(a+1)(a+7)(a+3)(a+5)+15

=(a^2+8a+7)(a^2+8a+15)+15

=(a^2+8a+11-4)(a^2+8a+11+4)+15

=(a^2+8a+11)^2 -4^2+15

=(a^2+8a+11)^2 -1

=(a^2+8a+11-1)(a^2+8a+11+1)

=(a^2+8a+10)(a^2+8a+12)

(a+1)(a+7)(a+3)(a+5)+15

=(a^2+8a+7)(a^2+8a+15)+15

=(a^2+8a+11-4)(a^2+8a+11+4)+15

=(a^2+8a+11)^2 -4^2+15

=(a^2+8a+11)^2 -1

=(a^2+8a+11-1)(a^2+8a+11+1)

=(a^2+8a+10)(a^2+8a+12)

Với x, y thực dương áp dụng BĐT Cauchy ta có:

\(P=\frac{16\sqrt{xy}}{x+y}+\frac{x^2+y^2}{xy}\)

\(=\frac{16\sqrt{xy}}{x+y}+\frac{\left(x+y\right)^2-2xy}{xy}\)

\(=\frac{16\sqrt{xy}}{x+y}+\left(\frac{\left(x+y\right)^2}{xy}+4\right)-6\)

\(\ge\frac{16\sqrt{xy}}{x+y}+2\sqrt{\frac{4\left(x+y\right)^2}{xy}}-6\)

\(=\frac{16\sqrt{xy}}{x+y}+\frac{4\left(x+y\right)}{\sqrt{xy}}-6\)

\(\ge2\sqrt{\frac{16\sqrt{xy}}{x+y}.\frac{4\left(x+y\right)}{xy}}-6=2\sqrt{16.4}-6=10\)

Vậy P_min = 10 tại x = y.

áp dụng bđt cauchy ->x+y\(\supseteq\)2\(\sqrt{xy}\)

x²+y²\(\supseteq\)2xy

nên P\(\supseteq\)\(\frac{16\sqrt{xy}}{2\sqrt{xy}}\)+\(\frac{2xy}{xy}\)=8+2=10

dấu = xảy ra\(\Leftrightarrow\)x=y

Đặt \(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}=B\)  = B

   Xét tích \(AB=\left(\sqrt{x^2-7x+19}-\sqrt{x^2-7x+15}\right)\left(\sqrt{x^2-7x+19}+\sqrt{x^2-7x+15}\right)\)

                   \(=x^2-7x+19-\left(x^2-7x+15\right)=x^2-7x+19-x^2+7x-15\)

                     \(=4\)

     Mà \(B=2\Leftrightarrow A=2\)

mình k mình nha

kb vs mình nhé