Toán u1 + u2 + u3 + u4 + u5 = 49*(1/u1 +1/u2+…+1/u5) và u1+u3 = 35 12/07/2021 By Maria u1 + u2 + u3 + u4 + u5 = 49*(1/u1 +1/u2+…+1/u5) và u1+u3 = 35
Ta có: $u_1(1+q+q^2+q^3+q^4)=\frac{49}{u_1}(1+$ $\frac{1}{q}+$ $\frac{1}{q^2}+$ $\frac{1}{q^3}+$ $\frac{1}{q^4})$ ⇔$u_1(1+q+q^2+q^3+q^4)=\frac{49}{u_1}\frac{q^4+q^3+q^2+q+1}{q^4}$ ⇔$(u_1^2.q^4)=49$ ⇔\(\left[ \begin{array}{l}u_1q^2=7\\u_1q^2=-7\end{array} \right.\) TH1: $u_1q^2=7$ ⇒$u_1+u_1.q^2=35$ ⇔$u_1+7=35$ ⇔$u_1=28$ ⇒$q=±1/2$ TH2: $u_1q^2=-7$ ⇒$u_1+u_1.q^2=35$ ⇔$u_1-7=35$ ⇔$u_1=42$ ⇒$q^2=-1/6(loại)$ vậy $u_1=28; q=±1/2$ Trả lời
Đáp án: Giải thích các bước giải: u1(1+q+q2+q3+q4)=49u1(1+u1(1+q+q2+q3+q4)=49u1(1+ 1q+1q+ 1q2+1q2+ 1q3+1q3+ 1q4)1q4) ⇔u1(1+q+q2+q3+q4)=49u1q4+q3+q2+q+1q4u1(1+q+q2+q3+q4)=49u1q4+q3+q2+q+1q4 ⇔(u21.q4)=49(u12.q4)=49 ⇔[u1q2=7u1q2=−7[u1q2=7u1q2=−7 TH1: u1q2=7u1q2=7 ⇒u1+u1.q2=35u1+u1.q2=35 ⇔u1+7=35u1+7=35 ⇔u1=28u1=28 ⇒q=±1/2q=±1/2 TH2: u1q2=−7u1q2=−7 ⇒u1+u1.q2=35u1+u1.q2=35 ⇔u1−7=35u1−7=35 ⇔u1=42u1=42 ⇒q2=−1/6(loại)q2=−1/6(loại) vậy u1=28;q=±1/2 Trả lời
Ta có:
$u_1(1+q+q^2+q^3+q^4)=\frac{49}{u_1}(1+$ $\frac{1}{q}+$ $\frac{1}{q^2}+$ $\frac{1}{q^3}+$ $\frac{1}{q^4})$
⇔$u_1(1+q+q^2+q^3+q^4)=\frac{49}{u_1}\frac{q^4+q^3+q^2+q+1}{q^4}$
⇔$(u_1^2.q^4)=49$
⇔\(\left[ \begin{array}{l}u_1q^2=7\\u_1q^2=-7\end{array} \right.\)
TH1: $u_1q^2=7$
⇒$u_1+u_1.q^2=35$
⇔$u_1+7=35$
⇔$u_1=28$
⇒$q=±1/2$
TH2: $u_1q^2=-7$
⇒$u_1+u_1.q^2=35$
⇔$u_1-7=35$
⇔$u_1=42$
⇒$q^2=-1/6(loại)$
vậy $u_1=28; q=±1/2$
Đáp án:
Giải thích các bước giải:
u1(1+q+q2+q3+q4)=49u1(1+u1(1+q+q2+q3+q4)=49u1(1+ 1q+1q+ 1q2+1q2+ 1q3+1q3+ 1q4)1q4)
⇔u1(1+q+q2+q3+q4)=49u1q4+q3+q2+q+1q4u1(1+q+q2+q3+q4)=49u1q4+q3+q2+q+1q4
⇔(u21.q4)=49(u12.q4)=49
⇔[u1q2=7u1q2=−7[u1q2=7u1q2=−7
TH1: u1q2=7u1q2=7
⇒u1+u1.q2=35u1+u1.q2=35
⇔u1+7=35u1+7=35
⇔u1=28u1=28
⇒q=±1/2q=±1/2
TH2: u1q2=−7u1q2=−7
⇒u1+u1.q2=35u1+u1.q2=35
⇔u1−7=35u1−7=35
⇔u1=42u1=42
⇒q2=−1/6(loại)q2=−1/6(loại)
vậy u1=28;q=±1/2