Giuúp em nhé em cảm ơn ạ
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Bài 8:
$A=2^{2n+1}+3^{2n+1}=4^n.2+9^n.3$
$\equiv (-1)^n.2+(-1)^n.3\pmod 5$
Nếu $n$ chẵn:
$A\equiv (-1)^n.2+(-1)^n.3\equiv 2+3\equiv 5\equiv 0\pmod 5$
$\Rightarrow A\vdots 5$
Nếu $n$ lẻ:
$A\equiv (-1)^n.2+(-1)^n.3\equiv -2+(-3)\equiv -5\equiv 0\pmod 5$
$\Rightarrow A\vdots 5$
Vậy $A$ chia hết cho $5$
Bài 9:
Có: $2^5=32\equiv 1\pmod {31}$
$\Rightarrow 2^{2002}=(2^5)^{400}.2^2\equiv 1^{400}.2^2\equiv 4\pmod {31}$
$\Rightarrow 2^{2002}-4\equiv 0\pmod {31}$
$\Rightarrow 2^{2002}-4\vdots 31$
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a: Xét ΔBAD và ΔBMD có
BA=BM
góc ABD=góc MBD
BD chung
=>ΔBAD=ΔBMD
b: Xét ΔBME và ΔBAC có
góc BME=góc BAC
BM=BA
góc EBM chung
=>ΔBME=ΔBAC
=>BE=BC
=>ΔBEC cân tại B
Cho em hỏi với ạ: Tại sao lại khẳng định được BA = BM thế ạ;-;?
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2 He was writing a report
3 Tom was calling the employers at 8 o'clock yesterday morning
4 I was decorating my house when you came
5 My sister was eating a chocolate cake
6 I was doing my exercuse at 6 o'clock yesterday morning
7 I was writing a report when my friend called
8 Tom was taking a photo of mine when I was eating
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Câu 7:
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1.56}{10}.100\%=56\%\\\%m_{CuO}=44\%\end{matrix}\right.\)
c, \(n_{CuO}=\dfrac{10-0,1.56}{80}=0,055\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe}+n_{CuO}=0,155\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,155.98}{100}.100\%=15,19\%\)
d, Theo PT: \(\left\{{}\begin{matrix}n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,055\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeSO_4}=0,1.152=15,2\left(g\right)\\m_{CuSO_4}=0,055.160=8,8\left(g\right)\end{matrix}\right.\)
Câu 8:
a, \(CuCO_3+2HCl\rightarrow CuCl_2+CO_2+H_2O\)
b, \(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{CuCO_3}=n_{CO_2}=0,15\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CuCO_3}=\dfrac{0,15.124}{20}.100\%=93\%\\\%m_{CuCl_2}=7\%\end{matrix}\right.\)
c, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)
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Bài `13`
\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =\sqrt{9\cdot3}+\sqrt{16\cdot3}-\sqrt{36\cdot3}-\sqrt{4\cdot3}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =\left(3+4-6-2\right)\sqrt{3}\\ =-\sqrt{3}\\ b,\left(\sqrt{28}+\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\\ =\left(\sqrt{4\cdot7}+\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{4\cdot21}\\ =\left(2\sqrt{7}+2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =2\cdot7+2\sqrt{21}-7+2\sqrt{21}\\ =14+2\sqrt{21}-7+2\sqrt{21}\\ =7+4\sqrt{21}\)
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1.This work needs…………….by ten today.
A. finish B. finishing C. to finish D. to have finished
Bị động với need thì ta chỉ cần dùng V-ing sau need thôi nhé
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\(x:0,125+x:0,25=1156,8\\ x:\left(0,125+0,25\right)=1156,8\\ x:0,375=1156,8\\ x=1156,8\times0,375\\ x=433,8\)
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