x^2+x eching

$${ x }^{ 2 } +x$$

Omil

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

Yechim qadamlari

x omili.

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

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

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

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.

$$x=\frac{-1±\sqrt{1^{2}}}{2}$$

1^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-1±1}{2}$$

x=\frac{-1±1}{2} tenglamasini yeching, bunda ± musbat. -1 ni 1 ga qo'shish.

$$x=\frac{0}{2}$$

0 ni 2 ga bo'lish.

$$x=0$$

x=\frac{-1±1}{2} tenglamasini yeching, bunda ± manfiy. -1 dan 1 ni ayirish.

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

-2 ni 2 ga bo'lish.

$$x=-1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun 0 ga va x_{2} uchun -1 ga bo‘ling.

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

p-\left(-q\right) shaklining barcha amallarigani p+q ga soddalashtiring.

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

Baholash

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