Ta có:

\(\overrightarrow{AB}=\left(-3;-2;2\right)\)\(\Rightarrow AB=\sqrt{\left(-3\right)^2+\left(-2\right)^2+2^2}=\sqrt{17}\)

\(\overrightarrow{BC}=\left(4;4;1\right)\Rightarrow BC=\sqrt{4^2+4^2+1^2}=\sqrt{33}\)

\(\overrightarrow{AC}=\left(1;2;3\right)\)\(\Rightarrow AC=\sqrt{1^2+2^2+3^2}=\sqrt{14}\)

\(\Rightarrow C_{\Delta ABC}=\sqrt{17}+\sqrt{33}+\sqrt{14}\)

\(a\text{)}lim\dfrac{5n+1}{2n}=lim\dfrac{5}{2}+lim\dfrac{1}{2n}=\dfrac{5}{2}\)

\(b\text{)}lim\dfrac{6n^2+8n+1}{5n^2+3}=lim\dfrac{6+\dfrac{8}{n}+\dfrac{1}{n^2}}{5+\dfrac{3}{n^2}}=\dfrac{6}{5}\)

\(c\text{)}lim\dfrac{3^n+2^n}{4.3^n}=lim\dfrac{\left(\dfrac{3}{3}\right)^n+\left(\dfrac{2}{3}\right)^n}{4}=\dfrac{1}{4}\)

\(d\text{)}lim\dfrac{\sqrt{n^2+5n+3}}{6n+2}=lim\dfrac{n\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{n\left(6+\dfrac{2}{n}\right)}=lim\dfrac{\sqrt{1+\dfrac{5}{n}+\dfrac{3}{n^2}}}{6+\dfrac{2}{n}}=\dfrac{1}{6}\)

\(MinA=0\Leftrightarrow7+4x-4x^2=0\Leftrightarrow x=\dfrac{1\pm2\sqrt{2}}{2}\)

Bạn xem lại đề bài, n=2 không thoả mãn

Do \(\left(SC;\left(ABCD\right)\right)=45^0;SA\perp\left(ABCD\right)\)

nên \(\left\{{}\begin{matrix}\left(SC;AC\right)=45^0\\AS\perp AC\end{matrix}\right.\)\(\Rightarrow AS=AC=\sqrt{AB^2+BC^2}=\sqrt{a^2+a^2}=a\sqrt{2}\)

\(\Rightarrow V_{S.ABCD}=\dfrac{1}{6}.\left(AD+BC\right).AB.AS\)

\(=\dfrac{1}{6}\left(2a+a\right).a.a\sqrt{2}=\dfrac{\sqrt{2}}{2}a^3\)

Ta có:

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

Xét hiệu \(u_{n+1}-u_n\), ta có:

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

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

\(=\dfrac{\left(6n^2+5n-6\right)-\left(6n^2+5n+1\right)}{\left(3n+1\right)\left(3n-2\right)}\)

\(=\dfrac{-7}{\left(3n+1\right)\left(3n-2\right)}< 0,\forall n\in N\text{*}\)

\(\Rightarrow u_{n+1}< u_n,\forall n\in N\text{*}\)

Vậy dãy số \(\left(u_n\right)\) đã cho là dãy giảm

\(n_{Ba^{2+}}=0,1.0,1=0,01\left(mol\right)\)

\(n_{SO_4^{2-}}=0,4.0,0175=7.10 ^{-3}\left(mol\right)\)

\(Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4\downarrow\)

\(\Rightarrow m=m_{BaSO_4}=7.10^{-3}.233=1,631\left(g\right)\)

Ta có:

\(n_{H^+}=0,4.0,0175.2=0,014\left(mol\right)\)

\(n_{OH^-}=0,1.0,1.2+0,1.0,1=0,03\left(mol\right)\)

Trong dung dịch X:

\(n_{OH^-}=0,03-0,014=0,016\left(mol\right)\)\(\Rightarrow\left[OH^-\right]=\dfrac{0,016}{0,1+0,4}=0,032\left(M\right)\)

\(n_{Ba^{2+}}=0,01-7.10^{-3}=3.10^{-3}\left(mol\right)\Rightarrow\left[Ba^{2+}\right]=\dfrac{3.10^{-3}}{0,1+0,4}=6.10^{-3}\left(M\right)\)

\(n_{Na^+}=0,1.0,1=0,01\left(mol\right)\Rightarrow\left[Na^+\right]=0,02\)

\(pOH=-lg\left(0,032\right)\approx1,5\Rightarrow pH=14-1,5=12,5\)