1. The sun always (rise) rises in the east. Look, it (rise) is rising!

2. I don't think i (go) will go out tonight. I'm too tired.

3. Where did you (spend) spend your summer holiday last year, Tam?

4. Has your dog ever bitten anyoine? - Yes, he bit a policeman last week.

5. There were many volunteers in Sea Games 22^nd in Viet Nam in 2003.

6. By (work) working day and night he succeeded in (finish) finishing the job in time.

7. Mai (be) was very happy when she (receive) received a letter from her sister this morning. 

8. Would you (join) to join my class next Sunday? We (visit) will visit the local museum.

9. My father (read) is reading now. He always (read) reads a book after dinner.

10. A friend of my brother’s (call) called him last night, but he (not be) wasn't at home. So I (take) took a message for him.

I. Give the correct form of the verbs

1. The sun always (rise) rises in the east. Look, it (rise)! …is rising………………………..

2. I don’t think I (go) am going to go  out tonight. I’m too tired. ………………………..

3. Where did you  spend(spend) your summer holiday last year, Tam? ……………………………..

4. Has Your dog ever (bite) bitten anyone?   - Yes, he (bite) bit a policeman last week.

5. There (be)  were many volunteers in Sea Games 22^nd in Viet Nam in 2003. …………………

6. By (work) working day and night he succeeded in (finish) finishing the job in time. ………………………

7. Mai (be) was very happy when she (receive) received a letter from her sister this morning. ……………

8. Would you lik e (join) joining my class next Sunday? We (visit) will visit the local museum.

9. My father (read)  is reading now. He always (read) reads a book after dinner……………………………

10. A friend of my brother’s (call) called him last night, but he (not be) wasn't at home. So I (take) took a message for him. …………………………………………………..………………………..

Trl :

bạn kia làm đúng rồi nhé 

    hk tốt nhé bạn @

a)ĐKXĐ : x > 0 

P = \(\left(\frac{x-1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(1+\sqrt{x}\right)}\right)\)

    = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{1}{\sqrt{x}}.\left(\sqrt{x}-1+\frac{\sqrt{x}-1}{\sqrt{x}+1}\right)\)

    = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\sqrt{x}-1}{\sqrt{x}}.\left(1-\frac{1}{\sqrt{x}+1}\right)\)

     = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right).\sqrt{x}}{\sqrt{x}}\)

       = \(\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)

Vậy P = \(\frac{\sqrt{x}+1}{\sqrt{x}}\)

b) x = \(\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=\frac{2.\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{3}-1\)

=> P = \(\frac{\sqrt{x}+1}{\sqrt{x}}=\frac{\sqrt{3}-1+1}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\)

        = \(\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3+1}\right)}=\frac{3+\sqrt{3}}{3-1}=\frac{3+\sqrt{3}}{2}\)

c)\(P\sqrt{x}=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)\sqrt{x}}{\sqrt{x}}=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow\sqrt{x}+1=6\sqrt{x}-3-\sqrt{x-4}\)

\(\Leftrightarrow\sqrt{x-4}=5\sqrt{x-4}\)

Đặt \(\hept{\begin{cases}a=\sqrt{x}\\b=\sqrt{x-4}\end{cases}\Rightarrow a^2+b^2=x-\left(x-4\right)=4}\)

\(\Rightarrow\hept{\begin{cases}a^2-b^2=4\\b=5a-4\end{cases}\Rightarrow\hept{\begin{cases}a^2-\left(5a-4\right)^2=4\left(^∗\right)\\b=5a-4\end{cases}}}\)

Từ (*) <=> a² -(25a² -40a + 16 ) =4

        <=>  -24a² + 40a - 20        = 0

=> \(\Delta'=-80< 0\)

=> PT vô nghiệm 

=> ko tồn tại x thỏa mãn

bn lm sai đề bài r 

Bạn kiểm tra lại đề bài nhé

\(\frac{2\sqrt{x}+1}{3-\sqrt{x}}\)

thế này ms đúng ajk. xin lỗi bn mk ghi nhầm dấu

\(\Leftrightarrow C=\frac{\left(2+\sqrt{a}\right)^2-\left(2-\sqrt{a}\right)^2+4a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}:\frac{2\sqrt{a}-\sqrt{a}-3}{\sqrt{a}\left(2-\sqrt{a}\right)}\)

\(\Leftrightarrow C=\frac{2\sqrt{a}+2\sqrt{a}+4a}{\left(2-\sqrt{a}\right)\left(\sqrt{a}+2\right)}.\frac{\left(2-\sqrt{a}\right).\sqrt{a}}{\sqrt{a}-3}=\frac{\left(4\sqrt{a}+4a\right)\sqrt{a}}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)}\)

b) Để C>0 thì \(\frac{4\left(a-\sqrt{a}\right)\sqrt{a}}{\left(\sqrt{a}+2\right)\sqrt{a}+3}>0hay\left(a-\sqrt{a}\right)>0=>a>1\)

c) bổ sung ý c) tìm a để C=-1

để B=-1

\(\Leftrightarrow\left(4\sqrt{a}+4a\right)\sqrt{a}=-\left(\sqrt{a}+2\right)\left(\sqrt{a}-3\right)\)

\(\Leftrightarrow4a+4a\sqrt{a}=-a+3\sqrt{a}-2\sqrt{a}+6\)

\(\Leftrightarrow5a+4a\sqrt{a}-\sqrt{a}-6=0=>\orbr{\begin{cases}\sqrt{a}=1\\5\sqrt{a}+4a-1=0\left(zô\right)lý\end{cases}=>a=1}\)

Ta có: \(\Delta=b^2-4ac=1-4\left(1-m\right)=4m-3\)

Để pt có nghiệm x₁;x₂ thì \(\Delta\ge0\)

<=> 4m-3 >0 

<=> \(m\ge\frac{3}{4}\)(*)

Theo định lý Vi-et ta có: \(x_1+x_2=-\frac{b}{a}=1\) và \(x_1x_2=\frac{c}{a}=1-m\)

Ta có: \(5\left(\frac{1}{x_1}+\frac{1}{x_2}\right)-x_1x_2+4=5\left(\frac{x_1+x_2}{x_1x_2}\right)-x_1x_2+4=\frac{5}{1-m}-\left(1-m\right)+4=0\)

\(\Leftrightarrow\hept{\begin{cases}5-\left(1-m\right)^2+4\left(1-m\right)=0\\m\ne1\end{cases}\Leftrightarrow\hept{\begin{cases}m^2+2m-8=0\\m\ne1\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=2\\m=-4\end{cases}}}\)

Kết hợp với điều kiện (*) ta có m=2 là giá trị cần tìm