560div(x+10)+1=560divx eching

$$560 \div (x+10)+1=560 \div x$$

x uchun yechish

$$x=-80$$

$$x=70$$

Kvadrat tenglamalaridan foydalanish qoidalari

x qiymati -10,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x\left(x+10\right) ga, x+10,x ning eng kichik karralisiga ko‘paytiring.

$$x\times 560+x\left(x+10\right)=\left(x+10\right)\times 560$$

x ga x+10 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x\times 560+x^{2}+10x=\left(x+10\right)\times 560$$

570x ni olish uchun x\times 560 va 10x ni birlashtirish.

$$570x+x^{2}=\left(x+10\right)\times 560$$

x+10 ga 560 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$570x+x^{2}=560x+5600$$

Ikkala tarafdan 560x ni ayirish.

$$570x+x^{2}-560x=5600$$

10x ni olish uchun 570x va -560x ni birlashtirish.

$$10x+x^{2}=5600$$

Ikkala tarafdan 5600 ni ayirish.

$$10x+x^{2}-5600=0$$

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.

$$x^{2}+10x-5600=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 1 ni a, 10 ni b va -5600 ni c bilan almashtiring.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-5600\right)}}{2}$$

10 kvadratini chiqarish.

$$x=\frac{-10±\sqrt{100-4\left(-5600\right)}}{2}$$

-4 ni -5600 marotabaga ko'paytirish.

$$x=\frac{-10±\sqrt{100+22400}}{2}$$

100 ni 22400 ga qo'shish.

$$x=\frac{-10±\sqrt{22500}}{2}$$

22500 ning kvadrat ildizini chiqarish.

$$x=\frac{-10±150}{2}$$

x=\frac{-10±150}{2} tenglamasini yeching, bunda ± musbat. -10 ni 150 ga qo'shish.

$$x=\frac{140}{2}$$

140 ni 2 ga bo'lish.

$$x=70$$

x=\frac{-10±150}{2} tenglamasini yeching, bunda ± manfiy. -10 dan 150 ni ayirish.

$$x=-\frac{160}{2}$$

-160 ni 2 ga bo'lish.

$$x=-80$$

Tenglama yechildi.

$$x=70 x=-80$$

Kvadratni yakunlash bo'yicha qadamlar

x qiymati -10,0 qiymatlaridan birortasiga teng bo‘lmaydi, chunki nolga bo‘lish mumkin emas. Tenglamaning ikkala tarafini x\left(x+10\right) ga, x+10,x ning eng kichik karralisiga ko‘paytiring.

$$x\times 560+x\left(x+10\right)=\left(x+10\right)\times 560$$

x ga x+10 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$x\times 560+x^{2}+10x=\left(x+10\right)\times 560$$

570x ni olish uchun x\times 560 va 10x ni birlashtirish.

$$570x+x^{2}=\left(x+10\right)\times 560$$

x+10 ga 560 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$570x+x^{2}=560x+5600$$

Ikkala tarafdan 560x ni ayirish.

$$570x+x^{2}-560x=5600$$

10x ni olish uchun 570x va -560x ni birlashtirish.

$$10x+x^{2}=5600$$

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

$$x^{2}+10x=5600$$

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

$$x^{2}+10x+5^{2}=5600+5^{2}$$

5 kvadratini chiqarish.

$$x^{2}+10x+25=5600+25$$

5600 ni 25 ga qo'shish.

$$x^{2}+10x+25=5625$$

x^{2}+10x+25 omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+5\right)^{2}=5625$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+5\right)^{2}}=\sqrt{5625}$$

Qisqartirish.

$$x+5=75 x+5=-75$$

Tenglamaning ikkala tarafidan 5 ni ayirish.

$$x=70 x=-80$$