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

4:

a: Xét ΔABM và ΔACN có

AB=AC

góc A chung

AM=AN

=>ΔABM=ΔACN

=>BM=CN

b: Xét ΔNBC và ΔMCB có

NB=MC

góc NBC=góc MCB

BC chung

=>ΔNBC=ΔMCB

=>góc IBC=góc ICB

=>IB=IC

c: Xét ΔABI và ΔACI có

AB=AC

IB=IC

AI chung

=>ΔABI=ΔACI

=>góc BAI=góc CAI

=>AI là phân giác của góc BAC

1 A

2 C

3 A

4 A

5 D

6 C

7 B

8 C

9 B

10 C

11 D

12 A

13 C

14 C

15 D

16 B

17 D

18 B

19 D

20 D

21 D

22 A

23 C

24 A

25 A

26 D

27 C

28 B

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

bài 1:

bn lấy giá trị của √(4^2-3,9^2) là dc

bài 2

AB+BC=2√(3^2+4^2)=??

\(y^2=x\left(x+1\right)\left(x+7\right)\left(x+8\right)\)

\(=\left(x^2+8x\right)\left(x^2+8x+7\right)\)

\(\Rightarrow4y^2=\left(2x^2+16x\right)\left(2x^2+16x+14\right)\)

\(=\left(2x^2+16x+7-7\right)\left(2x^2+16x+7+7\right)\)

\(=\left(2x^2+16x+7\right)^2-49\)

\(\Leftrightarrow\left(2x^2+16x+7\right)^2-4y^2=49\)

\(\Leftrightarrow\left(2x^2+16x+7-2y\right)\left(2x^2+16x+7+2y\right)=49=1.49=7.7\)

Xét các trường hợp và thu được các nghiệm là: \(\left(-3,0\right),\left(0,0\right)\).

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

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

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

Vậy...

cún bị bán

cún bị bán

Bài 18:

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

\(=\dfrac{\left(\sqrt{a}-1\right)^2\cdot\left(\sqrt{a}+1\right)^2}{4a}\cdot\dfrac{a-2\sqrt{a}+1-a-2\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\dfrac{\left(a-1\right)\cdot\left(-4\right)\cdot\sqrt{a}}{4a}\)

\(=\dfrac{-a+1}{\sqrt{a}}\)

b: Để P<0 thì -a+1<0

\(\Leftrightarrow-a< -1\)

hay a>1

c: Để P=-2 thì \(-a+1=-2\sqrt{a}\)

\(\Leftrightarrow-a+1+2\sqrt{a}=0\)

\(\Leftrightarrow a-2\sqrt{a}+1=2\)

\(\Leftrightarrow\left(\sqrt{a}-1\right)^2=2\)

\(\Leftrightarrow\sqrt{a}-1=\sqrt{2}\)

hay \(a=3+2\sqrt{2}\)

Bài 17:

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

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

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

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