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

P6:

1. big

2. been

3. bought

4. people

P7:

1. T

2. F

3. F

4. F

P6:

1. biggest

dạ mk cảm ơn ạ. Nhưng mk muốn thuyết minh về CÂY RAU MUỐNG ko phải rau muống luộc ạ 

À, xl ạ, mk lm lại cho bn nha

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: \(S_{AMN}+S_{BMNP}=S_{ABP}\)

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

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

1/

1. although

2. around

3. where

4. forward

5. and

6. will

2/ 1B 2F 3E 4A 5D 6C

3/

1. daily

2. wearing

3. would do

4. haven't eaten

5. is being carried

6. rises

7. inexperienced

8. equality

4/

1. which -> who

2. pollute -> polluting

5/

1. not to touch that

2. Tom has met is

3. weren't ill

4. may be built

6/

1. environmentalists

2. affect

3. media

4. responsible

7/

1. T

2. T

3. F

4. F

Part 6 

1/ students

2/affect

3/media

4/make

\(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy...

1 C => hers

2 B => will use

3 D => their

4 A => will be pedaling

5 A => travel

6 C => less

7 D => have they

8 D => than

9 A => worked

10 B => fewer