Vế phải đâu bạn?

https://books.google.com.vn/books?id=dtSMDwAAQBAJ&pg=PA25&lpg=PA25&dq=Cho+tam+gi%C3%A1c+c%C3%A2n+t%E1%BA%A1i+A+n%E1%BB%99i+ti%E1%BA%BFp+%C4%91%C6%B0%E1%BB%9Dng+tr%C3%B2n+(O).+G%E1%BB%8Di+M+l%C3%A0+trung+%C4%91i%E1%BB%83m+c%E1%BB%A7a+AC.+G+l%C3%A0+tr%E1%BB%8Dng+t%C3%A2m+c%E1%BB%A7a+tam+gi%C3%A1c+ABM.+G%E1%BB%8Di+Q+l%C3%A0+giao+%C4%91i%E1%BB%83m+BM+v%C3%A0+GO.+X%C3%A1c+%C4%91%E1%BB%8Bnh+t%C3%A2m+%C4%91%C6%B0%E1%BB%9Dng+tr%C3%B2n+ngo%E1%BA%A1i+ti%E1%BA%BFp+BGQ&source=bl&ots=v_OvQw42FT&sig=ACfU3U2Iyh_PC6r428LOoBL9-qwlsEembg&hl=vi&sa=X&ved=2ahUKEwjNte_I-onjAhUPAogKHQg0C-AQ6AEwBHoECAkQAQ#v=onepage&q=Cho%20tam%20gi%C3%A1c%20c%C3%A2n%20t%E1%BA%A1i%20A%20n%E1%BB%99i%20ti%E1%BA%BFp%20%C4%91%C6%B0%E1%BB%9Dng%20tr%C3%B2n%20(O).%20G%E1%BB%8Di%20M%20l%C3%A0%20trung%20%C4%91i%E1%BB%83m%20c%E1%BB%A7a%20AC.%20G%20l%C3%A0%20tr%E1%BB%8Dng%20t%C3%A2m%20c%E1%BB%A7a%20tam%20gi%C3%A1c%20ABM.%20G%E1%BB%8Di%20Q%20l%C3%A0%20giao%20%C4%91i%E1%BB%83m%20BM%20v%C3%A0%20GO.%20X%C3%A1c%20%C4%91%E1%BB%8Bnh%20t%C3%A2m%20%C4%91%C6%B0%E1%BB%9Dng%20tr%C3%B2n%20ngo%E1%BA%A1i%20ti%E1%BA%BFp%20BGQ&f=false

Xem tại link này(mình gửi cho)

Học tốt!!!!!!!!!!!!!

Ta có: 

\(x=\sqrt[3]{\sqrt{2}-1}-\frac{1}{\sqrt[3]{\sqrt{2}-1}}=\sqrt[3]{\sqrt{2}-1}-\sqrt[3]{\frac{1}{\sqrt{2}-1}}\)

\(=\sqrt[3]{\sqrt{2}-1}-\sqrt[3]{\sqrt{2}+1}\)

\(\Rightarrow x^3=-2-3x\sqrt[3]{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=-2-3x\)

\(\Rightarrow x^3+3x+2=0\)

\(\frac{1}{\sqrt{1}+\sqrt{2}}+....\frac{1}{\sqrt{79}+\sqrt{80}}>\frac{1}{\sqrt{100}}+...+\frac{1}{\sqrt{100}}\) (40 số)

................................................................\(>\frac{40}{10}=4\) 

=>đpcm

hc tốt

ko chắc lắm :)

dhasuxbhfc;CX

Các đường trung tuyến BD và CE vuông góc với nhau nên tam giác ABC vuông tại A

=> BC^2 = AB^2 + AC^2

=> BC^2 = 6^2 + 8^2

=> BC = 10 cm

\(x^2-mx+m-6=0\)

\(\Delta=\left(-m\right)^2-4\left(m-6\right)=m^2-4m+24=\left(m-2\right)^2+20>0\) pt có 2 nghiệm phân biệt 

\(\left|x_1-x_2\right|=2\sqrt{5}\)\(\Leftrightarrow\)\(x_1^2+x_2^2-2x_1x_2=20\)\(\Leftrightarrow\)\(\left(x_1+x_2\right)^2-4x_1x_2=20\)

Vi-et ta có : \(\hept{\begin{cases}x_1+x_2=m\\x_1x_2=m-6\end{cases}}\)

(1) \(\Leftrightarrow\)\(m^2-4\left(m-6\right)=20\)\(\Leftrightarrow\)\(\left(m-2\right)^2=0\)\(\Leftrightarrow\)\(m=2\)

...