He may not be the captain of Star Trek.

possibly he isn't the captain of Star Trek.(may)

=>He may not be the captain of Star Trek.

`@` `\text {Ans}`

`\downarrow`

`j)`

\(x^{17}\div x^{12}=x^{17-12}=x^5\)

`k)`

\(x^8\div x^5=x^{8-5}=x^3\)

`r)`

\(a^5\div a^5=a^{5-5}=a^0=1\)

`l)`

\(x^4\div x=x^{4-1}=x^3\)

`m)`

\(x^7\div x^6=x^{7-6}=x\)

`n)`

\(x^9\div x^9=x^{9-9}=x^0=1\)

`o)`

\(a^{12}\div a^5=a^{12-5}=a^7\)

`p)`

\(a^8\div a^6=a^{8-6}=a^2\)

`q)`

\(a^{10}\div a^7=a^{10-7}=a^3\)

`r(2),`

\(1024\div4=2^{10}\div2^2=2^8\)

`t)`

\(512\div2^3=2^9\div2^3=2^6\)

Câu 67:

Đổi 5,1 tấn = 5100 kg

Ta có: \(n_{NH_3}=\dfrac{5100}{17}=300\left(kmol\right)\)

Bảo toàn Nitơ: \(n_{NH_3}=n_{HNO_3\left(lý.thuyết\right)}=300\left(kmol\right)\)

\(\Rightarrow m_{ddHNO_3\left(thực\right)}=\dfrac{300\cdot40\%\cdot63}{63\%}=12000\left(kg\right)=12\left(tấn\right)\)

Câu 68:

Ta có: \(n_{HNO_3\left(thực\right)}=\dfrac{400\cdot12,6\%}{63}=0,8\left(mol\right)\) \(\Rightarrow n_{HNO_3\left(lý.thuyết\right)}=\dfrac{0,8}{40\%}=2\left(mol\right)\)

Bảo toàn Nitơ: \(n_{NH_3}=n_{HNO_3\left(lý.thuyết\right)}=2\left(mol\right)\) \(\Rightarrow V_{NH_3}=2\cdot22,4=44,8\left(l\right)\)

Thứ tự sắp xếp các cụm từ: \(\left(1\right);\left(3\right);\left(2\right);\left(4\right)\)

Ta có ^A=^D;^A=^B⇒^B=^D(tính chất bắc cầu)

Mà ^B=^D là 2 góc so le trong⇒AD//BC

Bài 1:

a) \(\dfrac{7}{8}>\dfrac{11}{13}\)

b) \(\dfrac{2010}{2011}< \dfrac{2011}{2012}\)

c) \(-\dfrac{17}{16}< -\dfrac{2}{3}\)

d) \(-\dfrac{2010}{2011}< \dfrac{2011}{-2000}\)

e: \(\dfrac{a}{m}< \dfrac{a+b}{2m}< \dfrac{b}{m}\)