y = x-1+1/x
tính y' , xác định khoảng đồng biến và nghịch biến
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thay x=1 và A=0 vào biểu thức, ta được:
\(\dfrac{3}{2m+1}+\dfrac{5}{2m-1}=0\)
=>6m-3+10m+5=0
=>16m+2=0
hay m=-1/8
![](https://rs.olm.vn/images/avt/0.png?1311)
- T- A - X- G - A - T- X - A- G.
- Số nu A: 5
- Số nu T: 5
- Số nu G: 4
- Số nu X: 4
![](https://rs.olm.vn/images/avt/0.png?1311)
PTHH : `2Na + 2H_2O -> 2NaOH + H_2`
Dung dịch `X` là `NaOH`
Khí không màu là : `H_2`
`a)`
`n_{Na} = (4,6)/(23) = 0,2` `mol`
`n_{H_2} = 1/2 . n_{Na} = 0,1` `mol`
`V_{H_2} = 0,1 . 22,4 = 2,24` `l`
`b)`
`400ml = 0,4l`
`n_{NaOH} = n_{Na} = 0,2` `mol`
`C_{M_(NaOH)} = (0,2)/(0,4) = 0,5` `M`
`c)`
PTHH : `NaOH + HCl -> NaCl + H_2O`
Ta có `n_{NaOH} = 0,2` `mol`
`-> n_{HCl} = n_{NaOH} = 0,2` `mol`
`-> V_{HCl} = (0,2)/(0,5) = 0,4` `l`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{NO}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
\(\begin{matrix}\overset{0}{Mg}\rightarrow\overset{+2}{Mg}+2e\\\overset{0}{Fe}\rightarrow\overset{+3}{Fe}+3e\\\overset{+5}{N}+3e\rightarrow\overset{+2}{N}\end{matrix}\)
Bảo toàn e:
\(2n_{Mg}+3n_{Fe}=3n_{NO}=1,05\left(1\right)\)
Lại có \(24n_{Mg}+56n_{Fe}=15,6\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}n_{mg}=0,3\left(mol\right)\\n_{Fe}=0,15\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Mg}=0,3.24=7,2\left(g\right)\Rightarrow\%m_{Mg}=\dfrac{7,2}{15,6}.100\%=46,15\%\)
\(\Rightarrow m_{Mg}=0,15.56=8,4\left(g\right)\)
\(\%m_{Fe}=100\%-46,15\%=53,85\%\)
Bảo toàn N:
\(n_{HNO_3}=n_N=2n_{Mg\left(NO_3\right)_2}+3n_{Fe\left(NO_3\right)_3}+n_{NO}\)
\(=2n_{Mg}+3n_{Fe}+n_{NO}\)
\(=2.0,3+3.0,15+0,35=1,4\left(mol\right)\)
\(C_M=\dfrac{1,4}{0,2}=7M\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}-2=-1\\\dfrac{4}{x}+\dfrac{3}{y}-2=5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a-b-2=-1\\4a+3b-2=5\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{10}{7}\\b=\dfrac{3}{7}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{10}{7}\Rightarrow x=\dfrac{7}{10}\\\dfrac{1}{y}=\dfrac{3}{7}\Rightarrow y=\dfrac{7}{3}\end{matrix}\right.\)
\(b.\left\{{}\begin{matrix}\dfrac{2}{x}+\dfrac{5}{\left(x+y\right)}=2\\\dfrac{3}{x}+\dfrac{1}{\left(x+y\right)}=\dfrac{17}{10}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2a+5b=2\\3a+b=\dfrac{17}{10}\end{matrix}\right.\) (với \(\dfrac{1}{x}=a-\dfrac{1}{x+y}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{1}{5}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2}\Rightarrow x=2\\\dfrac{1}{x+y}=\dfrac{1}{5}\Rightarrow y=3\end{matrix}\right.\)
\(c.\left\{{}\begin{matrix}\dfrac{2}{x-1}+\dfrac{1}{y+1}=7\\\dfrac{5}{x-1}-\dfrac{2}{y+1}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a+b=7\\5a-2b=4\end{matrix}\right.\) (với \(\dfrac{1}{x-1}=a-\dfrac{1}{y+1}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=2\Rightarrow x=\dfrac{3}{2}\\\dfrac{1}{y+1}=3\Rightarrow y=-\dfrac{2}{3}\end{matrix}\right.\)
\(d.\left\{{}\begin{matrix}\dfrac{2}{\sqrt{x-1}}-\dfrac{1}{\sqrt{y-1}}=1\\\dfrac{1}{\sqrt{x-1}}+\dfrac{1}{\sqrt{y-1}}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2a-b=1\\a+b=2\end{matrix}\right.\) (với \(\dfrac{1}{\sqrt{x-1}}=a-\dfrac{1}{\sqrt{y-1}}=b\))
\(\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x-1}}=1\Rightarrow x=2\\\dfrac{1}{\sqrt{y-1}}=1\Rightarrow y=2\end{matrix}\right.\)
TXĐ: \(D=R\backslash\left\{0\right\}\)
\(y=x-1+\dfrac{1}{x}\Rightarrow y'=1-\dfrac{1}{x^2}=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
BBT:
Hàm đồng biến trên các khoảng \(\left(-\infty;-1\right)\) ; \(\left(1;+\infty\right)\)
Hàm nghịch biến trên các khoảng \(\left(-1;0\right)\) và \(\left(0;1\right)\)