Gọi đường sinh là l, bán kính đáy R, chiều cao SO là h

Do thiết diện qua trục là tam giác vuông nên thiết diện là tam giác vuông cân

\(\Rightarrow SO=R\Rightarrow h=R\)

Áp dụng định lý cos: \(AB=\sqrt{OA^2+OB^2-2OA.OB.cos120^0}=R\sqrt{3}\)

Gọi H là trung điểm AB \(\Rightarrow OH\perp AB\) ; \(AH=\dfrac{AB}{2}=\dfrac{R\sqrt{3}}{2}\)

\(OH=\sqrt{OA^2-AH^2}=\dfrac{R}{2}\)

Kẻ \(OK\perp SH\Rightarrow OK\perp\left(SAB\right)\Rightarrow OK=d\left(O;\left(P\right)\right)\)

\(\dfrac{1}{SO^2}+\dfrac{1}{OH^2}=\dfrac{1}{OK^2}\Rightarrow\dfrac{1}{R^2}+\dfrac{4}{R^2}=\dfrac{5}{3a^2}\Rightarrow R=a\sqrt{3}\)

\(V=\dfrac{1}{3}\pi R^2h=\dfrac{1}{3}\pi R^3=\pi a^3\sqrt{3}\)

PTHH: 2Al + 6HCl → 2AlCl₃ + 3H₂

PTHH: Mg(OH)₂ + 2HCl → MgCl₂ + 2H₂O

PTHH: Fe₂O₃ + 6HCl → 2FeCl₃ + 3H₂O

PTHH: K₂O + 2HCl → 2KCl + H₂O

Lần sau em đăng câu hỏi đúng box để mọi người hỗ trợ nha

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

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

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

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

\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}\)

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

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

b) Ta có: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2\left(a-\sqrt{a}+1\right)}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ

hay P>6

Part 1

1 into

2 which

3 happily

4 aspect

5 if

6 compulsory

Part 2

1c 2f 3b 4g 5a 6e

Part 3

1 opened

2 to be finished

3 repairing

4 driving

5 harmful

6 modernize

7 environmentalists

8 effectively

Part 4

1 disappointing -> disappointed

2 come -> came

Part 5

1 succeed although he tried 

2 my father could speak

3 seen such an interesting 

4 being made to protect

Part 6

1 forest

2 climate

3 in

4 fortunatelt

Part 7

1T 2F 3F 4F

Bài 27: Ta có: \(CP=\frac13\times CA\)

=>\(S_{BPC}=\frac13\times S_{CBA}\)

Ta có: \(S_{BPA}+S_{BPC}=S_{ABC}\)

=>\(S_{BPA}=S_{ABC}-S_{BPC}=S_{ABC}-\frac13\times S_{ABC}=\frac23\times S_{ABC}\)

Ta có: \(AN=NP=\frac12\times AP\)

=>\(S_{BNA}=\frac12\times S_{BAP}=\frac12\times\frac23\times S_{ABC}=\frac13\times S_{ABC}\)

Ta có: \(MA=MB=\frac12\times AB\)

=>\(S_{AMN}=\frac12\times S_{ABN}=\frac12\times\frac13\times S_{ABC}=\frac16\times S_{ABC}\)

Ta có: \(S_{AMN}+S_{BMNP}=S_{ABP}\)

=>\(S_{BMNP}=S_{ABP}-S_{AMN}=\frac23\times S_{ABC}-\frac16\times S_{ABC}=\frac12\times S_{ABC}\)

=>\(S_{ABC}=12:\frac12=24\left(\operatorname{cm}^2\right)\)

4 phút 18 giây

4phút 18 giây