\(\dfrac{x+\sqrt{x}}{\sqrt{x}}\left(dk:x\ge0\right)\\ =\dfrac{\sqrt{x^2}+\sqrt{x}}{\sqrt{x}}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ =\sqrt{x}+1\)

Cách 1: \(\tan^2\alpha+\cot^2\alpha=\left(\tan\alpha+\cot\alpha\right)^2-2\tan\alpha\cot\alpha\) \(=2^2-2=2\) 

 \(\tan^3\alpha+\cot^3\alpha=\left(\tan\alpha+\cot\alpha\right)^3-3\tan\alpha\cot\alpha\left(\tan\alpha+\cot\alpha\right)\) \(=2^3-3.1.2=2\)

Cách 2: Ta thấy \(\cot\alpha=\dfrac{1}{\tan\alpha}\) nên ta có \(\tan\alpha+\dfrac{1}{\tan\alpha}=2\) (*). Áp dụng BDT AM-GM, ta có \(\tan\alpha+\dfrac{1}{\tan\alpha}\ge2\sqrt{\tan\alpha.\dfrac{1}{\tan\alpha}}=2\), do đó (*) xảy ra khi và chỉ khi \(\tan\alpha=\dfrac{1}{\tan\alpha}\Leftrightarrow\tan^2\alpha=1\Leftrightarrow\tan\alpha=1\) \(\Rightarrow\cot\alpha=1\). Từ đó dễ dàng tính được \(\tan^2\alpha+\cot^2\alpha=\tan^3\alpha+\cot^3\alpha=2\). 

(Tuyệt đối không được dùng cách 2 khi \(\tan\alpha\) hoặc \(\cot\alpha\) âm nhé, vì bất đẳng thức AM-GM chỉ dùng cho số dương thôi.)

Chỗ này phải sửa thành 2 mới đúng nhé.

1. B

2. D

3. C

4. B

1B

2D

3C

4B

```cpp
#include <iostream>
#include <vector>
#include <algorithm>

int main() {
int n;
std::cout << "Enter the number of integers (n < 10^5): ";
std::cin >> n;

std::vector&lt;int&gt; numbers(n); std::cout &lt;&lt; &quot;Enter &quot; &lt;&lt; n &lt;&lt; &quot; integers: &quot;; for (int i = 0; i &lt; n; ++i) { std::cin &gt;&gt; numbers[i]; } int sumOfOdd = 0; for (int i = 0; i &lt; n; ++i) { if (numbers[i] % 2 != 0) { sumOfOdd += numbers[i]; } } std::sort(numbers.begin(), numbers.end()); std::cout &lt;&lt; &quot;Sum of odd integers: &quot; &lt;&lt; sumOfOdd &lt;&lt; std::endl; std::cout &lt;&lt; &quot;Sorted sequence: &quot;; for (int i = 0; i &lt; n; ++i) { std::cout &lt;&lt; numbers[i] &lt;&lt; &quot; &quot;; } std::cout &lt;&lt; std::endl; return 0;

}
```

5. He said that he wouldn't have enough time to finish the job.

6.She asked me if Linda helped me with my luggage

7.She told him not to worrry

8.They asked us what time the party started

9.Lily accused Amanda of stealing her money

10. The doctor advised Mr.Ken to take more exercise

` @ H A N `

. He said that he wouldn't have enough time to finish the job.

6.She asked me if Linda helped me with my luggage

7.She told him not to worrry

8.They asked us what time the party started

9.Lily accused Amanda of stealing her money

10. The doctor advised Mr.Ken to take more exercise

Bạn nên viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề hơn.

Ta có : \(x^4+2x^3-10x^2+10x-3=y^2\)

\(\Leftrightarrow\left(x^4+2x^3-3\right)-\left(10x^2-10x\right)=y^2\)

\(\Leftrightarrow\left(x-1\right).\left(x^3+3x^2-7x+3\right)=y^2\)

\(\Leftrightarrow\left(x-1\right)^2.\left(x^2+4x-3\right)=y^2\)

Vì \(x,y\inℤ\) nên y² là số chính phương khi 

x² + 4x - 3 là số chính phương

Đặt x² + 4x - 3 = t²

\(\Leftrightarrow\left(x+t+2\right).\left(x-t+2\right)=7\)

Ta có bảng 

Ta được x = 2 ; x = -6 thỏa 

Với x = 2 <=> y = \(\pm3\)

Với x = -6 <=> y = \(\pm21\)

1 do

2 was traveling

3 was driving

4  read - has become

5 has loved - was

6 has eaten

7  have done

8 have cleaned

9 was watching

10 saw

#\(Errink \times Cream\)

#\(yGLinh\)

1. do

2. was traveling

3. was driving

4.  read - has become

5. has loved - was

6. has eaten

7.  have done

8. have cleaned

9. was watching

10. saw

a: \(\sin18^0-cos72^0\)

\(=\sin18^0-\sin\left(90^0-72^0\right)\)

\(=\sin18^0-\sin18^0=0\)

b: \(\frac{\tan46^0}{\cot44^0}\)

\(=\frac{\tan46^0}{\tan46^0}=1\)

c: \(\tan37^0\cdot\tan53^0\)

\(=\tan37^0\cdot\cot\left(90^0-53^0\right)\)

\(=\tan37^0\cdot\cot37^0=1\)

\(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)​

\(=\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{8}}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{2}+1+\sqrt[3]{4}\right)}{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}}=\sqrt[3]{2}\)