\(u_2=u_1+d=-2+d\) ; \(v_2=v_1q=-2q\)

\(u_2=v_2\Rightarrow-2+d=-2q\Rightarrow d=2-2q\)

\(u_3=v_3+8\Leftrightarrow-2+2d=-2q^2+8\)

\(\Leftrightarrow-2+2\left(2-2q\right)=-2q^2+8\)

\(\Leftrightarrow2q^2-4q-6=0\Rightarrow\left[{}\begin{matrix}q=-1\Rightarrow d=4\\q=3\Rightarrow d=-4\end{matrix}\right.\)

Câu 2: 

\(\left(x+1\right)^2+\left(y-2\right)^2=9\)

=>R=3 và I(-1;2)

Tọa độ I' là:

x=-1+1=0 và y=2-2=0

=>Phương trình (C') là: x^2+y^2=9

Câu 3: 

\(V_{\left(O;-2\right)}\left(C\right)=\left(C'\right)\)

\(x^2+y^2-2x-8=0\)

=>x^2-2x+1+y^2=9

=>(x-1)^2+y^2=9

=>R=3 và I(1;0)

Tọa độ I' là:

\(\left\{{}\begin{matrix}x=1\cdot\left(-2\right)=-2\\y=0\cdot\left(-2\right)=0\end{matrix}\right.\)

Độ dài R' là:

\(R=3\cdot\left|-2\right|=6\)

Tọa độ (C') là:

\(\left(x+2\right)^2+y^2=36\)

\(f\left(x\right)=\sqrt{4+3u\left(x\right)}\)

\(\Leftrightarrow f'\left(x\right)=\dfrac{\left(4+3u\left(x\right)\right)'}{2\sqrt{4+3u\left(x\right)}}=\dfrac{3u'\left(x\right)}{2\cdot\sqrt{4+3u\left(x\right)}}\)

\(f'\left(1\right)=\dfrac{3\cdot u'\left(1\right)}{2\cdot\sqrt{4+3u\left(1\right)}}=\dfrac{3\cdot10}{2\cdot\sqrt{4+3\cdot7}}=3\)

=>Chọn C

\(u_2=u_1.q,u_5=u_1.q^4,u_6=u_1.q^5\) nên

\(u_1(1+q^4)=51,u_1q(1+q^4)=102\)

chia 2 vế ta được q=2, suy ra u₁=3

a/

\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(u_n=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)

\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)

b/ \(u_n=\dfrac{1}{1^2+3}+\dfrac{1}{2^2+6}+...+\dfrac{1}{n^2+3n}=\dfrac{1}{1.4}+\dfrac{1}{2.5}+...+\dfrac{1}{n\left(n+3\right)}\)

\(u_n=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\right)\)

\(u_n=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\)

\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\right)\)

\(\Rightarrow lim\left(u_n\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{11}{18}\)