a: ĐKXĐ: x>=0; x<>9

Ta có: \(P=\frac{10}{\sqrt{x}+7}\left(\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{\sqrt{x}-21}{9-x}\right)\)

\(=\frac{10}{\sqrt{x}+7}\left(\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}-21}{x-9}\right)\)

\(=\frac{10}{\sqrt{x}+7}\left(\frac{\sqrt{x}}{\sqrt{x}-3}+\frac{\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{10}{\sqrt{x}+7}\left(\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\)

\(=\frac{10}{\sqrt{x}+7}\cdot\frac{x+4\sqrt{x}-21}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{10}{\sqrt{x}+7}\cdot\frac{\left(\sqrt{x}+7\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{10}{\sqrt{x}+3}\)

b: Để P là số nguyên thì 10⋮\(\sqrt{x}+3\)

=>\(\sqrt{x}+3\in\left\lbrace5;10\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace2;7\right\rbrace\)

=>x∈{4;49}

ĐKXĐ : -10 - 5x ≥ 0 ⇔5x ≤ -10 ⇔ x ≤ -2 

I. Supply the correct form of the verbs in brackets

1. I wish he ………were……. here now. (be)

2. She wishes you ………felt……. better. (feel)

3. I wish that you ………had been……. here yesterday. (be)

4. I wish he ………would finish……. he work tonight. (finish)

5. We wish you ………would come……. tomorrow. (come)

6. She wishes she ………had opened……. the window last night. (open)

7. I wish you ………had left……. earlier yesterday. (leave)

8. We wish they ………had come……. with us last weekend. (come)

9. They wish he ………would come……. with them the next day. (come)

10. They wish we ……had given………. them some food yesterday. (give)

11. We wish you ………had arrived……. yesterday. (arrive)

12. I wish I ……didn't lose………. the answers. (not lose)

13. You wish you ………had known……. what to do last year. (know)

14. I wish that he ……would visit………. us next year. (visit)

15. She wishes that she ………were……. at home now. (be)

16. I wish I ………heard……. the news. (hear)

17. You wish that he ………had helped……. you last week. (help)

18. I wish I ………found……. the subject more interesting. (find)

19. He always wishes he ……were………. rich. (be)

20. The boy wishes that he …would win…… the competition the next day. (win)

21. I wish the weather ……were………. warmer now. (be)

22. They wish he ………would telephone……. them next week. (telephone)

23. He wishes you ………would help……. him in the future. (help)

24. She wishes the mail ………would come……. soon. (come)

25. They wish she ……would make….. the arrangements for the meeting next week. (make)

26. We wish they ……hurried………. or we will miss the bus. (hurry)

27. You wish the door ………opened…….. (open)

28. He wishes he ………showed……. us the book. (show)

29. They wish we ………waited……. for them. (wait)

30. I wish you ………wrote……. to me. (to write)

Ta có: \(E=\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x}+1}+4\sqrt{x}\right):\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)\)

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

\(=\left(\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+4\sqrt{x}\right):\frac{x-1}{\sqrt{x}}\)

\(=4\sqrt{x}\left(\frac{1}{x-1}+1\right):\frac{x-1}{\sqrt{x}}\)

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

1 Because of my friend's absence, I had to copy all the lessons for him

2 Because of her sickness, Mary could not go to school

3 Due to the bad weather, she couldn't go to school on time

4 Because of my sickness, my mother told me to stay at home

5 Because of the big storm, I stayed at home

6 Because of the perfect weather, she went on holiday

a: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(BC^2=3^2+4^2=25=5^2\)

=>BC=5

Xét ΔABC vuông tại A có AH là đường cao

nên \(CH\cdot CB=CA^2\)

=>\(CH=\frac{4^2}{5}=3,2\)

b:

Xét ΔABC vuông tại A có AH là đường cao

nên \(AH\cdot BC=AB\cdot AC\)

=>\(AH\cdot5=3\cdot4=12\)

=>AH=2,4

Xét tứ giác AMHN có \(\hat{AMH}=\hat{ANH}=\hat{MAN}=90^0\)

nên AMHN là hình chữ nhật

=>AH=MN

=>MN=2,4

Xét ΔHAC vuông tại H có HN là đường cao

nên \(HN\cdot AC=HA\cdot HC\)

=>\(HN\cdot4=2,4\cdot3,2\)

=>\(HN=2,4\cdot0,8=1,92\)

Gọi AD là đường kính của (O). Gọi H là trung điểm của BC

Ta có: AB=AC

=>A nằm trên đường trung trực của BC(1)

Ta có: OB=OC

=>O nằm trên đường trung trực của BC(2)

Ta có: HB=HC

=>H nằm trên đường trung trực của BC(3)

Từ (1),(2),(3) suy ra A,O,H thẳng hàng

=>A,O,H,D thẳng hàng

ΔABC đều

mà AH là đường trung tuyến

nên AH⊥BC tại H

H là trung điểm của BC

=>\(HB=HC=\frac{a}{2}\)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA^2=a^2-\left(\frac{a}{2}\right)^2=\frac{3a^2}{4}\)

=>\(HA=\frac{a\sqrt3}{2}\)

Xét (O) có

ΔABD nội tiếp

AD là đường kính

Do đó; ΔABD vuông tại B

Xét ΔBAD vuông tại B có BH là đường cao

nên \(AH\cdot AD=AB^2\)

=>\(AD=a^2:\frac{a\sqrt3}{2}=a^2\cdot\frac{2}{a\sqrt3}=\frac{2a}{\sqrt3}\)

=>Bán kính là \(R=\frac{2a}{\sqrt3}:2=\frac{a}{\sqrt3}\)