1. learn

2. went

3. will do

4. is playing

5. doesn't like

6. Will we meet

7. Did you finish

8. aren't taking

9. Do - get

10. didn't eat

11. know

12. didn't arrive

13. was - didn't have

14. waters - forgot

15. feels - goes

16. didn't do - was - does

17. will -see

1 learn

2 went

3 will do

4 is playing

5 doesn't like

6 Will we meet

7 Did you finish

8 aren't talking

9 Do you often get

10 didn't eat

11 know

12 didn't arrive

13 was/didn't have

14 waters/forgot

15 feels/goes

16 didn't do/was/does

17 will you see

- Hợp chất vô cơ: CO₂, Na₂CO₃ 

- Hợp chất hữu cơ: 

+ Hiđrocacbon: C₄H₁₀, C₆H₆, C₃H₄ 

+ Dẫn xuất của hiđrocacbon: C₃H₈O, CH₃Cl, C₆H₆Cl₆

Câu 1.

Tờ vé số có dạng \(\overline{a_1a_2a_3a_4a_5a_6}\in A=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

\(;a_i\ne a_j\)

Chọn \(a_1\ne0\) nên \(a_1\) có 9 cách chọn.

5 số còn lại là chỉnh hợp chập 5 của 8 số còn lại \(\in A\backslash\left\{a_1\right\}\)

\(\Rightarrow\)Có \(A_8^5\) cách.

Vậy có tất cả \(A_8^5\cdot9=60480\) vé số.

c

Mai mình thi rồi các bạn giúp mình đi mà😭😭

a) \(\dfrac{1}{\sqrt[]{x}-1}+\dfrac{1}{1+\sqrt[]{x}}+1\left(x\ge0;x\ne1\right)\)

\(=\dfrac{\sqrt[]{x}+1+\sqrt[]{x}-1+x-1}{\left(\sqrt[]{x}-1\right)\left(\sqrt[]{x}+1\right)}\)

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

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

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

b) \(\dfrac{1}{\sqrt[]{x}+2}-\dfrac{2}{\sqrt[]{x}-2}-\dfrac{4}{4-x}\left(x\ge0;x\ne4\right)\)

\(=\dfrac{\sqrt[]{x}-2-2\left(\sqrt[]{x}+2\right)+4}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{\sqrt[]{x}-2-2\sqrt[]{x}-4+4}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{-\sqrt[]{x}-2}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{-\left(\sqrt[]{x}+2\right)}{\left(\sqrt[]{x}+2\right)\left(\sqrt[]{x}-2\right)}\)

\(=\dfrac{-1}{\sqrt[]{x}-2}\)

a)Xét ΔAKB và ΔCAB có:

\(\widehat{AKB}=\widehat{BAC}=90^o\left(gt\right)\)

       \(\widehat{C}chung\)

⇒ΔAKB ~ ΔCAB(g-g)

b)Xét ΔABC có:OB=OC(O là trung điểm BC);BI=AI(I là trung điểm AB)

⇒OI là đường TB ΔABC(đ/n)

⇒OI//AC(t/c)

Mà AC⊥AB(gt) ⇒OI⊥AB(t/c)

Xét ΔBOI và ΔBAK có:

\(\widehat{BIO}=\widehat{BKA}=90^o\)

\(\widehat{B}\) chung

⇒ΔBOI ~ ΔBAK(g-g)

⇒\(\dfrac{BI}{BK}=\dfrac{BO}{BA}\Rightarrow BI.BA=BK.BO\)(đpcm)