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,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)

=> \(\frac{A B}{H B} = \frac{B C}{B A}\) ( các cạnh tưng ứng )

​=> AB.BA=HB.BC \(A B^{2}\) = BC.BH

​Vậy ΔABC ~ ΔHBA ; \(A B^{2}\)= BC.BH

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)

=> \(\frac{A B}{H B} = \frac{B C}{B A}\) ( các cạnh tưng ứng )

​=> AB.BA=HB.BC \(A B^{2}\) = BC.BH

​Vậy ΔABC ~ ΔHBA ; \(A B^{2}\)= BC.BH

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 \(A = \frac{2}{3}\)