Ai giúp vs cần gấp
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$a)2Mg+O_2\xrightarrow{t^o}2MgO$
$b)4P+5O_2\xrightarrow{t^o}2P_2O_5$
$c)4Al+3O_2\xrightarrow{t^o}2Al_2O_3$
$d)2Na+S\xrightarrow{t^o}Na_2S$
$e)2H_2O\xrightarrow{đp}2H_2+O_2$
$f)2KClO_3\xrightarrow{t^o,MnO_2}2KCl+3O_2$
$g)Cu+Cl_2\xrightarrow{t^o}CuCl_2$
$h)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$
$i)Mg+2HCl\to MgCl_2+H_2$
$j)2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2$
$k)H_2+CuO\xrightarrow{t^o}Cu+H_2O$
$l)CaO+H_2O\to Ca(OH)_2$
Hóa hợp: $a,b,c,d,g,l$
Phân hủy: $e,f,h$
Thế: $i,j,k$
\(1,A=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\left(x>0;x\ne1;x\right)\\ A=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-1+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\left(\sqrt{x}-1\right)=\dfrac{x-1}{\sqrt{x}}\)
\(2,x=2\sqrt{2}+3=\left(\sqrt{2}+1\right)^2\\ \Leftrightarrow A=\dfrac{2\sqrt{2}+3}{\sqrt{2}+1}=\dfrac{\left(2\sqrt{2}+3\right)\left(\sqrt{2}-1\right)}{1}\\ =4\sqrt{2}-2\sqrt{2}+3\sqrt{2}-3=5\sqrt{2}-3\)
1. Where is the teacher right now?
2. Where is Dr Green staying?
3. How do you get to your new office?
4. What do the contruction workers do in the park?
5. When does she leave for work?
6. Where did he go yesterday?
7. Who will she have dinner with tomorrow night?
8. Why will she sleep in her sofa?
9. Who are studying C in the laboratory?
10. How long does it take you to go to school?
Câu 3:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{90}{5}=18\)
Do đó: x=54; y=36
1,\(\sqrt{\left(x-1\right)^2}=\left|x-1\right|=-\left(x-1\right)=1-x\)
2,\(\sqrt{\left(a-2b\right)^2}=\left|a-2b\right|=-\left(a-2b\right)=2b-a\)
3,\(\sqrt{\left(2x-1\right)^2}=\left|2x-1\right|=2x-1\)
Bài 5:
Gọi số sách 7A,7B,7C,7D lần lượt là \(a,b,c,d\in \mathbb{N^*},sách\)
Áp dụng tc dtsbn:
\(\dfrac{a}{37}=\dfrac{b}{37}=\dfrac{c}{40}=\dfrac{d}{36}=\dfrac{c-d}{40-36}=\dfrac{12}{4}=3\\ \Rightarrow\left\{{}\begin{matrix}a=111\\b=111\\c=120\\d=108\end{matrix}\right.\)
Vậy ...
Bài 6:
Gọi cd, cr theo thứ tự là \(a,b>0;m\)
\(\Rightarrow a:b=5:4\Rightarrow\dfrac{a}{5}=\dfrac{b}{4}\)
Đặt \(\dfrac{a}{5}=\dfrac{b}{4}=k\Rightarrow a=5k;b=4k\)
Mà \(ab=500\Rightarrow20k^2=500\Rightarrow k^2=25\Rightarrow k=5\left(k>0\right)\)
\(\Rightarrow\left\{{}\begin{matrix}a=5\cdot5=25\\b=5\cdot4=20\end{matrix}\right.\\ \Rightarrow\text{Chu vi là }2\left(a+b\right)=2\left(25+20\right)=90\left(m\right)\)
dài quá !😅😅😅