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

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

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

\(\Leftrightarrow y=\frac{\left(2+2\sqrt{x}\right).\sqrt{x}.\left(x\sqrt{x}-1\right)}{\sqrt{x}\left(x+\sqrt{x}+1\right)}=\frac{\left(2+2\sqrt{x}\right)\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)

Kẻ AE⊥AN⇒ˆEAN=90o⇒ˆDAE=15o,AB=AD,ˆB=ˆD⇒ΔADE=ΔABM⇒AE=AMAE⊥AN⇒EAN^=90o⇒DAE^=15o,AB=AD,B^=D^⇒ΔADE=ΔABM⇒AE=AM

Theo hệ thức..... ⇒1AH2=1AE2+1AN2⇒1AH2=1AM2+1AN2⇒1AH2=1AE2+1AN2⇒1AH2=1AM2+1AN2

Lại có AH2+HD2=AD2⇒AH2=AD2−HD2=AD2−AD24⇒AH2=34AD2⇒1AH2=43AB2AH2+HD2=AD2⇒AH2=AD2−HD2=AD2−AD24⇒AH2=34AD2⇒1AH2=43AB2

Kẻ AE⊥AN⇒ˆEAN=90o⇒ˆDAE=15o,AB=AD,ˆB=ˆD⇒ΔADE=ΔABM⇒AE=AMAE⊥AN⇒EAN^=90o⇒DAE^=15o,AB=AD,B^=D^⇒ΔADE=ΔABM⇒AE=AM

Theo hệ thức..... ⇒1AH2=1AE2+1AN2⇒1AH2=1AM2+1AN2⇒1AH2=1AE2+1AN2⇒1AH2=1AM2+1AN2

Lại có AH2+HD2=AD2⇒AH2=AD2−HD2=AD2−AD24⇒AH2=34AD2⇒1AH2=43AB2AH2+HD2=AD2⇒AH2=AD2−HD2=AD2−AD24⇒AH2=34AD2⇒1AH2=43AB2

Vậy....

Bạn xem lại đề