2.5-bo'lim : Kvadrat tenglamalar - I qism

1 – 7 masalalar uchun kvadrat tenglamani ko'paytuvchilarga ajratish yordamida yeching.

1. \({u^2} - 5u - 14 = 0\) Yechim
2. \({x^2} + 15x = - 50\) Yechim
3. \({y^2} = 11y - 28\) Yechim
4. \(19x = 7 - 6{x^2}\) Yechim
5. \(6{w^2} - w = 5\) Yechim
6. \({z^2} - 16z + 61 = 2z - 20\) Yechim
7. \(12{x^2} = 25x\) Yechim

8 va 9-masalalar uchun tenglamani yechish uchun ko'paytuvchilarga ajratishdan foydalaning.

1. \({x^4} - 2{x^3} - 3{x^2} = 0\) Yechim
2. \({t^5} = 9{t^3}\) Yechim

10-12-masalalar uchun tenglamani yechish uchun ko'paytuvchilarga ajratishdan foydalaning.

1. \(\displaystyle \frac{{{w^2} - 10}}{{w + 2}} + w - 4 = w - 3\) Yechim
2. \(\displaystyle \frac{{4z}}{{z + 1}} + \frac{5}{z} = \frac{{6z + 5}}{{{z^2} + z}}\) Yechim
3. \(\displaystyle x + 1 = \frac{{2x - 7}}{{x + 5}} - \frac{{5x + 8}}{{x + 5}}\) Yechim

13-16-masalalar uchun tenglamani yechish uchun Kvadrat ildiz xususiyatidan foydalaning.

1. \(9{u^2} - 16 = 0\) Yechim
2. \({x^2} + 15 = 0\) Yechim
3. \({\left( {z - 2} \right)^2} - 36 = 0\) Yechim
4. \({\left( {6t + 1} \right)^2} + 3 = 0\) Yechim