1. entrance

2. improving

1. exactly

2. cultural

3. speaker

4. advertised

5. national

6. agreement

7. speakers

8. unable

9. qualification

10. edition

11. editor

12. possibility

13. reputation

Ta có: \(24< 25\)

\(\Leftrightarrow\sqrt[3]{24}< \sqrt[3]{25}\)

\(\Leftrightarrow\sqrt[3]{8.3}< \sqrt[3]{25}\)

\(\Leftrightarrow\sqrt[3]{8}.\sqrt[3]{3}< \sqrt[3]{25}\)

\(\Leftrightarrow2\sqrt[3]{3}< \sqrt[3]{25}\)

a) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne0\\\sqrt{x}+2\ne0\end{matrix}\right.\Leftrightarrow x>0\)

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

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

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

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

b) Điều kiện xác định: \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}+1\ne0\\\sqrt{x}-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

\(B=\left(5-\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right).\left(5+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)\)

\(=\left[5-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right].\left[5+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)

\(=\left(5-\sqrt{x}\right).\left(5+\sqrt{x}\right)\)

\(=25-x\)

25. friendliness

26. interesting

27. pray

28. peacefully

29. differences

30. Unluckily

B

\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)

\(=\sqrt{5}.\sqrt{5}.\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{5.4}+\sqrt{5}\)

\(=\sqrt{5}.\sqrt{5.\dfrac{1}{5}}+\dfrac{1}{2}.\sqrt{4}.\sqrt{5}+\sqrt{5}\)

\(=\sqrt{5}.\sqrt{1}+\dfrac{1}{2}.2.\sqrt{5}+\sqrt{5}\)

\(=\sqrt{5}+\sqrt{5}+\sqrt{5}\)

\(=3\sqrt{5}\)

1. Điều kiện xác định: \(x-1\ne0\Leftrightarrow x\ne1\)

Đặt \(\dfrac{1}{x-1}=a,\) khi đó hệ phương trình trở thành:

\(\left\{{}\begin{matrix}a+2y=6\\2a-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6-2y\\2\left(6-2y\right)-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6-2y\\12-4y-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=6-2y\\7y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=4\\y=1\end{matrix}\right.\)

Với a = 4, ta có: \(\dfrac{1}{x-1}=4\Rightarrow4\left(x-1\right)=1\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\) (nhận)

Vậy hệ phương trình có nghiệm là: \(\left(\dfrac{5}{4};1\right).\)

bỏ chữ nhận chỗ m = 0 nữa

Sửa lại

(2m + 5)² - 4(2m + 1) > 0

<=> (2m + 5)² - 4(2m + 5 - 4) > 0

<=> (2m + 5)² - 4(2m + 5) + 16 > 0

<=> [(2m + 5)² - 4(2m + 5) + 4] + 12 > 0

<=> (2m + 3)² + 12 > 0