1. What are you going to do next weekend?
2. There was a large round wooden table in the kitchen.
3. We didn’t have enough money to buy that car.
4. I strongly believe that friends play an important role in our life.
5. Children spend more time playing video games than reading books nowadays.
6. Although we have quite different characters, we are very close friends.
7. His instructions weren't clear enough  to understand for me.
8. Generosity is part of the American character.
9. I would like to study abroad but I don't have enough money.
10. She is wearing a pretty long new green dress.
11. He spends his free time doing volunteer work at a local orphanage near his house.
12. Of all my friends, Peter and Jack are the ones I spend most of my time with.

Xét đoạn thẳng AI có: C nằm giữa A và I (do C thuộc đoạn AI)

=> CA + CI = AI => CA = AI - CI

Xét đoạn thẳng CB có: I nằm giữa C và B

=> CI + BI = CB

Vì I là trung điểm của đoạn thẳng AB (giả thiết)

=> AI = BI

Do đó: 

\(\dfrac{CB-CA}{2}\)

\(=\dfrac{\left(CI+BI\right)-\left(AI-CI\right)}{2}\)

\(=\dfrac{CI+BI-AI+CI}{2}\)

\(=\dfrac{CI+AI-AI+CI}{2}\)

\(=\dfrac{2CI}{2}\)

= CI (điều phải chứng minh)

sần sùi

xù xì

\(-\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=-\dfrac{1}{99}-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\left(\dfrac{1}{97}-\dfrac{1}{98}\right)-\left(\dfrac{1}{96}-\dfrac{1}{97}\right)-...-\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{2}\right)\)

\(=-\dfrac{1}{99}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{97}+\dfrac{1}{98}-\dfrac{1}{96}+\dfrac{1}{97}-...-\dfrac{1}{2}+\dfrac{1}{3}-1+\dfrac{1}{2}\)

\(=-1\)

1. Điều kiện: \(x-1\ge0\Leftrightarrow x\ge1\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)

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

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

\(\Leftrightarrow\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|=2\)

\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\) (vì \(\sqrt{x-1}+1>0\))

\(\Leftrightarrow\left|\sqrt{x-1}-1\right|=1-\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{x-1}-1< 0\)

\(\Leftrightarrow\sqrt{x-1}< 1\)

\(\Leftrightarrow x-1< 1\)

\(\Leftrightarrow x< 2\)

So với điều kiện ta được: \(1\le x< 2.\)

Vậy \(1\le x< 2.\)

2. Điều kiện: \(x-1\ge0\Leftrightarrow x\ge1\)

\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)

\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)

Trường hợp 1: \(x< 5,\) khi đó phương trình trở thành:

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

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

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=5\) (loại)

Trường hợp 2: \(5\le x< 10,\) khi đó phương trình trở thành:

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

\(\Leftrightarrow1=1\) (hiển nhiên)

Trường hợp 3: \(x\ge10,\) khi đó phương trình trở thành:

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

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

\(\Leftrightarrow x-1=9\)

\(\Leftrightarrow x=10\) (nhận)

Vậy \(5\le x\le10\)

14) Điều kiện: \(\left\{{}\begin{matrix}10+2x\ge0\\15-2x\ge0\end{matrix}\right.\Leftrightarrow-5\le x\le\dfrac{15}{2}\)

\(\sqrt{10+2x}=7-\sqrt{15-2x}\)

\(\Leftrightarrow\sqrt{10+2x}+\sqrt{15-2x}=7\)

\(\Leftrightarrow10+2x+15-2x+2\sqrt{\left(10+2x\right)\left(15-2x\right)}=49\)

\(\Leftrightarrow\sqrt{\left(10+2x\right)\left(15-2x\right)}=12\)

\(\Leftrightarrow\left(10+2x\right)\left(15-2x\right)=144\)

\(\Leftrightarrow150-20x+30x-4x^2=144\)

\(\Leftrightarrow6+10x-4x^2=0\)

\(\Leftrightarrow3+5x-2x^2=0\)

\(\Leftrightarrow\left(3-x\right)+\left(6x-2x^2\right)=0\)

\(\Leftrightarrow\left(3-x\right)+2x\left(3-x\right)=0\)

\(\Leftrightarrow\left(1+2x\right)\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

So với điều kiện thấy \(x=-\dfrac{1}{2}\) và \(x=3\) thỏa mãn.

Vậy phương trình có tập nghiệm là: \(S=\left\{-\dfrac{1}{2};3\right\}.\)

15) Điều kiện: \(\left\{{}\begin{matrix}x+3\ge0\\x-4\ge0\end{matrix}\right.\Leftrightarrow x\ge4\)

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

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

\(\Leftrightarrow x+3=x-4+1+2\sqrt{x-4}\)

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

\(\Leftrightarrow x-4=9\)

\(\Leftrightarrow x=13\) (nhận)

Vậy phương trình có tập nghiệm là: \(S=\left\{13\right\}.\)

16) Điều kiện: \(\left\{{}\begin{matrix}2x-1\ge0\\x-2\ge0\\x+1\ge0\end{matrix}\right.\Leftrightarrow x\ge2\)

\(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)

\(\Leftrightarrow\left(\sqrt{2x-1}-\sqrt{x+1}\right)+\sqrt{x-2}=0\)

\(\Leftrightarrow\dfrac{\left(2x-1\right)-\left(x+1\right)}{\sqrt{2x-1}+\sqrt{x+1}}+\sqrt{x-2}=0\)

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

\(\Leftrightarrow\sqrt{x-2}\left(\dfrac{\sqrt{x-2}}{\sqrt{2x-1}+\sqrt{x+1}}+1\right)=0\)

\(\Leftrightarrow\sqrt{x-2}=0\) (vì \(\dfrac{\sqrt{x-2}}{\sqrt{2x-1}+\sqrt{x+1}}+1>0\))

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\) (nhận)

Vậy phương trình có tập nghiệm là: \(S=\left\{2\right\}.\)

17) Điều kiện: \(\left\{{}\begin{matrix}2x-1\ge0\\4-3x\ge0\\x+3\ge0\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}\le x\le\dfrac{4}{3}\)

\(\sqrt{2x-1}+\sqrt{4-3x}=\sqrt{x+3}\)

\(\Leftrightarrow2x-1+4-3x+2\sqrt{\left(2x-1\right)\left(4-3x\right)}=x+3\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(4-3x\right)}=x\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(2x-1\right)\left(4-3x\right)=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\8x-6x^2-4+3x=x^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\11x-7x^2-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(11x-11x^2\right)-\left(4-4x^2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\11x\left(1-x\right)-4\left(1-x^2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\11x\left(1-x\right)-4\left(1-x\right)\left(1+x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(1-x\right)\left[11x-4\left(1+x\right)\right]=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(1-x\right)\left(7x-4\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{7}\end{matrix}\right.\)

So với điều kiện ban đầu thấy \(x=1\) và \(x=\dfrac{4}{7}\) thỏa mãn.

Vậy phương trình có tập nghiệm là: \(S=\left\{1;\dfrac{4}{7}\right\}.\)

18) Điều kiện: \(\left\{{}\begin{matrix}x-1\ge0\\5x-1\ge0\\3x-2\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)

\(\sqrt{x-1}-\sqrt{5x-1}=\sqrt{3x-2}\)

\(\Leftrightarrow\sqrt{x-1}=\sqrt{5x-1}+\sqrt{3x-2}\)

\(\Leftrightarrow x-1=5x-1+3x-2+2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)

\(\Leftrightarrow2-7x=2\sqrt{\left(5x-1\right)\left(3x-2\right)}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2-7x\ge0\\49x^2-28x+4=4\left(5x-1\right)\left(3x-2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{7}\\49x^2-28x+4=4\left(15x^2-13x+2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{7}\\49x^2-28x+4=60x^2-42x+8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{2}{7}\\11x^2-14x+4=0\end{matrix}\right.\) \(\left(1\right)\)

Ta có: \(\Delta'=49-44=5>0\Rightarrow\) phương trình có hai nghiệm phân biệt là: \(\left[{}\begin{matrix}x=\dfrac{7-\sqrt{5}}{11}\\x=\dfrac{7+\sqrt{5}}{11}\end{matrix}\right.\)

Khi đó, \(\left(1\right)\Leftrightarrow\) Phương trình vô nghiệm.

Vậy phương trình vô nghiệm.

It is 4 years since I wore uniform.

1. A. organize          B. being held              C. held                D. hold

        2. A. see                   B. commemorate       C. watch             D. love

        3. A. However         B. Nevertheless         C. Moreover        D. Therefore

        4. A. Therefore        B. However               C. Because          D. Although

        5. A. to                     B. about                     C. for                   D. of

        6. A. take the place   B. take places           C. take part          D. take place

III. Choose the best answer. Help

 3. The Gong Festivals is held ____________ in the Central Highlands.

      A. year                    B. yearly                       C. annual                     D. annually

 4.Teenagers enjoy listening to music and____________ out with friends.

      A. hang                     B. to hang                   C. hanging                     D. hangs

5. There are a lot of __________ at Mid-Autumn Festival.

     A. lanterns     B. offerings      C. worships     D. archways

6. We like Flower Festival in Da Lat __________ it’s joyful.

     A. although     B. however    C. because     D. therefore

7. Hung King Temple is the place where all Hung Emperors are__________.

    A. worship     B. worshipped     C. worshipping    D. to worship

8. You should listen to __________ court music when going on a cruise in Huong river.

   A. national     B. royal      C. principal     D. animal

9. Nick want to take part __________ the dragon boat race.

    A. on     B. in        C. at     D. for