x/2x+1+2/1-2x=3 eching

$$\frac { x } { 2 x + 1 } + \frac { 2 } { 1 - 2 x } = 3$$

x uchun yechish

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4}\approx 0,153112887$$

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4}\approx -0,653112887$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -\frac{1}{2},\frac{1}{2} qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(2x-1\right)\left(2x+1\right) ga, 2x+1,1-2x ning eng kichik karralisiga ko‘paytiring.

$$\left(2x-1\right)x+\left(-1-2x\right)\times 2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-x+\left(-1-2x\right)\times 2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-x-2-4x=3\left(2x-1\right)\left(2x+1\right)$$

-5x ni olish uchun -x va -4x ni birlashtirish.

$$2x^{2}-5x-2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-5x-2=\left(6x-3\right)\left(2x+1\right)$$

6x-3 ga 2x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$2x^{2}-5x-2=12x^{2}-3$$

Ikkala tarafdan 12x^{2} ni ayirish.

$$2x^{2}-5x-2-12x^{2}=-3$$

-10x^{2} ni olish uchun 2x^{2} va -12x^{2} ni birlashtirish.

$$-10x^{2}-5x-2=-3$$

3 ni ikki tarafga qo’shing.

$$-10x^{2}-5x-2+3=0$$

1 olish uchun -2 va 3'ni qo'shing.

$$-10x^{2}-5x+1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -10 ni a, -5 ni b va 1 ni c bilan almashtiring.

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

-5 kvadratini chiqarish.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-10\right)}}{2\left(-10\right)}$$

-4 ni -10 marotabaga ko'paytirish.

$$x=\frac{-\left(-5\right)±\sqrt{25+40}}{2\left(-10\right)}$$

25 ni 40 ga qo'shish.

$$x=\frac{-\left(-5\right)±\sqrt{65}}{2\left(-10\right)}$$

-5 ning teskarisi 5 ga teng.

$$x=\frac{5±\sqrt{65}}{2\left(-10\right)}$$

2 ni -10 marotabaga ko'paytirish.

$$x=\frac{5±\sqrt{65}}{-20}$$

x=\frac{5±\sqrt{65}}{-20} tenglamasini yeching, bunda ± musbat. 5 ni \sqrt{65} ga qo'shish.

$$x=\frac{\sqrt{65}+5}{-20}$$

5+\sqrt{65} ni -20 ga bo'lish.

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4}$$

x=\frac{5±\sqrt{65}}{-20} tenglamasini yeching, bunda ± manfiy. 5 dan \sqrt{65} ni ayirish.

$$x=\frac{5-\sqrt{65}}{-20}$$

5-\sqrt{65} ni -20 ga bo'lish.

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Tenglama yechildi.

$$x=-\frac{\sqrt{65}}{20}-\frac{1}{4} x=\frac{\sqrt{65}}{20}-\frac{1}{4}$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -\frac{1}{2},\frac{1}{2} qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(2x-1\right)\left(2x+1\right) ga, 2x+1,1-2x ning eng kichik karralisiga ko‘paytiring.

$$\left(2x-1\right)x+\left(-1-2x\right)\times 2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-x+\left(-1-2x\right)\times 2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-x-2-4x=3\left(2x-1\right)\left(2x+1\right)$$

-5x ni olish uchun -x va -4x ni birlashtirish.

$$2x^{2}-5x-2=3\left(2x-1\right)\left(2x+1\right)$$

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

$$2x^{2}-5x-2=\left(6x-3\right)\left(2x+1\right)$$

6x-3 ga 2x+1 ni ko‘paytirish orqali distributiv xususiyatdan foydalaning va ifoda sifatida birlashtiring.

$$2x^{2}-5x-2=12x^{2}-3$$

Ikkala tarafdan 12x^{2} ni ayirish.

$$2x^{2}-5x-2-12x^{2}=-3$$

-10x^{2} ni olish uchun 2x^{2} va -12x^{2} ni birlashtirish.

$$-10x^{2}-5x-2=-3$$

2 ni ikki tarafga qo’shing.

$$-10x^{2}-5x=-3+2$$

-1 olish uchun -3 va 2'ni qo'shing.

$$-10x^{2}-5x=-1$$

Ikki tarafini -10 ga bo‘ling.

$$\frac{-10x^{2}-5x}{-10}=-\frac{1}{-10}$$

-10 ga bo'lish -10 ga ko'paytirishni bekor qiladi.

$$x^{2}+\left(-\frac{5}{-10}\right)x=-\frac{1}{-10}$$

\frac{-5}{-10} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{1}{2}x=-\frac{1}{-10}$$

-1 ni -10 ga bo'lish.

$$x^{2}+\frac{1}{2}x=\frac{1}{10}$$

\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{4} olish uchun. Keyin, \frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{1}{10}+\left(\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{4} kvadratini chiqarish.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{10}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{10} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{13}{80}$$

x^{2}+\frac{1}{2}x+\frac{1}{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.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{13}{80}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{13}{80}}$$

Qisqartirish.

$$x+\frac{1}{4}=\frac{\sqrt{65}}{20} x+\frac{1}{4}=-\frac{\sqrt{65}}{20}$$

Tenglamaning ikkala tarafidan \frac{1}{4} ni ayirish.

$$x=\frac{\sqrt{65}}{20}-\frac{1}{4} x=-\frac{\sqrt{65}}{20}-\frac{1}{4}$$