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I. \(\left\{{}\begin{matrix}n_{HCl}=0,5.0,2=0,1\left(mol\right)\\n_{NaOH}=0,3.0,5=0,15\left(mol\right)\end{matrix}\right.\)
PTHH: NaOH + HCl ---> NaCl + H2O
bđ 0,15 0,1
sau pư 0,05 0
`=>` \(C_{M\left(NaOH\right)}=\dfrac{0,05}{0,2+0,3}=0,1M\)
`=>` \(\left[OH^-\right]=0,1M=10^{-1}M\)
`=>` \(pOH=1\Rightarrow pH=14-1=13\)
II.
| CTHH của chất | PT điện ly |
| HNO3 | \(HNO_3\rightarrow H^++NO_3^-\) |
| Ba(OH)2 | \(Ba\left(OH\right)_2\rightarrow Ba^{2+}+2OH^-\) |
| NaOH | \(NaOH\rightarrow Na^++OH^-\) |
| H2SO4 | \(H_2SO_4\rightarrow2H^++SO_4^{2-}\) |
| Ca(OH)2 | \(Ca\left(OH\right)_2\rightarrow Ca^{2+}+2OH^-\) |
| Na2CO3 | \(Na_2CO_3\rightarrow2Na^++CO_3^{2-}\) |
| BaCl2 | \(BaCl_2\rightarrow Ba^{2+}+2Cl^-\) |
| NaHCO3 | \(NaHCO_3\rightarrow Na^++HCO_3^-\\ HCO_3^-⇌H^++CO_3^{2-}\) |
| H2S | \(H_2S⇌H^++HS^-\\ HS^-\rightarrow H^++S^{2-}\) |
2.
| CTHH của chất | PT điện ly |
| CuSO4 | \(CuSO_4\rightarrow Cu^{2+}+SO_4^{2-}\) |
| Na2SO4 | \(Na_2SO_4\rightarrow2Na^++SO_4^{2-}\) |
| Fe2(SO4)3 | \(Fe_2\left(SO_4\right)_3\rightarrow2Fe^{3+}+3SO_4^{2-}\) |
| \(Na_2HPO_4\) | \(Na_2HPO_4\rightarrow2Na^++HPO_4^{2-}\\ HPO_4^{2-}⇌H^++PO_4^{3-}\) |
| Mg(OH)2 | \(Mg\left(OH\right)_2⇌Mg^{2+}+2OH^-\) |
| CH3COOH | \(CH_3COOH⇌CH_3COO^-+H^+\) |
| H3PO4 | \(H_3PO_4⇌H^++H_2PO_4^-\\ H_2PO_4⇌H^++HPO_4^{2-}\\ HPO_4^{2-}⇌H^++PO_4^{3-}\) |
| HF | \(HF⇌H^++F^-\) |
Ta có: \(u_n>0\)
Mặt khác:
\(u_n=\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{n+n}< \dfrac{1}{n}+\dfrac{1}{n}+...+\dfrac{1}{n}\\ \Leftrightarrow u_n< n.\dfrac{1}{n}=1\)
\(0< u_n< 1\) nên \(u_n\) bị chặn
Xét tính đơn điệu dãy \(u_n\)
\(u_{n+1}=\dfrac{1}{\left(n+1\right)+1}+\dfrac{1}{\left(n+1\right)+2}+...+\dfrac{1}{\left(n+1\right)+\left(n+1-1\right)}+\dfrac{1}{\left(n+1\right)+\left(n+1\right)}\\ =\dfrac{1}{n+2}+\dfrac{1}{n+3}+...+\dfrac{1}{2n+1}+\dfrac{1}{2n+2}\)
\(u_{n+1}-u_n=\dfrac{1}{2n+1}+\dfrac{1}{2n+2}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}+\dfrac{1}{2\left(n+1\right)}-\dfrac{1}{n+1}\\ =\dfrac{1}{2n+1}-\dfrac{1}{2\left(n+1\right)}\\ =\dfrac{2\left(n+1\right)-\left(2n+1\right)}{2\left(2n+1\right)\left(n+1\right)}=\dfrac{1}{2\left(2n+1\right)\left(n+1\right)}>0\)
\(\Rightarrow u_{n+1}>u_n\)
Dãy \(u_n\) là dãy tăng.
Vậy \(u_n\) bị chặn và tăng nghiêm ngặt nên \(u_n\) hội tụ. đpcm
1: \(cos\left(2x+15^0\right)=\frac{\sqrt2}{2}\)
=>\(\left[\begin{array}{l}2x+15^0=45^0+k\cdot360^0\\ 2x+15^0=-45^0+k\cdot360^0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=30^0+k\cdot360^0\\ 2x=-60^0+k\cdot360^0\end{array}\right.\)
=>\(\left[\begin{array}{l}x=15^0+k\cdot180^0\\ x=-30^0+k\cdot180^0\end{array}\right.\)
2:
ĐKXĐ: \(2x<>\frac{\pi}{2}+k\pi\)
=>\(x<>\frac{\pi}{4}+\frac{k\pi}{2}\)
\(\sqrt3\cdot\tan2x-3=0\)
=>\(\sqrt3\cdot\tan2x=3\)
=>\(\tan2x=\sqrt3\)
=>\(2x=\frac{\pi}{3}+k\pi\)
=>\(x=\frac{\pi}{6}+\frac{k\pi}{2}\) (nhận)
3: \(\sin\left(3x+45^0\right)=\frac{\sqrt3}{2}\)
=>\(\left[\begin{array}{l}3x+45^0=60^0+k\cdot360^0\\ 3x+45^0=180^0-60^0+k\cdot360^0=120^0+k\cdot360^0\end{array}\right.\)
=>\(\left[\begin{array}{l}3x=15^0+k\cdot360^0\\ 3x=75^0+k\cdot360^0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5^0+k\cdot120^0\\ x=25^0+k\cdot120^0\end{array}\right.\)
