x^2+5/25-x^2=3/x+5+x/x-5 eching

$$\frac { x ^ { 2 } + 5 } { 25 - x ^ { 2 } } = \frac { 3 } { x + 5 } + \frac { x } { x - 5 }$$

x uchun yechish

$$x=1$$

Guruhlash orqali faktorlash qadamlari

x qiymati -5,5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(x+5\right) ga, 25-x^{2},x+5,x-5 ning eng kichik karralisiga ko‘paytiring.

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

x^{2}+5 teskarisini topish uchun har birining teskarisini toping.

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

x-5 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-x^{2}-5=3x-15+\left(x+5\right)x$$

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

$$-x^{2}-5=3x-15+x^{2}+5x$$

8x ni olish uchun 3x va 5x ni birlashtirish.

$$-x^{2}-5=8x-15+x^{2}$$

Ikkala tarafdan 8x ni ayirish.

$$-x^{2}-5-8x=-15+x^{2}$$

Ikkala tarafdan -15 ni ayirish.

$$-x^{2}-5-8x-\left(-15\right)=x^{2}$$

-15 ning teskarisi 15 ga teng.

$$-x^{2}-5-8x+15=x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

$$-x^{2}-5-8x+15-x^{2}=0$$

10 olish uchun -5 va 15'ni qo'shing.

$$-x^{2}+10-8x-x^{2}=0$$

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

$$-2x^{2}+10-8x=0$$

Ikki tarafini 2 ga bo‘ling.

$$-x^{2}+5-4x=0$$

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

$$-x^{2}-4x+5=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon -x^{2}+ax+bx+5 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-4 ab=-5=-5$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=1 b=-5$$

-x^{2}-4x+5 ni \left(-x^{2}+x\right)+\left(-5x+5\right) sifatida qaytadan yozish.

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

Birinchi guruhda x ni va ikkinchi guruhda 5 ni faktordan chiqaring.

$$x\left(-x+1\right)+5\left(-x+1\right)$$

Distributiv funktsiyasidan foydalangan holda -x+1 umumiy terminini chiqaring.

$$\left(-x+1\right)\left(x+5\right)$$

Tenglamani yechish uchun -x+1=0 va x+5=0 ni yeching.

$$x=1 x=-5$$

x qiymati -5 teng bo‘lmaydi.

$$x=1$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -5,5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(x+5\right) ga, 25-x^{2},x+5,x-5 ning eng kichik karralisiga ko‘paytiring.

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

x^{2}+5 teskarisini topish uchun har birining teskarisini toping.

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

x-5 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-x^{2}-5=3x-15+\left(x+5\right)x$$

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

$$-x^{2}-5=3x-15+x^{2}+5x$$

8x ni olish uchun 3x va 5x ni birlashtirish.

$$-x^{2}-5=8x-15+x^{2}$$

Ikkala tarafdan 8x ni ayirish.

$$-x^{2}-5-8x=-15+x^{2}$$

Ikkala tarafdan -15 ni ayirish.

$$-x^{2}-5-8x-\left(-15\right)=x^{2}$$

-15 ning teskarisi 15 ga teng.

$$-x^{2}-5-8x+15=x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

$$-x^{2}-5-8x+15-x^{2}=0$$

10 olish uchun -5 va 15'ni qo'shing.

$$-x^{2}+10-8x-x^{2}=0$$

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

$$-2x^{2}+10-8x=0$$

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.

$$-2x^{2}-8x+10=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} -2 ni a, -8 ni b va 10 ni c bilan almashtiring.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

-8 kvadratini chiqarish.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-2\right)\times 10}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64+8\times 10}}{2\left(-2\right)}$$

8 ni 10 marotabaga ko'paytirish.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-2\right)}$$

64 ni 80 ga qo'shish.

$$x=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-2\right)}$$

144 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-8\right)±12}{2\left(-2\right)}$$

-8 ning teskarisi 8 ga teng.

$$x=\frac{8±12}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{8±12}{-4}$$

x=\frac{8±12}{-4} tenglamasini yeching, bunda ± musbat. 8 ni 12 ga qo'shish.

$$x=\frac{20}{-4}$$

20 ni -4 ga bo'lish.

$$x=-5$$

x=\frac{8±12}{-4} tenglamasini yeching, bunda ± manfiy. 8 dan 12 ni ayirish.

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

-4 ni -4 ga bo'lish.

$$x=1$$

Tenglama yechildi.

$$x=-5 x=1$$

x qiymati -5 teng bo‘lmaydi.

$$x=1$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -5,5 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini \left(x-5\right)\left(x+5\right) ga, 25-x^{2},x+5,x-5 ning eng kichik karralisiga ko‘paytiring.

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

x^{2}+5 teskarisini topish uchun har birining teskarisini toping.

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

x-5 ga 3 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$-x^{2}-5=3x-15+\left(x+5\right)x$$

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

$$-x^{2}-5=3x-15+x^{2}+5x$$

8x ni olish uchun 3x va 5x ni birlashtirish.

$$-x^{2}-5=8x-15+x^{2}$$

Ikkala tarafdan 8x ni ayirish.

$$-x^{2}-5-8x=-15+x^{2}$$

Ikkala tarafdan x^{2} ni ayirish.

$$-x^{2}-5-8x-x^{2}=-15$$

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

$$-2x^{2}-5-8x=-15$$

5 ni ikki tarafga qo’shing.

$$-2x^{2}-8x=-15+5$$

-10 olish uchun -15 va 5'ni qo'shing.

$$-2x^{2}-8x=-10$$

Ikki tarafini -2 ga bo‘ling.

$$\frac{-2x^{2}-8x}{-2}=-\frac{10}{-2}$$

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

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

-8 ni -2 ga bo'lish.

$$x^{2}+4x=-\frac{10}{-2}$$

-10 ni -2 ga bo'lish.

$$x^{2}+4x=5$$

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

$$x^{2}+4x+2^{2}=5+2^{2}$$

2 kvadratini chiqarish.

$$x^{2}+4x+4=5+4$$

5 ni 4 ga qo'shish.

$$x^{2}+4x+4=9$$

x^{2}+4x+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.

$$\left(x+2\right)^{2}=9$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{9}$$

Qisqartirish.

$$x+2=3 x+2=-3$$

Tenglamaning ikkala tarafidan 2 ni ayirish.

$$x=1 x=-5$$

x qiymati -5 teng bo‘lmaydi.

$$x=1$$