish uchun 20 va 20'ni qo'shing.

$$33x-5x^{2}=40$$

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$-5x^{2}+33x=40$$

Ikki tarafini -5 ga bo‘ling.

$$\frac{-5x^{2}+33x}{-5}=\frac{40}{-5}$$

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

$$x^{2}+\frac{33}{-5}x=\frac{40}{-5}$$

33 ni -5 ga bo'lish.

$$x^{2}-\frac{33}{5}x=\frac{40}{-5}$$

40 ni -5 ga bo'lish.

$$x^{2}-\frac{33}{5}x=-8$$

-\frac{33}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{33}{10} olish uchun. Keyin, -\frac{33}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}-\frac{33}{5}x+\left(-\frac{33}{10}\right)^{2}=-8+\left(-\frac{33}{10}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{33}{10} kvadratini chiqarish.

$$x^{2}-\frac{33}{5}x+\frac{1089}{100}=-8+\frac{1089}{100}$$

-8 ni \frac{1089}{100} ga qo'shish.

$$x^{2}-\frac{33}{5}x+\frac{1089}{100}=\frac{289}{100}$$

x^{2}-\frac{33}{5}x+\frac{1089}{100} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x-\frac{33}{10}\right)^{2}=\frac{289}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x-\frac{33}{10}\right)^{2}}=\sqrt{\frac{289}{100}}$$

Qisqartirish.

$$x-\frac{33}{10}=\frac{17}{10} x-\frac{33}{10}=-\frac{17}{10}$$

\frac{33}{10} ni tenglamaning ikkala tarafiga qo'shish.

$$x=5 x=\frac{8}{5}$$