XI

1 That book was published a few years ago

2 The magazines are put on the shelf in the corner

3 These toys are sold on Disneyland and in Hong Kong

4 My house was built in 2001

5 This computer was made in China

6 These old clothes are collected for the poor children.

7 This reports had been finished by five o'clock

8 Nam said he would attend the lecture last night

XI.1. That book/ publish/ a few years ago

--> That book was published a few years ago

2. The magazines/ put/ shelf/ the corner 

--> The magazines are put on the sheft in the corner

3. These toys/ sell/ Disneyland/ Hong Kong

--> These toys are sold at Disneyland in Hong Kong

4. My house/ build/ 2001

--> My house was built in 2001

5. This computer/ make/ China

--> This computer is made in China

6. These old clothes/ collect/ the poor children

--> These old clothes are collected for the poor children

7. These reports/ finish/ by / 5 o'clock

--> These reports are finish by him/her at 5 o'clock

8. Nam/ said/ he/ attend/ the lecture/ that night

--> Nam said he attended the lecture that night

                                 Chúc cậu học tốt:3

1 about - in

2 In - to

3 from - of - in

4 in - at - during

5 in - on

6 about

7 from

8 as

9 by - in - in

10 to - in

1 Where did you go?

2 Who did you go with?

3 How did you get there?

4 What did you do during the day?

5 Did you have a good time?

1. Where did you go?

Where was you going?

2. Who did you go with?

Who was you going with?

3.  How did you get there?

How was you getting there?

a) 1 dm = 1/10 m

3 dm = 3/10 m

9 dm = 9/10 m

b) 1 g = 1/1000 kg

8 g = 8/1000 kg

25 g = 25/1000 kg

c) 1 phút = 1/60 giờ

6 phút = 1/10 giờ

12 phút = 1/5 giờ

em ơi, đề khó đọc quá

khó đọc quá bn ơi mik ko cs đọc đc á

Bài 14:

a)

Sửa đề: \(AE\cdot AB=AD\cdot AC\)

Xét ΔADB vuông tại D và ΔAEC vuông tại E có 

\(\widehat{BAD}\) chung

Do đó: ΔADB\(\sim\)ΔAEC(g-g)

Suy ra: \(\dfrac{AD}{AE}=\dfrac{AB}{AC}\)

hay \(AE\cdot AB=AD\cdot AC\)(đpcm)

b) Ta có: \(\dfrac{AD}{AE}=\dfrac{AB}{AC}\)(cmt)

nên \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)

Xét ΔADB vuông tại D có 

\(\cos\widehat{A}=\dfrac{AD}{AB}\)

Xét ΔAED và ΔACB có 

\(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)(cmt)

\(\widehat{A}\) chung

Do đó: ΔAED∼ΔACB(c-g-c)

Suy ra: \(\dfrac{AD}{AB}=\dfrac{ED}{CB}\)(Các cặp cạnh tương ứng tỉ lệ)

hay \(\dfrac{AD}{AB}\cdot BC=DE\)

\(\Leftrightarrow DE=BC\cdot\cos\widehat{A}\)(đpcm)

c) Ta có: \(DE=BC\cdot\cos\widehat{A}\)(cmt)

nên \(DE=BC\cdot\cos60^0=\dfrac{1}{2}BC\)(1)

Ta có: ΔEBC vuông tại E(gt)

mà EM là đường trung tuyến ứng với cạnh huyền BC(M là trung điểm của BC)

nên \(EM=\dfrac{1}{2}BC\)(2)

Ta có: ΔDBC vuông tại D(gt)

mà DM là đường trung tuyến ứng với cạnh huyền BC(M là trung điểm của BC)

nên \(DM=\dfrac{1}{2}BC\)(3)

Từ (1), (2) và (3) suy ra ME=MD=DE

hay ΔMDE đều(đpcm)

Câu 10:

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\notin\left\{2;-1\right\}\\y\ne-5\end{matrix}\right.\)

\(A=\dfrac{y+5}{x^2-4x+4}\cdot\dfrac{x^2-4}{x+1}\cdot\dfrac{x-2}{y+5}\)

\(=\dfrac{y+5}{y+5}\cdot\dfrac{\left(x^2-4\right)}{x^2-4x+4}\cdot\dfrac{x-2}{x+1}\)

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

\(=\dfrac{\left(x+2\right)\left(x-2\right)\cdot\left(x-2\right)}{\left(x+1\right)\left(x-2\right)^2}=\dfrac{x+2}{x+1}\)

b: \(A=\dfrac{x+2}{x+1}\)

=>A không phụ thuộc vào biến y

Khi x=1/2 thì \(A=\left(\dfrac{1}{2}+2\right):\left(\dfrac{1}{2}+1\right)=\dfrac{5}{2}:\dfrac{3}{2}=\dfrac{5}{2}\cdot\dfrac{2}{3}=\dfrac{5}{3}\)

Câu 12:

a: \(A=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{x^2-9}\)

\(=\dfrac{x}{x+3}+\dfrac{2x}{x-3}+\dfrac{9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{x\left(x-3\right)+2x\left(x+3\right)+9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{x^2-3x+2x^2+6x+9-3x^2}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x+9}{\left(x+3\right)\left(x-3\right)}=\dfrac{3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3}{x-3}\)

b: Khi x=1 thì \(A=\dfrac{3}{1-3}=\dfrac{3}{-2}=-\dfrac{3}{2}\)

\(x+\dfrac{1}{3}=\dfrac{10}{3}\)

=>\(x=\dfrac{10}{3}-\dfrac{1}{3}\)

=>\(x=\dfrac{9}{3}=3\left(loại\right)\)

Vậy: Khi x=3 thì A không có giá trị

c: \(B=A\cdot\dfrac{x-3}{x^2-4x+5}\)

\(=\dfrac{3}{x-3}\cdot\dfrac{x-3}{x^2-4x+5}\)

\(=\dfrac{3}{x^2-4x+5}\)

\(x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1>=1\forall x\) thỏa mãn ĐKXĐ

=>\(B=\dfrac{3}{x^2-4x+5}< =\dfrac{3}{1}=3\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-2=0

=>x=2

\(\dfrac{2\left(5x+2\right)}{9}-1=\dfrac{4\left(33+2x\right)}{5}-\dfrac{5\left(1-11x\right)}{9}\)

\(\dfrac{10\left(5x+2\right)}{45}-\dfrac{45}{45}=\dfrac{36\left(33+2x\right)}{45}-\dfrac{25\left(1-11x\right)}{45}\)

\(50x-20-45=1188+72x-25+275x\)

\(50x-25=347x+1163\)

\(50x-347x=25+1163\)

\(-297x=1188\)

\(x=4\\ \)

d) 

\(\dfrac{2\left(x-4\right)}{3}+\dfrac{3x+13}{8}=\dfrac{2\left(2x-3\right)}{5}+12\)

\(\dfrac{80\left(x-4\right)}{120}+\dfrac{15\left(3x+13\right)}{120}=\dfrac{40\left(2x-3\right)}{120}+\dfrac{1440}{120}\)

\(80x-320+45x+195=80x-120+1440\)

\(125x-125=80x+1320\)

\(125x-80x=125+1320\)

\(45x=1445\)

   \(x=\dfrac{1445}{45}\) \(=\dfrac{289}{9}\)