16. to finish painting

17. to stop working

18. to pay

19. to give up taking

1. B

2. D

3. A

4. C

5. C

6. D

7. D

8. C

9. B

10. B

11. A

11. A

12. D

13. B

14. B

15. A

16. C

17. A

18. A

19. C

20. A

21. C

22. B

23. A

24. A

25. B

26. B

27. A

28. B

29. A

30. B

Điều kiện xác định: \(\left\{{}\begin{matrix}1-x\ge0\\6-x\ge0\\-5-2x\ge0\end{matrix}\right.\Leftrightarrow x\le-\dfrac{5}{2}\)

\(\sqrt{1-x}=\sqrt{6-x}-\sqrt{-5-2x}\)

\(\Leftrightarrow\sqrt{1-x}+\sqrt{-5-2x}=\sqrt{6-x}\)

\(\Leftrightarrow1-x-5-2x+2\sqrt{\left(1-x\right)\left(-5-2x\right)}=6-x\)

\(\Leftrightarrow2\sqrt{\left(1-x\right)\left(-5-2x\right)}=2x+10\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\2x^2+3x-5=x^2+10x+25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^2-7x-30=0\end{matrix}\right.\) \(\left(1\right)\)

Ta có: \(\Delta=7^2-4.1.\left(-30\right)=49+120=169>0\Rightarrow\) phương trình có hai nghiệm phân biệt là: \(\left[{}\begin{matrix}x=10\\x=-3\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-3\end{matrix}\right.\)

So với điều kiện xác định thấy \(x=-3\) thỏa mãn.

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

\(\dfrac{x^9+x^8+x^7+...+1}{x^2-1}=\dfrac{\left(x^9+x^8\right)+\left(x^7+x^6\right)+...+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^8\left(x+1\right)+x^6\left(x+1\right)+...+\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x^8+x^6+x^4+x^2+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^8+x^6+x^4+x^2+1}{x-1}\)

a) Điều kiện: \(\left\{{}\begin{matrix}x-1\ge0\\4x-4\ge0\\25x-25\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)

\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

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

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

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\) (nhận)

Vậy \(x=2.\)

b) Điều kiện: \(\left\{{}\begin{matrix}x-1\ge0\\9x-9\ge0\\\dfrac{x-1}{64}\ge0\end{matrix}\right.\Leftrightarrow x\ge1\)

\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+24.\dfrac{\sqrt{x-1}}{8}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow-\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

\(\Leftrightarrow x=290\) (nhận)

Vậy \(x=290.\)

a) \(A=\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)

\(=\sqrt{\dfrac{10-4\sqrt{6}}{2}}-\sqrt{\dfrac{10+4\sqrt{6}}{2}}\)

\(=\sqrt{\dfrac{4-4\sqrt{6}+6}{2}}-\sqrt{\dfrac{4+4\sqrt{6}+6}{2}}\)

\(=\sqrt{\dfrac{2^2-2.2\sqrt{6}+\left(\sqrt{6}\right)^2}{2}}-\sqrt{\dfrac{2^2+2.2\sqrt{6}+\left(\sqrt{6}\right)^2}{2}}\)

\(=\sqrt{\dfrac{\left(2-\sqrt{6}\right)^2}{2}}-\sqrt{\dfrac{\left(2+\sqrt{6}\right)^2}{2}}\)

\(=\dfrac{\left|2-\sqrt{6}\right|}{\sqrt{2}}-\dfrac{\left|2+\sqrt{6}\right|}{\sqrt{2}}\)

\(=\dfrac{\sqrt{6}-2}{\sqrt{2}}-\dfrac{2+\sqrt{6}}{\sqrt{2}}\) (vì \(2-\sqrt{6}< 0,\) \(2+\sqrt{6}>0\))

\(=\dfrac{\sqrt{6}-2-2-\sqrt{6}}{\sqrt{2}}\)

\(=\dfrac{-4}{\sqrt{2}}\)

\(=-2\sqrt{2}\)

b) \(B=\sqrt{\left(5x\right)^2}+\sqrt{36x^2}-\sqrt{\left(-7x\right)^2}\)

\(=\sqrt{\left(5x\right)^2}+\sqrt{\left(6x\right)^2}-\sqrt{\left(-7x\right)^2}\)

\(=\left|5x\right|+\left|6x\right|-\left|-7x\right|\)

\(=5x+6x-7x\) (vì \(x\ge0\))

\(=4x\)

c) \(C=3-x+\sqrt{x^2-4x+4}\)

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

\(=3-x+\left|x-2\right|\)

\(=3-x+2-x\) (vì \(x\le2\) nên \(x-2\le0\))

\(=5-2x\)

Theo giả thiết, ta có: \(AB:AC=3:4\Rightarrow\dfrac{AB}{AC}=\dfrac{3}{4}\Leftrightarrow\dfrac{3}{AB}=\dfrac{4}{AC}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{3}{AB}=\dfrac{4}{AC}=\dfrac{3+4}{AB+AC}=\dfrac{7}{21}=\dfrac{1}{3}\)

\(\Rightarrow AB=3.3=9\) \(\left(cm\right)\) và \(AC=3.4=12\) \(\left(cm\right)\)

Áp dụng định lí Pytago vào tam giác ABC vuông tại A, có:

\(BC^2=AB^2+AC^2=9^2+12^2=81+144=225\Rightarrow BC=15\) \(\left(cm\right)\)

Vì tam giác ABC vuông tại A có đường cao AH nên áp dụng hệ thức lượng trong tam giác vuông, ta có:

- \(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}=\dfrac{1}{9^2}+\dfrac{1}{12^2}=\dfrac{9^2+12^2}{9^2.12^2}=\dfrac{15^2}{108^2}\Leftrightarrow AH^2=\left(\dfrac{108}{15}\right)^2\Rightarrow AH=\dfrac{108}{15}\)\(\left(cm\right)\)

- \(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{9^2}{15}=\dfrac{27}{5}\) \(\left(cm\right)\)

- \(AC^2=CH.BC\Rightarrow CH=\dfrac{AC^2}{BC}=\dfrac{12^2}{15}=\dfrac{48}{5}\) \(\left(cm\right)\)

Ta có:

\(\sqrt{2x+yz}\)

\(=\sqrt{\left(x+y+z\right)x+yz}\)

\(=\sqrt{\left(x^2+xy\right)+\left(xz+yz\right)}\)

\(=\sqrt{x\left(x+y\right)+z\left(x+y\right)}\)

\(=\sqrt{\left(x+y\right)\left(x+z\right)}\)

Áp dụng bất đẳng thức Cô-si, ta có: 

\(\sqrt{\left(x+y\right)\left(x+z\right)}\le\dfrac{\left(x+y\right)+\left(x+z\right)}{2}=\dfrac{2x+y+z}{2}\Leftrightarrow\sqrt{2x+yz}\le\dfrac{2x+y+z}{2}\)

Tương tự, ta có: \(\sqrt{2y+xz}\le\dfrac{2y+x+z}{2}\) và \(\sqrt{2z+xy}\le\dfrac{2z+x+y}{2}\)

Từ đó: \(P\le\dfrac{2x+y+z}{2}+\dfrac{2y+x+z}{2}+\dfrac{2z+x+y}{2}\)

\(\Leftrightarrow P\le\dfrac{2x+y+z+2y+x+z+2z+x+y}{2}\)

\(\Leftrightarrow P\le\dfrac{4x+4y+4z}{2}\)

\(\Leftrightarrow P\le4\)

Đẳng thức xảy ra \(\Leftrightarrow x=y=z=\dfrac{2}{3}.\)

Vậy \(MaxP=4\Leftrightarrow x=y=z=\dfrac{2}{3}.\)