Quá sai.

1B; 2B; 3A; 4D; 5A

a) ĐKXD: \(\left\{{}\begin{matrix}a>0\\a\ne1\\a\ne4\end{matrix}\right.\)

b) Với \(a>0;a\ne1;a\ne4\), ta có:

\(B=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\\ =\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{3}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

c)\(B\le\dfrac{1}{3}\rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}\le\dfrac{1}{3}\rightarrow\dfrac{-2}{\sqrt{a}}\le0\) (đúng với mọi a thoả ĐKXĐ).

\(\lim\limits_{x\rightarrow2}f\left(x\right)=\lim\limits_{x\rightarrow2}\dfrac{x^2+x-6}{x-2}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+3\right)}{x-2}=\lim\limits_{x\rightarrow2}\left(x+3\right)=5\\ f\left(2\right)=5\\ \rightarrow\lim\limits_{x\rightarrow2}f\left(x\right)=f\left(2\right)\)

Suy ra f(x) liên tục tại x = 2.

\(\dfrac{u_2}{q\left(1-q\right)}=32\rightarrow\dfrac{8}{q\left(1-q\right)}=32\rightarrow q=\dfrac{1}{2}\rightarrow u_1=16\\ \rightarrow u_n=16\left(\dfrac{1}{2}\right)^{n-1}=2^{5-n}\)

C

Thiếu \(\lim\limits_{x\rightarrow1^+}\left(x-1\right)=0\)

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

(Do \(lim\left(1-\dfrac{1}{3^n}\right)=1>0\\ lim\left[\left(\dfrac{2}{3}\right)^{n+1}-\dfrac{2}{3^{n+1}}\right]=0\\ \left(\dfrac{2}{3}\right)^{n+1}-\dfrac{2}{3^{n+1}}>0\))

Ta sẽ chứng minh \(\left(u_n\right)\) giảm, tức \(u_{n+1}< u_n\) (*) bằng phương pháp quy nạp.

Với n = 1: \(u_2-u_1=\dfrac{u_1^2+1}{4}-u_1=\dfrac{2^2+1}{4}-2=\dfrac{-3}{4}< 0\)

Giả sử (*) đúng với n = k (\(k\in N;k>1\)), tức \(u_{k+1}< u_k\)

Ta sẽ chứng minh (*) đúng với n = k + 1, tức \(u_{k+2}< u_{k+1}\)

\(u_{k+2}=\dfrac{\left(u_{k+1}\right)^2+1}{4}< \dfrac{u_k^2+1}{4}=u_{k+1}\)

Theo nguyên lí quy nạp, ta được đpcm.

Vậy \(\left(u_n\right)\) giảm.

Xét dãy \(\left(u_n\right)\) là cấp số nhân có \(\left\{{}\begin{matrix}u_1=x^2\\q=-x\end{matrix}\right.\)

\(S=x^2-x^3+x^4-x^5+...+\left(-1\right)^nx^n+...=\dfrac{x^2}{1-\left(-x\right)}=\dfrac{x^2}{x+1}\)