13/9x^2+1leqx^2+4/3x eching

$$\frac{ 13 }{ 9 } { x }^{ 2 } +1 \leq { x }^{ 2 } + \frac{ 4 }{ 3 } x$$

x uchun yechish

$$x = \frac{3}{2} = 1\frac{1}{2} = 1,5$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan x^{2} ni ayirish.

$$\frac{13}{9}x^{2}+1-x^{2}\leq \frac{4}{3}x$$

\frac{4}{9}x^{2} ni olish uchun \frac{13}{9}x^{2} va -x^{2} ni birlashtirish.

$$\frac{4}{9}x^{2}+1\leq \frac{4}{3}x$$

Ikkala tarafdan \frac{4}{3}x ni ayirish.

$$\frac{4}{9}x^{2}+1-\frac{4}{3}x\leq 0$$

Tengsizlikni yechish uchun chap tomon faktorini hisoblang. 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.

$$\frac{4}{9}x^{2}+1-\frac{4}{3}x=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun \frac{4}{9} ni, b uchun -\frac{4}{3} ni va c uchun 1 ni ayiring.

$$x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times \frac{4}{9}\times 1}}{\frac{4}{9}\times 2}$$

Hisoblarni amalga oshiring.

$$x=\frac{\frac{4}{3}±0}{\frac{8}{9}}$$

Yechimlar bir xil.

$$x=\frac{3}{2}$$

Yechimlardan foydalanib tengsizlikni qaytadan yozing.

$$\frac{4}{9}\left(x-\frac{3}{2}\right)^{2}\leq 0$$

Tengsizlikda x=\frac{3}{2} bor.

$$x=\frac{3}{2}$$