Question

Найдите значение дроби y^2 -11y+26/9y+18 при y=5 y=31 y=112

Answer 1

$displaystyle frac{y^{2}-11y-26}{9y+18}= frac{(y-13)(y+2)}{9(y+2)}= frac{y-13}{9} \ \ npu:y=5 \ \ frac{y-13}{9}= frac{5-13}{9}=- frac{8}{9} \ \ npu:y=31 \ \ frac{y-13}{9}= frac{31-13}{9}=2 \ \ npu:y=112 \ \ frac{y-13}{9}= frac{112-13}{9}=11 \ \ \y^{2}-11y-26=0 \ D=b^{2}-4ac=121+104=225=15^{2} \ \ x_{1,2}= frac{-bб sqrt{D}}{2a} \ x_{1}=13 \ x_{2}=-2 \ \y^{2}-11y-26=(y-13)(y+2)$