(5x-1)(5x+1)=-1-5x eching

$$(5x-1)(5x+1)=-1-5x$$

x uchun yechish

$$x=-\frac{1}{5}=-0,2$$

$$x=0$$

Kvadrat tenglamalaridan foydalanish qoidalari

Hisoblang: \left(5x-1\right)\left(5x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

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

\left(5x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$25x^{2}-1=-1-5x$$

Ikkala tarafdan -1 ni ayirish.

$$25x^{2}-1-\left(-1\right)=-5x$$

-1 ning teskarisi 1 ga teng.

$$25x^{2}-1+1=-5x$$

5x ni ikki tarafga qo’shing.

$$25x^{2}-1+1+5x=0$$

0 olish uchun -1 va 1'ni qo'shing.

$$25x^{2}+5x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 25 ni a, 5 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-5±\sqrt{5^{2}}}{2\times 25}$$

5^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-5±5}{2\times 25}$$

2 ni 25 marotabaga ko'paytirish.

$$x=\frac{-5±5}{50}$$

x=\frac{-5±5}{50} tenglamasini yeching, bunda ± musbat. -5 ni 5 ga qo'shish.

$$x=\frac{0}{50}$$

0 ni 50 ga bo'lish.

$$x=0$$

x=\frac{-5±5}{50} tenglamasini yeching, bunda ± manfiy. -5 dan 5 ni ayirish.

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

\frac{-10}{50} ulushini 10 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

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

Tenglama yechildi.

$$x=0 x=-\frac{1}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

Hisoblang: \left(5x-1\right)\left(5x+1\right). Ko‘paytirish qoida yordamida turli kvadratlarga aylantirilishi mumkin: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. 1 kvadratini chiqarish.

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

\left(5x\right)^{2} ni kengaytirish.

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

2 daraja ko‘rsatkichini 5 ga hisoblang va 25 ni qiymatni oling.

$$25x^{2}-1=-1-5x$$

5x ni ikki tarafga qo’shing.

$$25x^{2}-1+5x=-1$$

1 ni ikki tarafga qo’shing.

$$25x^{2}+5x=-1+1$$

0 olish uchun -1 va 1'ni qo'shing.

$$25x^{2}+5x=0$$

Ikki tarafini 25 ga bo‘ling.

$$\frac{25x^{2}+5x}{25}=\frac{0}{25}$$

25 ga bo'lish 25 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{5}{25}x=\frac{0}{25}$$

\frac{5}{25} ulushini 5 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x^{2}+\frac{1}{5}x=\frac{0}{25}$$

0 ni 25 ga bo'lish.

$$x^{2}+\frac{1}{5}x=0$$

\frac{1}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{1}{10} olish uchun. Keyin, \frac{1}{10} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

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

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{1}{10} kvadratini chiqarish.

$$x^{2}+\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

x^{2}+\frac{1}{5}x+\frac{1}{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.

$$\left(x+\frac{1}{10}\right)^{2}=\frac{1}{100}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Qisqartirish.

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

Tenglamaning ikkala tarafidan \frac{1}{10} ni ayirish.

$$x=0 x=-\frac{1}{5}$$