\(\Leftrightarrow\left\{{}\begin{matrix}u_1-u_1-2q+u_1+4q=65\\u_1+u_1+6q=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u_1+2q=65\\2u1+6q=325\end{matrix}\right.\)

=>u1=-130; q=195/2

`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):}` 

`u_n=u_1 . q^(n-1)`

`{(u_1-u_3+u_5=65),(u_1+u_7=325):}`

`<=>{(u_1 -u_1 .q^2 +u_1 .q^4=65),(u_1+u_1 .q^6=325):}`

`<=>{(u_1(1- q^2 + q^4)=65 \(1)),(u_1 .(1+q^6=325 \(2)):}`

Lấy (2) : (1) được: `(q^6+1)/(q^4-q^2+1)=5`

`<=>q=+-2`

TH1: `q=2=>u_1=5`

TH2: `q=-2=> u_1=5`

Vậy `(u_1;q)=(5;2),(5;-2)` 

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

=>\(u1=\dfrac{325}{\left(-2\right)^6+1}=5\)

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

=>\(u1=-\dfrac{60}{-3-27}=\dfrac{60}{30}=2\)

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.

\(\left\{{}\begin{matrix}u_1+d+u_1+3d=5\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}4d=5-2u_1\\u_1^2+\left(u_1+4d\right)^2=25\end{matrix}\right.\)

\(\Rightarrow u_1^2+\left(u_1+5-2u_1\right)^2=25\)

\(\Rightarrow u_1^2+u_1^2-10u_1+25=25\)

\(\Rightarrow\left[{}\begin{matrix}u_1=0\Rightarrow d=\dfrac{5}{4}\\u_1=5\Rightarrow d=-\dfrac{5}{4}\end{matrix}\right.\)

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) \(\left\{{}\begin{matrix}u_4-u_2=25\\u_3-u_1=50\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q^3-u_1.q=25\\u_1.q^2-u_1=50\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}u_1.q\left(q^2-1\right)=25\left(1\right)\\u_1.\left(q^2-1\right)=50\left(2\right)\end{matrix}\right.\)

\(\left(1\right):\left(2\right)\Leftrightarrow q=\dfrac{25}{50}=\dfrac{1}{2}\)

\(\left(2\right)\Leftrightarrow u_1=\dfrac{50}{q^2-1}=\dfrac{50}{\dfrac{1}{4}-1}=-\dfrac{200}{3}\)

a) Từ hệ thức đã cho ta có:

hay

.Giải hệ này tìm u₁ và d. Đáp số u₁ = 16, d = -3.

b) Từ hệ đã cho ta có:

hay

Giải hệ này để tìm u₁ và d. Đáp số u₁ = 3 và d = 2 hoặc u₁ = -17 và d = 2

u₁ = 3 và d = 2 hoặc u₁ = -17 và d = 2.

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