1 She has lived in HN for 2 years

2 He has studied E since he was young

3 This is the first time I have seen such a beautiful girl

4 I have never eaten such a good novel before

Lời giải:

$Q=\frac{80}{81}a^2+\frac{a^2}{81}+\frac{1}{3a}+\frac{1}{3a}$

$\geq \frac{80}{81}a^2+3\sqrt[3]{\frac{a^2}{81}.\frac{1}{3a}.\frac{1}{3a}}$

$\geq \frac{80}{81}.3^2+\frac{1}{3}=\frac{83}{9}$

Vậy $Q_{\min}=\frac{83}{9}$. 

Giá trị này đạt tại $a=3$

a)Đk \(x\ge0,x\ne1\)

\(\Rightarrow P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

b)\(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\Rightarrow\sqrt{x}=2-\sqrt{3}\)

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

c)\(TacoP>\dfrac{1}{2}\)

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

\(\Leftrightarrow2\sqrt{x}-2>\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)

Vậy\(x>9\)Thì \(P>\dfrac{1}{2}\)

2.D

3.A

Câu 2: D

Câu 3: A

2.D

3.A

1 D

2 C

3 C

1D 2C 3B

1D 2B 3B

ảnh có lỗi ko mn