Topic 1: Letter about Flood Safety

Dear Minh,

I’m writing to share some tips on staying safe during flood season. Before a flood, you should prepare an emergency kit with food and water. During the flood, stay updated via local news and move to higher ground immediately; never try to swim or drive through floodwaters. After the water recedes, be careful with electricity and avoid drinking tap water until it's declared safe. Stay alert and keep yourself protected!

Best,

Chi

1 .Planting trees help reduce pollution and provide homes for many animals

2 . The supermarket provides a wide range of products at reasonable prices

1 It took John an hour to browse for new clothes on Lazada

2 The flood destroyed many houses; however, no one was badly injured

3 If communities don't work together, the forest won’t recover quickly

Ta có \(V T = \frac{\frac{4 x^{2}}{y^{2}}}{\left(\left(\right. \frac{x^{2}}{y^{2}} + 1 \left.\right)\right)^{2}} + \frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}}\)

Đặt \(\frac{x^{2}}{y^{2}} = t \left(\right. t > 0 \left.\right)\) thì VT thành

\(\frac{4 t}{\left(\left(\right. t + 1 \left.\right)\right)^{2}} + t + \frac{1}{t}\)

\(= \frac{4 t}{\left(\left(\right. t + 1 \left.\right)\right)^{2}} + \frac{t^{2} + 1}{t}\)

\(= \frac{4 t}{\left(\left(\right. t + 1 \left.\right)\right)^{2}} + \frac{\left(\left(\right. t + 1 \left.\right)\right)^{2}}{t} - 2\)

Đặt \(\frac{\left(\left(\right. t + 1 \left.\right)\right)^{2}}{t} = u \left(\right. u \geq 4 \left.\right)\) (vì BĐT \(\left(\left(\right. a + b \left.\right)\right)^{2} \geq 4 a b\))

Khi đó \(V T = u + \frac{4}{u} - 2\)

 \(= \frac{4}{u} + \frac{u}{4} + \frac{3 u}{4} - 2\)

\(\geq 2 \sqrt{\frac{4}{u} . \frac{u}{4}} + \frac{3.4}{4} - 2\)

\(= 2 + 3 - 2\)

\(= 3\)

\(\Rightarrow V T \geq 3\)

Dấu "=" xảy ra \(\Leftrightarrow u = 4\) \(\Leftrightarrow t = 1\) \(\Leftrightarrow x = \pm y\)

Vậy ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow x = \pm y\)

a) Chứng minh ΔABC ∼ ΔHBA và \(A B^{2} = B C * B H\) Vì ΔABC vuông tại A ⇒ ∠A = 90° AH là đường cao ⇒ AH ⟂ BC ⇒ ∠H = 90° trong ΔHBA Xét hai tam giác ΔABC và ΔHBA: ∠A = ∠H = 90° ∠ABC chung ⇒ ΔABC ∼ ΔHBA (g.g) Từ tính chất đồng dạng:

\(\frac{A B}{B C} = \frac{B H}{A B}\)

⇒ \(A B^{2} = B C * B H\) \(\)

Gọi quãng đường AB là: �(��,�>0)x(km,x>0)

Vận tốc trung bình là 15km/h nên vận tốc lúc về là: 2⋅15−12=18(��/ℎ)

2⋅15−12=18(km/h)

Thời gian đi là:\(\frac{x}{12}\) �12(ℎ)

(h)

Thời gian về là: \(\frac{x}{18}\)�18(ℎ)
​(h)

Lúc về nhiều hơn lúc đi 45 phút ta có phương trình:

�12−�18=34
​


​ \(\frac{x}{12}\)\(- \frac{x}{18}\)=\(\frac{3}{4}\)

x(\(\frac{1}{12} - \frac{1}{18}\))=\(\frac{3}{4}\)

x.\(\frac{1}{36}\)=\(\frac{3}{4}\)

x=27(km)

câu a

\(\frac{3 x + 15}{x^{2} - 9} + \frac{1}{x + 3} - \frac{2}{x - 3} = \frac{3 \cdot \left(\right. x + 5 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} + \frac{1}{x + 3} - \frac{2}{x - 3} = \frac{3 \cdot \left(\right. x + 5 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} + \frac{x - 3}{\left(\right. x + 3 \left.\right) \cdot \left(\right. x - 3 \left.\right)} - \frac{2 \cdot \left(\right. x + 3 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)}\)\(= \frac{3 \cdot \left(\right. x + 5 \left.\right) + x - 3 - 2 \cdot \left(\right. x + 3 \left.\right)}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} = \frac{3 x + 15 + x - 3 - 2 x - 6}{\left(\right. x - 3 \left.\right) \cdot \left(\right. x + 3 \left.\right)} = \frac{2 x + 6}{\left(\right. x + 3 \left.\right) \cdot \left(\right. x - 3 \left.\right)} = \frac{2 \cdot \left(\right. x + 3 \left.\right)}{\left(\right. x + 3 \left.\right) \cdot \left(\right. x - 3 \left.\right)} = \frac{2}{x - 3}\)

câu b

để \(\frac{2}{x - 3} = \frac{2}{3}\) thì \(x - 3 = 3\)

\(\Rightarrow x = 3 + 3 = 6\)

vậy  \(x = 6\) thì \(A = \frac{2}{3}\)

a) 

 và 

.
b) 

.
c) 

 và 

.

Ta có 𝑥2−4𝑥+9=(𝑥−2)2+5⩾5x² -4x +9 = (x-2)²+5 ≥ 5

=> B= 1/x²-4x+9=1/(x-2)²+5 ≤1/5

Dấu "=" xảy ra khi x=2

a) 7x + 2 = 0

7x = 0 - 2

7x = -2

x = -2/7

Vậy S = {-2/7}

b) 18 - 5x = 7 + 3x

3x + 5x = 18 - 7

8x = 11

x = 11/8

Vậy S = {11/8}