Question

написать первых четыре члена геометрической прогрессии
{b1+b4=27
{b2*b3=72

Answer 1

$left { {{b_1+b_4=27} atop {b_2cdot b+3=72}} right.; left { {{b_1+b_1q^3=27} atop {b_1qcdot b_1q^2=72}} right. ; left { {{b_1cdot (1+q^3)=27} atop {b_1^2cdot q^3=72}} right. ; left { {{b_1cdot (1+frac{72}{b_1^2})=27} atop {q^3=frac{72}{b_1^2}}} right. \\\b_1cdot (1+frac{72}{b_1^2})=27; ; to ; ; ; b_1+frac{72}{b_1}=27; ,; ; frac{b_1^2-27b_1+71}{b_1}=0; ,\\b_1^2-27b_1+72=0; ,; ; D=441=21^2; ,; ; b_{1,2}=frac{27pm 21}{2}\\b_1=3; ; ; ili; ; ; b_1=24$

$q^3=frac{72}{3^2}=8; ; ; ili; ; ; q^3=frac{72}{24^3}=0,125\\q=2qquad ; iliqquad q=0,5\\a); ; b_1=3; ,; q=2:; quad 3; ;; 6; ;; 12; ;; 24; ;; 48; ;, ...\\b); ; b_1=24; ,; q=0,5:quad ; 24; ;; 12; ;; 6; ;; 3; ;; 1,5; ;, ...$