Làm giúp mình câu 9 với ạ
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Bài 8:
a: Ta có: \(\sqrt{4x}=\sqrt{5}\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
b: Ta có: \(\sqrt{4\cdot\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow2\left|x-1\right|=6\)
\(\Leftrightarrow\left|x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
c: Ta có: \(\sqrt{2x-3}=\sqrt{7}\)
\(\Leftrightarrow2x-3=7\)
hay x=5
d: Ta có: \(\sqrt{\left(3x-2\right)^2}=4\)
\(\Leftrightarrow\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
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9. NaHCO3 + HCl → NaCl + H2O + CO2
=> PT ion : \(HCO_3^-+H^+\rightarrow CO_2+H_2O\)
10. NaHCO3 + NaOH → Na2CO3 + H2O
=> PT ion : \(HCO_3^-+OH^-\rightarrow CO_3^{2-}+H_2O\)
11. NH4Cl + NaOH → NH3 + H2O + NaCl
=> PT ion :\(NH_4^++OH^-\rightarrow NH_3+H_2O\)
12. Na2SO4 + FeCl3 -----//---->
![](https://rs.olm.vn/images/avt/0.png?1311)
8.
Gọi \(A\left(x_0;y_0\right)\) là điểm cố định mà đt luôn đi qua với mọi m
\(\Leftrightarrow mx_0+2y_0-3my_0+m-1=0\\ \Leftrightarrow m\left(x_0-3y_0+1\right)+\left(2y_0-1\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x_0-3y_0+1=0\\2y_0-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_0=\dfrac{1}{2}\\y_0=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow A\left(\dfrac{1}{2};\dfrac{1}{2}\right)\)
Vậy đt luôn đi qua \(A\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) với mọi m
9.
PT giao Ox là \(y=0\Leftrightarrow mx+m-1=0\Leftrightarrow x=\dfrac{1-m}{m}\Leftrightarrow A\left(\dfrac{1-m}{m};0\right)\Leftrightarrow OA=\left|\dfrac{1-m}{m}\right|\)
PT giao Oy là \(x=0\Leftrightarrow\left(2-3m\right)y+m-1=0\Leftrightarrow y=\dfrac{1-m}{2-3m}\Leftrightarrow B\left(0;\dfrac{1-m}{2-3m}\right)\Leftrightarrow OB=\left|\dfrac{1-m}{2-3m}\right|\)
Để \(\Delta OAB\) cân thì \(OA=OB\Leftrightarrow\left|\dfrac{1-m}{m}\right|=\left|\dfrac{1-m}{2-3m}\right|\)
\(\Leftrightarrow\left|m\right|=\left|2-3m\right|\Leftrightarrow\left[{}\begin{matrix}m=2-3m\\m=3m-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{1}{2}\\m=1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}m=\dfrac{1}{2}\\m=1\end{matrix}\right.\) thỏa mãn đề
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tất cả k dưới đây đều là \(k\in Z\)
6.
\(\Leftrightarrow\sqrt{3}cot\left(3x-\dfrac{\pi}{3}\right)=1\)
\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=\dfrac{1}{\sqrt{3}}\)
\(\Leftrightarrow cot\left(3x-\dfrac{\pi}{3}\right)=cot\left(\dfrac{\pi}{3}\right)\)
\(\Leftrightarrow3x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k\pi\)
\(\Leftrightarrow3x=\dfrac{2\pi}{3}+k\pi\)
\(\Leftrightarrow x=\dfrac{2\pi}{9}+\dfrac{k\pi}{3}\)
7.
\(\Leftrightarrow\sqrt{3}tan\left(3x-15^0\right)=-1\)
\(\Leftrightarrow tan\left(3x-15^0\right)=-\dfrac{1}{\sqrt{3}}\)
\(\Leftrightarrow tan\left(3x-15^0\right)=tan\left(-30^0\right)\)
\(\Leftrightarrow3x-15^0=-30^0+k180^0\)
\(\Leftrightarrow3x=-15^0+k180^0\)
\(\Leftrightarrow x=-3^0+k60^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c, \(2H_2+O_2 \rightarrow2H_2O\)
\(n_{H_2}=\dfrac{33,6}{22,4}=1,5(mol) \Rightarrow n_{O_2}=0,75(mol)\)
\(V_{O_2}=22,4.0,75=16,8(l)\)
\(n_{H_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)
a. PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PTHH: \(n_{Fe}=n_{H_2}=1,5\left(mol\right)\)
\(\Rightarrow m_{Fe}=56\cdot1,5=84\left(g\right)\)
b. Đổi: \(500ml=0,5l\)
\(CM_{H_2SO_4}=\dfrac{1,5}{0,5}=3M\)
c. \(2H_2+O_2\rightarrow2H_2O\)
Theo PTHH: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}\cdot1,5=0,75\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,75\cdot22,4=16,8\left(l\right)\)
Mình cần gấp lắm ạ , mọi người giúp mình với
\(9,=\dfrac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{\sqrt{2}-\sqrt{3}}+\dfrac{5\left(1-\sqrt{6}\right)}{-5}-2\sqrt{6}\\ =\sqrt{6}-1+\sqrt{6}-2\sqrt{6}=-1\)