36. voluntarily

37.industrial

38. breathless

39. .intentionally

40.breakdown

41. pleasantly

42. preference

43. economical

44. ko biết :v

45. misunderstanding

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: $2^5=32\equiv 1\pmod {31}$

$\Rightarrow 2^{2002}=(2^5)^{400}.2^2\equiv 1^{400}.2^2\equiv 4\pmod {31}$

$\Rightarrow 2^{2002}-4\equiv 0\pmod {31}$

$\Rightarrow 2^{2002}-4\vdots 31$

Bài 10:
Có:

$3^5\equiv 1\pmod {11}$

$\Rightarrow 3^{2005}=(3^5)^{401}\equiv 1^{401}\equiv 1\pmod {11}$
$4^5\equiv 1\pmod {11}$

$\Rightarrow 4^{2005}=(4^5)^{401}\equiv 1^{401}\equiv 1\pmod {11}$

$\Rightarrow A=3^{2005}+4^{2005}\equiv 1+1\equiv 2\pmod {11}$

Vậy $A$ chia 11 dư 2.

----------------

$3^3\equiv 1\pmod {13}$

$\Rightarrow 3^{2005}=(3^3)^{668}.3\equiv 1^{334}.3\equiv 3\pmod {13}$

$4^3\equiv -1\pmod {13}$

$\Rightarrow 4^{2005}=(4^3)^{668}.4\equiv (-1)^{668}.4\equiv 4\pmod {13}$

$\Rightarrow A=3^{2005}+4^{2005}\equiv 3+4\equiv 7\pmod {13}$

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

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, \(n_{HCl}=2n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

Bài 6: 

a, \(SO_2+Ca\left(OH\right)_2\rightarrow CaSO_3+H_2O\)

b, \(n_{SO_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)

Theo PT: \(n_{Ca\left(OH\right)_2}=n_{CaSO_3}=n_{SO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{0,05.74}{100}.100\%=3,7\%\)

c, \(m_{CaSO_3}=0,05.120=6\left(g\right)\)

thế  kỉ XV-XVII, chúc bạn học tốt!!!

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

\(x:0,125+x:0,25=1156,8\\ x:\left(0,125+0,25\right)=1156,8\\ x:0,375=1156,8\\ x=1156,8\times0,375\\ x=433,8\)

=>x:0,375=1156,8

hay x=433,8

Bài `13`

\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =\sqrt{9\cdot3}+\sqrt{16\cdot3}-\sqrt{36\cdot3}-\sqrt{4\cdot3}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =\left(3+4-6-2\right)\sqrt{3}\\ =-\sqrt{3}\\ b,\left(\sqrt{28}+\sqrt{12}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{84}\\ =\left(\sqrt{4\cdot7}+\sqrt{4\cdot3}-\sqrt{7}\right)\cdot\sqrt{7}+\sqrt{4\cdot21}\\ =\left(2\sqrt{7}+2\sqrt{3}-\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\\ =2\cdot7+2\sqrt{21}-7+2\sqrt{21}\\ =14+2\sqrt{21}-7+2\sqrt{21}\\ =7+4\sqrt{21}\)

17:
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)

Để A là số nguyên thì \(\sqrt{x}-1⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2+1⋮\sqrt{x}-2\)

=>\(\sqrt{x}-2\in\left\{1;-1\right\}\)

=>\(\sqrt{x}\in\left\{3;1\right\}\)

=>\(x\in\left\{9;1\right\}\)

16:

a: BC=BH+CH

=9+16

=25(cm)

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH^2=HB\cdot HC\)

=>\(AH=\sqrt{9\cdot16}=12\left(cm\right)\)

Xét ΔABC vuông tại A có AH là đường cao

nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AB=\sqrt{9\cdot25}=15\left(cm\right)\\AC=\sqrt{16\cdot25}=20\left(cm\right)\end{matrix}\right.\)

b: M là trung điểm của AC

=>AM=AC/2=10(cm)

Xét ΔAMB vuông tại A có

\(tanAMB=\dfrac{AB}{AM}=\dfrac{15}{10}=\dfrac{3}{2}\)

nên \(\widehat{AMB}\simeq56^0\)

  9C