Ta có:

\(\frac{4 x^{2} y^{2}}{\left(\right. x^{2} + y^{2} \left.\right)^{2}} + \frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}} \geq 3\)

\(\Leftrightarrow \frac{4 x^{2} y^{2}}{\left(\right. x^{2} + y^{2} \left.\right)^{2}} - 1 + \frac{x^{2}}{y^{2}} + \frac{y^{2}}{x^{2}} - 2 \geq 0\)

\(\Leftrightarrow \frac{4 x^{2} y^{2} - \left(\right. x^{2} + y^{2} \left.\right)^{2}}{\left(\right. x^{2} + y^{2} \left.\right)^{2}} + \left(\left(\right. \frac{x}{y} - \frac{y}{x} \left.\right)\right)^{2} \geq 0\)

\(\Leftrightarrow \frac{- \left(\right. x^{2} - y^{2} \left.\right)^{2}}{\left(\right. x^{2} + y^{2} \left.\right)^{2}} + \frac{\left(\right. x^{2} - y^{2} \left.\right)^{2}}{x^{2} y^{2}} \geq 0\)

\(\Leftrightarrow \left(\right. x^{2} - y^{2} \left.\right)^{2} \left(\right. \frac{1}{x^{2} y^{2}} - \frac{1}{\left(\right. x^{2} + y^{2} \left.\right)^{2}} \left.\right) \geq 0\)

\(\Leftrightarrow \frac{\left(\right. x^{2} - y^{2} \left.\right)^{2} \left(\right. x^{4} + y^{4} + x^{2} y^{2} \left.\right)}{x^{2} y^{2} \left(\right. x^{2} + y^{2} \left.\right)^{2}} \geq 0\) (luôn đúng)

Do đó ta có đpcm.

Dấu bằng xảy ra khi \(x = y\)

a) xét tám giác ABC và tam giác HBA 

góc A= góc H (=90 độ)

góc A :chung

=> tam giác ABC ~ tam giác HBA (g-g)

Gọi x (km) là độ dài quãng đường AB (x > 0)

Thời gian đi là: x/15 (h)

Thời gian về là: x/12 (h)

45 phút = 3/4 (h)

Theo đề bài ta có phương trình:

x/12 - x/15 = 3/4

5x - 4x = 45

x = 45 (nhận)

Vậy quãng đường AB dài 45 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 \(A = \frac{2}{3}\)