\(\dfrac{2}{xy}-\dfrac{2}{y\left(x+y\right)}-\dfrac{2}{x\left(x+y\right)}=\dfrac{2\left(x+y\right)-2x-2y}{xy\left(x+y\right)}=0\)

\(A=\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{\left(x+y\right)^2}}\)

\(=\sqrt{\left(\dfrac{1}{x}\right)^2+\left(\dfrac{1}{y}\right)^2+\left(\dfrac{1}{x+y}\right)^2+2\times\dfrac{1}{x}\times\dfrac{1}{y}-2\times\dfrac{1}{y}\times\dfrac{1}{x+y}-2\times\dfrac{1}{x}\times\dfrac{1}{x+y}}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right)}\)

\(=\left|\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{x+y}\right|\left(\text{đ}pcm\right)\)

a)

\(\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-\sqrt{24}\right)}{\sqrt{75}-5\sqrt{2}}\)

\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(5-2\sqrt{6}\right)}{5\sqrt{3}-5\sqrt{2}}\)

\(=\dfrac{\left(3\sqrt{3}+5\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)^2}{5\left(\sqrt{3}-\sqrt{2}\right)}\)

\(=\dfrac{9-3\sqrt{6}+5\sqrt{6}-10}{5}=\dfrac{-1+2\sqrt{6}}{5}\)

b)

\(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)=6-121=-115\)

\(A=\sqrt{x+\sqrt{2x-1}}-\sqrt{x+\sqrt{2x+1}}\) (ĐKXĐ: \(x\ge\dfrac{1}{2}\))

\(=\sqrt{\dfrac{2x+2\sqrt{2x-1}}{2}}-\sqrt{\dfrac{2x-2\sqrt{2x-1}}{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2x-1}\right)^2+2\sqrt{2x-1}+1^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{2x-1}\right)^2-2\sqrt{2x-1}+1^2}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{2x-1}+1\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{2x-1}-1\right)^2}}{\sqrt{2}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{2x-1}+1\right|-\left|\sqrt{2x-1}-1\right|\right)\)

TH1: \(\sqrt{2x-1}-1\ge0\Leftrightarrow\sqrt{2x-1}\ge1\Leftrightarrow2x-1\ge1\Leftrightarrow x\ge1\)

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{2x-1}+1-\sqrt{2x-1}+1\right)=\sqrt{2}\)

TH2: \(\sqrt{2x-1}-1< 0\Leftrightarrow\sqrt{2x-1}< 1\Leftrightarrow2x-1< 1\Leftrightarrow\dfrac{1}{2}\le x< 1\)

\(A=\dfrac{1}{\sqrt{2}}\left(\sqrt{2x-1}+1+\sqrt{2x-1}-1\right)=\sqrt{4x-2}\)

Copy như thế xong các bạn hiểu gì k :V giúp mình vài bài lý 12 với :V

P/s : I'm not me :) :V

Ức nhất là lúc vắt óc ra làm mà chỉ nhận đc 1 tick :v còn lại copy thì đc hơn chục tick :V