:D?

Nãy máy mk lag nên ảnh ko lên:,)) 

tạm ổn

Theo mình nghĩ nha, là she was sitting hơi sai pk hong?

Chắc z là đc

I/ supply the correct verb form

1. She asked me where i (buy)....BOUGHT........that car

2. She asked him where he (be)......WAS.....from.

3. They said that they (buy)....WOULD BUY.......that house the following month.

4. They said that they (sell).........HAD SOLD..their house

5. I asked them if they (pass).....HAD PASSED.....the final exam

6. I asked him if he (can).....COULD....help me

1. She asked me where i (buy).......bought.....that car

2. She asked him where he (be)....was.......from.

3. They said that they (buy).....would buy......that house the following month.

4. They said that they (sell)......had sold.....their house

5. I asked them if they (pass)...had passed......the final exam

6. I asked him if he (can)...could......help me

a: Xét ΔAHB vuông tại H và ΔCHA vuông tại H có

góc HAB=góc HCA

=>ΔAHB đồng dạng với ΔCHA

b,c: góc FAE+góc FHE=180 độ

=>FAEH nội tiếp

=>góc HFE=góc HAE=góc C

Xét ΔHFE vuông tại H và ΔHCA vuông tại H có

góc HFE=góc HCA

=>ΔHFE đồng dạng với ΔHCA

=>HF/HC=HE/HA

=>HF*HA=HC*HE

Chuyển các câu sau về câu bị động:

1)She didn't do her homework last night

-> Her homework wasn't done by her last night?

2)What did she read yesterday?

-> What was read by her yesterday?

3)They teach me English

-> I am taught English by them.

4)She doesn't plant trees in the garden

-> Trees aren't planted in the gerden by her.

5)They donated money to street children

-> Money was donated to street children by them.

ĐK: \(x\ge\dfrac{5}{3}\)

Ta có: \(\sqrt{2x+5}=2+\sqrt{3x-5}\)

      \(\Leftrightarrow2x+5=4+3x-5+4\sqrt{3x-5}\)

      \(\Leftrightarrow6-x=4\sqrt{3x-5}\)                    ĐK: x≤6

      \(\Leftrightarrow36-12x+x^2=48x-80\)

      \(\Leftrightarrow x^2-60x+116=0\)

      \(\Leftrightarrow\left(x-2\right)\left(x-58\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=58\end{matrix}\right.\)

So với điều kiện thì phương trình có nghiệm duy nhất là x = 2

\(ĐK:x\ge\dfrac{5}{3}\\ PT\Leftrightarrow\left(\sqrt{2x+5}-3\right)-\left(\sqrt{3x-5}-1\right)=0\\ \Leftrightarrow\dfrac{2x-4}{\sqrt{2x+5}+3}-\dfrac{3x-6}{\sqrt{3x-5}+1}=0\\ \Leftrightarrow\left(x-2\right)\left(\dfrac{2}{\sqrt{2x+5}+3}-\dfrac{3}{\sqrt{3x-5}+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\\dfrac{2}{\sqrt{2x+5}+3}=\dfrac{3}{\sqrt{3x-5}+1}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2\sqrt{3x-5}+2=3\sqrt{2x+5}+9\\ \Leftrightarrow2\sqrt{3x-5}=7+3\sqrt{2x+5}\\ \Leftrightarrow4\left(3x-5\right)=49+9\left(2x+5\right)+42\sqrt{2x+5}\\ \Leftrightarrow12x-20=49+18x+45+42\sqrt{2x+5}\\ \Leftrightarrow-6x-144=42\sqrt{2x+5}\)

Vì \(x\ge\dfrac{5}{3}>0\Leftrightarrow-6x-144< 0< 42\sqrt{2x+5}\)

Do đó (1) vô nghiệm

Vậy PT có nghiệm \(x=2\)