1. C

2. A

3. C

4. B

5. C

6. B

17. C

Giải thích: find + something + adjective: cảm thấy cái gì, như thế nào

18. B

Giải thích: be fond of: thích

19. B

Giải thích: take up + tên thể thao: bắt đầu chơi môn thể thao nào đó

20. A

Giải thích: entertaining (adj) giải trí

Áp dụng bất đẳng thức Cô-si, ta có:

\(P=5x+3y+\dfrac{10}{x}+\dfrac{8}{y}\)

\(=\left(\dfrac{5x}{2}+\dfrac{10}{x}\right)+\left(\dfrac{3y}{6}+\dfrac{8}{y}\right)+\left(\dfrac{5x}{2}+\dfrac{15y}{6}\right)\)

\(=\left(\dfrac{5x}{2}+\dfrac{10}{x}\right)+\left(\dfrac{3y}{6}+\dfrac{8}{y}\right)+\dfrac{15\left(x+y\right)}{6}\)

\(\ge2.\sqrt{\dfrac{5x}{2}.\dfrac{10}{x}}+2.\sqrt{\dfrac{3y}{6}.\dfrac{8}{y}}+\dfrac{15.6}{6}\)

\(=2\sqrt{25}+2\sqrt{4}+15\)

\(=2.5+2.2+15\)

\(=10+4+15\)

\(=29\)

Đẳng thức xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5x}{2}=\dfrac{10}{x}\\\dfrac{3y}{6}=\dfrac{8}{y}\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

Vậy \(MinP=29\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)

5) \(x^3+x^2-3x-27\)

\(=\left(x^3-27\right)+\left(x^2-3x\right)\)

\(=\left(x-3\right)\left(x^2+3x+9\right)+x\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2+4x+9\right)\)

6) \(x^3+x^2-4x-4\)

\(=\left(x^3-4x\right)+\left(x^2-4\right)\)

\(=x\left(x^2-4\right)+\left(x^2-4\right)\)

\(=\left(x+1\right)\left(x^2-4\right)\)

\(=\left(x+1\right)\left(x-2\right)\left(x+2\right)\)

7) \(x^2-9x-9y-y^2\)

\(=\left(x^2-y^2\right)-\left(9x+9y\right)\)

\(=\left(x-y\right)\left(x+y\right)-9\left(x+y\right)\)

\(=\left(x-y-9\right)\left(x+y\right)\)

8) \(y^4-3y^3+3y^2-y\)

\(=y\left(y^3-3y^2+3y-1\right)\)

\(=y\left(y-1\right)^3\)

Sửa đề: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(=6-121\)

\(=-115\)

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