a,Ta có: đường cao AH 

=> AH vuông góc BC => AHB = 90`

Tam giác ABC vuông tại A

=> ABC = 90`

Xét hai tam giác ABC và HBA có:

• AHB = ABC (=90`)
• chung góc B

=>    Δ ABC ~ Δ HBA (g-g)

=> ����=����HBAB​=BABC​ ( các cạnh tưng ứng )

​Vậy ΔABC ~ ΔHBA ; ��2AB2= BC.BH

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 \(\Leftrightarrow x = \pm y\)

Gọi quãng đường AB là: \(x \left(\right. k m , x > 0 \left.\right)\)

Vận tốc trung bình là 15km/h nên vận tốc lúc về là: \(2 \cdot 15 - 12 = 18 \left(\right. k m / h \left.\right)\)

Thời gian đi là: \(\frac{x}{12} \left(\right. h \left.\right)\)

Thời gian về là: \(\frac{x}{18} \left(\right. h \left.\right)\)

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

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

\(\Leftrightarrow x \left(\right. \frac{1}{12} - \frac{1}{18} \left.\right) = \frac{3}{4}\)

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

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

\(\Leftrightarrow x = 27 \left(\right. k m \left.\right)\)

vậy quãng đường AB là 27km

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 \(A = \frac{2}{3}\)