left(1-9/4right)x^2+x+1-k=0 eching

$$\left( 1- \frac{ 9 }{ 4 } \right) { x }^{ 2 } +x+1-k = 0$$

k uchun yechish

$$k=-\frac{5x^{2}}{4}+x+1$$

Chiziqli tenglamalarni yechish bo'yicha qadamlar

-\frac{5}{4} olish uchun 1 dan \frac{9}{4} ni ayirish.

$$-\frac{5}{4}x^{2}+x+1-k=0$$

\frac{5}{4}x^{2} ni ikki tarafga qo’shing. Har qanday songa nolni qo‘shsangiz, o‘zi chiqadi.

$$x+1-k=\frac{5}{4}x^{2}$$

Ikkala tarafdan x ni ayirish.

$$1-k=\frac{5}{4}x^{2}-x$$

Ikkala tarafdan 1 ni ayirish.

$$-k=\frac{5}{4}x^{2}-x-1$$

Tenglama standart shaklda.

$$-k=\frac{5x^{2}}{4}-x-1$$

Ikki tarafini -1 ga bo‘ling.

$$\frac{-k}{-1}=\frac{\frac{5x^{2}}{4}-x-1}{-1}$$

-1 ga bo'lish -1 ga ko'paytirishni bekor qiladi.

$$k=\frac{\frac{5x^{2}}{4}-x-1}{-1}$$

\frac{5x^{2}}{4}-x-1 ni -1 ga bo'lish.

$$k=-\frac{5x^{2}}{4}+x+1$$

x uchun yechish (complex solution)

$$x=\frac{-2\sqrt{6-5k}+2}{5}$$

$$x=\frac{2\sqrt{6-5k}+2}{5}$$

x uchun yechish

$$x=\frac{-2\sqrt{6-5k}+2}{5}$$

$$x=\frac{2\sqrt{6-5k}+2}{5}\text{, }k\leq \frac{6}{5}$$