5/x-3/2=x/5 eching

$$\frac { 5 } { x } - \frac { 3 } { 2 } = \frac { x } { 5 }$$

x uchun yechish

$$x = \frac{5}{2} = 2\frac{1}{2} = 2,5$$

$$x=-10$$

Guruhlash orqali faktorlash qadamlari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 10x ga, x,2,5 ning eng kichik karralisiga ko‘paytiring.

$$10\times 5+10x\left(-\frac{3}{2}\right)=2xx$$

50 hosil qilish uchun 10 va 5 ni ko'paytirish.

$$50+10x\left(-\frac{3}{2}\right)=2xx$$

10\left(-\frac{3}{2}\right) ni yagona kasrga aylantiring.

$$50+\frac{10\left(-3\right)}{2}x=2xx$$

-30 hosil qilish uchun 10 va -3 ni ko'paytirish.

$$50+\frac{-30}{2}x=2xx$$

-15 ni olish uchun -30 ni 2 ga bo‘ling.

$$50-15x=2xx$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$50-15x=2x^{2}$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$50-15x-2x^{2}=0$$

Polinomni standart shaklga keltirish uchun uni qayta tartiblang. Shartlarni eng yuqoridan eng pastki qiymat ko'rsatgichiga joylashtirish.

$$-2x^{2}-15x+50=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -2x^{2}+ax+bx+50 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-15 ab=-2\times 50=-100$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. -100-mahsulotni beruvchi bunday butun juftliklarni roʻyxat qiling.

$$1,-100 2,-50 4,-25 5,-20 10,-10$$

Har bir juftlik yigʻindisini hisoblang.

$$1-100=-99 2-50=-48 4-25=-21 5-20=-15 10-10=0$$

Yechim – -15 yigʻindisini beruvchi juftlik.

$$a=5 b=-20$$

-2x^{2}-15x+50 ni \left(-2x^{2}+5x\right)+\left(-20x+50\right) sifatida qaytadan yozish.

$$\left(-2x^{2}+5x\right)+\left(-20x+50\right)$$

Birinchi guruhda -x ni va ikkinchi guruhda -10 ni faktordan chiqaring.

$$-x\left(2x-5\right)-10\left(2x-5\right)$$

Distributiv funktsiyasidan foydalangan holda 2x-5 umumiy terminini chiqaring.

$$\left(2x-5\right)\left(-x-10\right)$$

Tenglamani yechish uchun 2x-5=0 va -x-10=0 ni yeching.

$$x=\frac{5}{2} x=-10$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 10x ga, x,2,5 ning eng kichik karralisiga ko‘paytiring.

$$10\times 5+10x\left(-\frac{3}{2}\right)=2xx$$

50 hosil qilish uchun 10 va 5 ni ko'paytirish.

$$50+10x\left(-\frac{3}{2}\right)=2xx$$

10\left(-\frac{3}{2}\right) ni yagona kasrga aylantiring.

$$50+\frac{10\left(-3\right)}{2}x=2xx$$

-30 hosil qilish uchun 10 va -3 ni ko'paytirish.

$$50+\frac{-30}{2}x=2xx$$

-15 ni olish uchun -30 ni 2 ga bo‘ling.

$$50-15x=2xx$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$50-15x=2x^{2}$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$50-15x-2x^{2}=0$$

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.

$$-2x^{2}-15x+50=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -2 ni a, -15 ni b va 50 ni c bilan almashtiring.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-2\right)\times 50}}{2\left(-2\right)}$$

-15 kvadratini chiqarish.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-2\right)\times 50}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-15\right)±\sqrt{225+8\times 50}}{2\left(-2\right)}$$

8 ni 50 marotabaga ko'paytirish.

$$x=\frac{-\left(-15\right)±\sqrt{225+400}}{2\left(-2\right)}$$

225 ni 400 ga qo'shish.

$$x=\frac{-\left(-15\right)±\sqrt{625}}{2\left(-2\right)}$$

625 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-15\right)±25}{2\left(-2\right)}$$

-15 ning teskarisi 15 ga teng.

$$x=\frac{15±25}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{15±25}{-4}$$

x=\frac{15±25}{-4} tenglamasini yeching, bunda ± musbat. 15 ni 25 ga qo'shish.

$$x=\frac{40}{-4}$$

40 ni -4 ga bo'lish.

$$x=-10$$

x=\frac{15±25}{-4} tenglamasini yeching, bunda ± manfiy. 15 dan 25 ni ayirish.

$$x=-\frac{10}{-4}$$

\frac{-10}{-4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=-10 x=\frac{5}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati 0 teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini 10x ga, x,2,5 ning eng kichik karralisiga ko‘paytiring.

$$10\times 5+10x\left(-\frac{3}{2}\right)=2xx$$

50 hosil qilish uchun 10 va 5 ni ko'paytirish.

$$50+10x\left(-\frac{3}{2}\right)=2xx$$

10\left(-\frac{3}{2}\right) ni yagona kasrga aylantiring.

$$50+\frac{10\left(-3\right)}{2}x=2xx$$

-30 hosil qilish uchun 10 va -3 ni ko'paytirish.

$$50+\frac{-30}{2}x=2xx$$

-15 ni olish uchun -30 ni 2 ga bo‘ling.

$$50-15x=2xx$$

x^{2} hosil qilish uchun x va x ni ko'paytirish.

$$50-15x=2x^{2}$$

Ikkala tarafdan 2x^{2} ni ayirish.

$$50-15x-2x^{2}=0$$

Ikkala tarafdan 50 ni ayirish. Har qanday sonni noldan ayirsangiz, o‘zining manfiyi chiqadi.

$$-15x-2x^{2}=-50$$

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

$$-2x^{2}-15x=-50$$

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}-15x}{-2}=-\frac{50}{-2}$$

-2 ga bo'lish -2 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{15}{-2}\right)x=-\frac{50}{-2}$$

-15 ni -2 ga bo'lish.

$$x^{2}+\frac{15}{2}x=-\frac{50}{-2}$$

-50 ni -2 ga bo'lish.

$$x^{2}+\frac{15}{2}x=25$$

\frac{15}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{15}{4} olish uchun. Keyin, \frac{15}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{15}{2}x+\left(\frac{15}{4}\right)^{2}=25+\left(\frac{15}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{15}{4} kvadratini chiqarish.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=25+\frac{225}{16}$$

25 ni \frac{225}{16} ga qo'shish.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=\frac{625}{16}$$

x^{2}+\frac{15}{2}x+\frac{225}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{15}{4}\right)^{2}=\frac{625}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{15}{4}\right)^{2}}=\sqrt{\frac{625}{16}}$$

Qisqartirish.

$$x+\frac{15}{4}=\frac{25}{4} x+\frac{15}{4}=-\frac{25}{4}$$

Tenglamaning ikkala tarafidan \frac{15}{4} ni ayirish.

$$x=\frac{5}{2} x=-10$$