\(\dfrac{AB^2\sqrt3}{4}=\dfrac{9^2\sqrt3}{4}=\dfrac{81\sqrt3}{4}\)

\(\dfrac{-2}{9}=\dfrac{-16}{72}\\\dfrac{3}{4}=\dfrac{54}{72}\\\dfrac{-1}{12}=\dfrac{-6}{72}\\\dfrac{5}{6}=\dfrac{60}{72}\\\dfrac{-5}{8}=\dfrac{-45}{72} \\Do\ -45<-16<-6<54<60\ nên\ \dfrac{-5}{8}<\dfrac{-2}{9}<\dfrac{-1}{12}<\dfrac{3}{4}<\dfrac{5}{6}\)

Dịch hạch

như nhau nhé

322245

8m; 15m; 17m

\(1) 3x-2=6\\<=>3x=8\\<=>x=\dfrac{8}{3}\\ 2)x(x+8)-12=x^2+6x\\<=>2x-12=0\\<=>2x=12\\<=>x=6 \\3)\dfrac{3x+1}{6}=\dfrac{5x-2}{9} \)

\(<=>27x+9=30x-12 \\<=>3x=21\\<=>x=7 \\4)(x-2)(2x+3)-x^2+4=0 \\<=>(x-2)(x+1)=0\\<=>x=2\ hoặc\ x=-1\)

A

a)\(x^4-5x^2-6=0 \\<=>(x^2+1)(x^2-6)=0 \\Mà\ x^2+1>0 \\<=>x^2-6=0 \\<=>x^2=6 \\<=>x=\sqrt6\ hoặc\ x=-\sqrt6\)

b) \(x^3-4x^2+5x=0 \\<=>x(x^2-4x+5)=0 \\Mà\ x^2-4x+5>0 \\<=>x=0\)

c)\(x^3-3x^2+4x+8=0 \\<=>(x+1)(x^2-4x+8)=0 \\Mà\ x^2-4x+8>0 \\<=>x+1=0\\<=>x=-1\)

D