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Lỗi

:v

ngu quá

ê ku giải đc hông mà bảo tui ngu 

\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\)

\(\Leftrightarrow\dfrac{1105-1100}{x+5}=2\)

\(\Leftrightarrow\dfrac{5}{x-5}=2\)

\(\Leftrightarrow5=2\left(x-5\right)\)

\(\Leftrightarrow5=2x-10\)

\(\Leftrightarrow2x=15\)

\(\Leftrightarrow x=\dfrac{15}{2}=7,5\)

\(\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\left(ĐK:x\ne0;x\ne-5\right)\\ \Leftrightarrow\dfrac{1100\left(x+5\right)-1100x}{x\left(x+5\right)}=\dfrac{2x\left(x+5\right)}{x\left(x+5\right)}\\ \Leftrightarrow2x^2+10x-5500=0\\ \Leftrightarrow2x^2-100x+110x-5500=0\\ \Leftrightarrow2x.\left(x-50\right)+110.\left(x-50\right)=0\\ \Leftrightarrow\left(2x+110\right).\left(x-50\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+110=0\\x-50=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-55\left(TM\right)\\x=50\left(TM\right)\end{matrix}\right.\)

Vậy: S={-55;50}

\(ZnO,BaO,K_2O\)

\(ZnCl_2,BaCl_2,KCl\)

\(Zn\left(OH\right)_2,Ba\left(OH\right)_2,KOH\)

\(ZnSO_4,BaSO_4,K_2SO_4\)

\(Zn\left(NO_3\right)_2,Ba\left(NO_3\right)_2,KNO_3\)

\(ZnCO_3,BaCO_3,K_2CO_3\)

\(Zn_3\left(PO_4\right)_2,Ba_3\left(PO_4\right)_2,K_3PO_4\)

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

1A

2A

3D

4C

5B

\(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)

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

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

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

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

\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)

\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)

BPTT: Liệt kê

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