(x^2-12x+10)^2=(3x+10)^2 tenglamaning barcha haqiqiy yechimlari yigʻindisini toping.

$\left(x^2-12x+10\right)^2=\left(3x+10\right)^2$ ni $\left(x^2-12x+10\right)^2-\left(3x+10\right)^2=0$ ko‘rinishda ifodalab olamiz. $a^2-b^2=(a+b)(a-b)$ qisqa ko‘paytirish formulasidan foydalanib, ko‘paytuvchilarga ajratamiz. $\left(x^2-12x+10\right)^2-\left(3x+10\right)^2=\left(x^2-9x+20\right)\left(x^2-15x\right)=0.$ Bu tenglama $\left[\begin{array}{l} x^2-9x+20=0 \\ x^2-15x=0 \end{array}\right.$ tenglamaga teng kuchli. $x^2-9x+20=0$ tenglamaning yechimlari 4 va 5 ga, $x^2-15x=0$ tenglamaning yechimlari esa 0 va 15 ga teng. Demak, $\left(x^2-12x+10\right)^2=\left(3x+10\right)^2$ tenglamaning yechimlari 0; 4; 5 va 15. Ularning yig‘indisi 24 ga teng.

Manba: UZBMB namunaviy test topshiriqlari va ularni yechish bo‘yicha tavsiyalar (2020). M.A.Mirzaahmedov va boshqalar. Matematika, o‘rta ta’lim muassasalarining 10-sinfi va o‘rta maxsus, kasb-hunar ta’limi muassasalari o‘quvchilari uchun darslik. “EXTREMUM PRESS”, Toshkent – 2017