3:

1: =>15x-9x+6=45-10x+25

=>6x+6=-10x+70

=>16x=64

=>x=4

2: =>x^2+4x-16-16=0

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

3: ĐKXĐ: x<>4; x<>-4

\(PT\Leftrightarrow\dfrac{x+4+\left(x+2\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x-4}{\left(x-4\right)\left(x+4\right)}\)

=>x+4+x^2-2x-8=5x-4

=>x^2-x-4=5x-4

=>x^2-6x=0

=>x(x-6)=0

=>x=0 hoặc x=6

4: \(\Leftrightarrow5\left(4x+1\right)-x+2>=3\left(2x-3\right)\)

=>20x+5-x+2>=6x-9

=>19x+7>=6x-9

=>13x>=-16

=>x>=-16/13

Xét ΔABC có 

E là trung điểm của AB

N là trung điểm của AC

Do đó: EN là đường trung bình của ΔABC

Suy ra: EN//BC và \(EN=\dfrac{BC}{2}\left(1\right)\)

Xét ΔBDC có

M là trung điểm của BD

F là trung điểm của CD

Do đó: MF là đường trung bình của ΔBDC

Suy ra: MF//BC và \(MF=\dfrac{BC}{2}\left(2\right)\)

Xét ΔABD có 

E là trung điểm của AB

M là trung điểm của BD

Do đó: EM là đường trung bình của ΔABD

Suy ra: \(EM=\dfrac{AD}{2}=\dfrac{BC}{2}\left(3\right)\)

Từ (1) và (2) suy ra EN//MF và EN=MF

Từ (1) và (3) suy ra EN=EM

Xét tứ giác ENFM có

EN//MF

EN=MF

Do đó: ENFM là hình bình hành

mà EN=EM

nên ENFM là hình thoi

1. Lan said she was working in Hanoi then

2. Binh said he needed a new bicycle

3. Mary said her English teacher was very humorous

4. My younger brother said he didn't want to stay at home then

5. Binh said Mr.Thanh could speak two languages well

6. Miss Hang said she would go on a picnic to Nha Trang the next day

7. Tom said he didn't know who she was

8. Mai told us that it was raining heavily outside

9. He said to his friend that he had to go home then

10. Hoa said she couldn't go out after 8 p.m

________

II

1. Lan asked Hoa if she could speak French well

2. I asked Mai if she was free that night

3. Nhi asked Mai if she liked listening to pop music

4. She asked me if she lived near there

5. A tourist asked me if I could tell her the way to the nearest post office

1: \(=\dfrac{4\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{3\left(x+1\right)}{-20\left(x-1\right)}=\dfrac{-12}{20}\cdot\dfrac{1}{x+1}=\dfrac{-3}{5x+5}\)

2: \(=\dfrac{x^2-xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}\)

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

3: \(=\dfrac{1-4x^2-1}{1-2x}:\dfrac{4x^2-2x-4x^2}{2x-1}\)

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

=-2x

1. I'd rather you didn't ask me that question

2. I haven't seen Bob seen I was in HCM City

3. He would rather read books than watch TV

4. It took Peter three hours to repaint his house

5. He asked me if I knew to speak English

6. We haven't met each other for ten years

7. The film was so boring that she fell asleep

8. The furniture was too expensive for me to buy

9. The weather is so good that they are going for a picnic

10. The coffee is too hot for me to drink

1: AD=8-2=6cm

AD/AB=6/8=3/4

AE/AC=9/12=3/4

=>AD/AB=AE/AC

2: Xét ΔADE và ΔABC có

AD/AB=AE/AC
góc A chung

=>ΔADE đồng dạng với ΔABC

3: AI là phân giác

=>IB/IC=AB/AC

=>IB/IC=AD/AE

=>IB*AE=AD*IC

1: Sửa đề: Qua N kẻ đường song song với PC cắt AB tại F

Xét tứ giác CNFP có NF//PC

nên CNFP là hình thang

3:

1: \(A=\dfrac{x\left(x^2+3\right)-\left(x^2+3\right)}{x^3\left(x+3\right)+3\left(x+3\right)}=\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x+3\right)\left(x^2+3\right)}\)

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

2: A=-1

=>x-1=-x-3

=>2x=-2

=>x=-1(nhận)

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

1

Với \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)

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

2

Để M = 0 thì \(\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}=0\Rightarrow x+1=0\Rightarrow x=-1\) (loại)

Vậy không có giá trị x thỏa mãn M = 0

1) \(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\cdot\dfrac{x^2+2x+1}{4x^4-4x^2+1}\) (ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\))

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

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

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

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

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

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

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

2) Ta có: \(M=0\)

\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)

\(\Leftrightarrow-\left(x+1\right)=0\)

\(\Leftrightarrow-x=1\)

\(\Leftrightarrow x=-1\left(ktm\right)\)

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

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

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

2: M=0

=>x+1=0

=>x=-1(loại)