Question

Треба допомога, ......​

Answer 1

$\displaystyle\bf\\5)\\\\b_{1} =2x-3\\\\b_{2} =x-4\\\\b_{3} =x+2\\\\\\b_{2} ^{2} =b_{1} \cdot b_{2} \\\\(x-4)^{2} =(2x-3)\cdot(x+2)\\\\x^{2} -8x+16=2x^{2} +4x-3x-6\\\\x^{2} -8x+16=2x^{2} +x-6\\\\2x^{2} +x-6-x^{2} +8x-16=0\\\\x^{2} +9x-22=0\\\\Teorema \ Vieta \ :\\\\x_{1} +x_{2} =-9\\\\x_{1} \cdot x_{2} =-22\\\\x_{1} =-11 \ \ ; \ \ x_{2} =2\\\\1) \ x_{1} =-11\\\\b_{1} =2\cdot(-11)-3=-22-3=-25\\\\b_{2} =-11-4=-15\\\\b_{3} =-11+2=-9$

При значении x = - 11 прогрессия не является геометрической .

$\displaystyle\bf\\2) \ x_{2} =2\\\\b_{1} =2\cdot 2-3=4-3=1\\\\b_{2} =2-4=-2\\\\b_{3} =2+2=4\\\\Otvet \ : x = 2 \ \ ; \ \ b_{1} =1 \ \ ; \ \ b_{2} =-2 \ \ ; \ \ b_{3} =4\\\\\\b)\\\\\left \{ {{b_{4} - b_{2}=30 } \atop {b_{4} - b_{3}=24 }} \right. \\\\\\\left \{ {{b_{1}\cdot q^{3} - b_{1}\cdot q=30 } \atop {b_{1}\cdot q^{3} - b_{1}\cdot q^{2} =24 }} \right. \\\\\\\left \{ {{b_{1}q\cdot(q^{2} -1)=30 } \atop {b_{1} q^{2} \cdot(q-1)=24}} \right.$

$\displaystyle\bf\\:\left \{ {{b_{1}q\cdot(q-1)(q+1)=30 } \atop {b_{1}q^{2} \cdot(q-1)=24}} \right. \\-------------\\\frac{q+1}{q} =\frac{30}{24} \\\\\\\frac{q+1}{q} =\frac{5}{4} \\\\\\5q=4q+4\\\\\boxed{q=4}\\\\\\b_{1} =\frac{24}{q^{2} \cdot(q-1)} =\frac{24}{16\cdot3} =\frac{1}{2} \\\\\\\boxed{b_{1}=0,5 }$