xác định số hạng đầu và công bội của cấp số nhân :
\(\left\{{}\begin{matrix}u_1+u_3=3\\u^2_1+u_3^2=5\end{matrix}\right.\)
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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1 2022
\(\left\{{}\begin{matrix}u_1+u_3=10\\\left(u_1+u_3\right)^2-2u_1u_3=50\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u_1+u_3=10\\u_1u_3=25\end{matrix}\right.\)
Theo Viet đảo, \(u_1\) và \(u_3\) là nghiệm:
\(x^2-10x+25=0\Rightarrow x=5\)
\(\Rightarrow u_1=u_3=5\)
\(\Rightarrow\left\{{}\begin{matrix}u_1=5\\u_1q^2=5\end{matrix}\right.\) \(\Rightarrow q^2=1\Rightarrow q=\pm1\)
21 tháng 10 2023
a: \(\left\{{}\begin{matrix}u5-u1=15\\u4-u1=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}u1\cdot q^4-u1=15\\u1\cdot q^3-u1=6\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}u1\left(q^4-1\right)=15\\u1\left(q^3-1\right)=6\end{matrix}\right.\Leftrightarrow\dfrac{q^4-1}{q^3-1}=\dfrac{5}{2}\)
=>\(2\left(q^4-1\right)=5\left(q^3-1\right)\)
=>\(2q^4-2-5q^3+5=0\)
=>\(2q^4-5q^3+3=0\)
=>\(2q^4-2q^3-3q^3+3=0\)
=>\(2q^3\left(q-1\right)-3\left(q-1\right)\left(q^2+q+1\right)=0\)
=>\(\left(q-1\right)\left(2q^3-3q^2-3q-3\right)=0\)
=>\(\left[{}\begin{matrix}q=1\\q\simeq2,39\end{matrix}\right.\)
=>\(u1=\dfrac{6}{q^3-1}\simeq\dfrac{6}{2.39^3-1}\simeq0,47\)
b: \(\left\{{}\begin{matrix}u1-u3+u5=65\\u1+u7=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}u1-u1\cdot q^2+u1\cdot q^4=65\\u1+u1\cdot q^6=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}u1\cdot\left(1-q^2+q^4\right)=65\\u1\left(1+q^6\right)=325\end{matrix}\right.\)
=>\(\dfrac{1-q^2+q^4}{1+q^6}=\dfrac{65}{325}=\dfrac{1}{5}\)
=>\(\dfrac{1}{q^2+1}=\dfrac{1}{5}\)
=>\(q^2+1=5\)
=>q^2=4
=>q=2 hoặc q=-2
TH1: q=2
=>\(u1=\dfrac{325}{q^6+1}=5\)
TH2: q=-2
=>\(u1=\dfrac{325}{\left(-2\right)^6+1}=5\)
2 tháng 1 2023
\(\Leftrightarrow\left\{{}\begin{matrix}u_1-u_1-2q+u_1+4q=65\\u_1+u_1+6q=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+2q=65\\2u1+6q=325\end{matrix}\right.\)
=>u1=-130; q=195/2
TA
3 tháng 1 2023
`u_n = u_1 + (n-1).d`
`{(u_1-u_3+u_5=65),(u_1+u_7=325):}`
`<=>{(u_1-u_1-2d+u_1+4d=65),(u_1+u_1+6d=325):}`
`<=>{(u_1+2d=65),(2u_1+6d=325):}`
`<=>{(u_1=-130),(u_2=195/2):}`
8 tháng 10 2023
a:
ĐKXĐ: \(q\notin\left\{0;1;-1\right\}\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^4-u1=15\\u1\cdot q^3-u1\cdot q=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-1}{q^3-q}=\dfrac{15}{6}=\dfrac{5}{2}\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2q^4-2=5q^3-5q\\u1\left(q^4-1\right)=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2q^4-5q^3+5q-2=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(q-2\right)\left(q-1\right)\left(q+1\right)\left(2q-1\right)=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}q=2\\q=\dfrac{1}{2}\end{matrix}\right.\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
TH1: q=2
=>\(u1=\dfrac{15}{2^4-1}=\dfrac{15}{15}=1\)
TH2: q=1/2
=>\(u1=\dfrac{15}{\dfrac{1}{16}-1}=15:\dfrac{-15}{16}=-16\)
b:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1-u1\cdot q^2+u1\cdot q^4=65\\u1+u1\cdot q^6=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-q^2+1}{q^6+1}=\dfrac{1}{5}\\u1\left(1+q^6\right)=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{q^2+1}=\dfrac{1}{5}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=4\\u1\left(q^6+1\right)=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}q\in\left\{2;-2\right\}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow u1=\dfrac{325}{65}=5\)
c: \(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^3+u1\cdot q^5=-540\\u1\cdot q+u1\cdot q^3=-60\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{q^5+q^3}{q^3+q}=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\)
TH1: q=3
\(u1=-\dfrac{60}{3+3^3}=-\dfrac{60}{30}=-2\)
TH2: q=-3
=>\(u1=-\dfrac{60}{-3-27}=\dfrac{60}{30}=2\)
ND
Nguyễn Đức Trí
VIP
17 tháng 9 2023
a) \(\left\{{}\begin{matrix}u_5=96\\u_7=384\end{matrix}\right.\)
\(u^2_6=u_5.u_7=96.384=36864\)
\(\Leftrightarrow u_6=192\)
\(q=\dfrac{u_7}{u_6}=\dfrac{384}{192}=2\)
\(u_5=u_1.q^4\)
\(\Leftrightarrow u_1=\dfrac{u_5}{q^4}=\dfrac{96}{2^4}=6\)
b) \(\left\{{}\begin{matrix}u_4-u_2=25\\u_3-u_1=50\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q^3-u_1.q=25\\u_1.q^2-u_1=50\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q\left(q^2-1\right)=25\left(1\right)\\u_1.\left(q^2-1\right)=50\left(2\right)\end{matrix}\right.\)
\(\left(1\right):\left(2\right)\Leftrightarrow q=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\left(2\right)\Leftrightarrow u_1=\dfrac{50}{q^2-1}=\dfrac{50}{\dfrac{1}{4}-1}=-\dfrac{200}{3}\)
10 tháng 9 2023
a: u4=4 và u6=8
=>u1+3d=4 và u1+5d=8
=>-2d=-4 và u1+3d=4
=>d=2 và u1=4-3d=-2
b: u1-u3+u5=10 và u1+u6=17
=>u1-u1-2d+u1+4d=10 và u1+u1+5d=17
=>u1+2d=10 và 2u1+5d=17
=>u1=16 và d=-3
c: u1+u2=5 và u3*u5=91
=>u1+u1+d=5 và (u1+2d)(u1+4d)=91
=>2u1+d=5 và (u1+2d)(u1+4d)=91
=>d=5-2u1 và (u1+10-4u1)(u1+20-8u1)=91
=>d=5-2u1 và (-3u1+10)(-7u1+20)=91
(-3u1+10)(-7u1+20)=91
=>21u1^2-60u1-70u1+200=91
=>21u1^2-130u1+109=0
=>u1=1 hoặc u1=109/21
Khi u1=1 thì d=5-2u1=5-2=3
Khi u1=109/21 thì d=5-2u1=5-218/21=-113/21
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1 2022
a.
\(\left\{{}\begin{matrix}u_1+\left(u_1+4d\right)-\left(u_1+2d\right)=10\\\left(u_1+d\right)+\left(u_1+4d\right)=7\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u_1+2d=10\\2u_1+5d=7\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=36\\d=-13\end{matrix}\right.\)
b.
\(\left\{{}\begin{matrix}u_1+d+u_1+3d=5\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}4d=5-2u_1\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)
\(\Rightarrow u_1^2+\left(u_1+5-2u_1\right)^2=25\)
\(\Rightarrow u_1^2+u_1^2-10u_1+25=25\)
\(\Rightarrow\left[{}\begin{matrix}u_1=0\Rightarrow d=\dfrac{5}{4}\\u_1=5\Rightarrow d=-\dfrac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}u_1+u_3=3\\\left(u_1+u_3\right)^2-2u_1u_3=5\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u_1+u_3=3\\u_1u_3=2\end{matrix}\right.\)
Theo Viet đảo, \(u_1\) và \(u_3\) là nghiệm: \(t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}u_1=1\\u_3=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=1\\u_1.q^2=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=1\\q=\pm\sqrt{2}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}u_1=2\\u_3=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=2\\u_1q^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u_1=2\\q=\pm\frac{1}{\sqrt{2}}\end{matrix}\right.\)