Muốn tick vô đây trl nha!

\(\dfrac{2}{5}< \dfrac{3}{7}\)

\(\dfrac{2}{7}< \dfrac{4}{9}\)

\(\dfrac{5}{8}< \dfrac{8}{5}\)

\(\dfrac{11}{18}< \dfrac{5}{6}\)

\(\dfrac{7}{8}< \dfrac{11}{12}\)

?

a) \(\dfrac{x+9}{x^2-9}\)-\(\dfrac{3}{x^2+3x}\) = \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}\)-\(\dfrac{3}{x\left(x+3\right)}\)

                                = \(\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

                                = \(\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}\)

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

                                = \(\dfrac{x+3}{x\left(x-3\right)}\)

\(2x-1-x^2\\ =x+x-1-x^2\\ =\left(x-x^2\right)+\left(x-1\right)\\ =-x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(1-x\right)\)

2x - 1 - x²

= -x² + 2x - 1

= -(x² - 2x + 1)

= -(x - 1)²

làm ơn ai bít giúp mik với nha

\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)

= \(\dfrac{x+9}{\left(x-3\right).\left(x+3\right)}-\dfrac{3}{x.\left(x+3\right)}\)

=\(\dfrac{\left(x+9\right).x}{\left(x-3\right).\left(x+3\right).x}-\dfrac{3.\left(x-3\right)}{x.\left(x+3\right).\left(x-3\right)}\)

=\(\dfrac{x^2+9x}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3x-9}{x\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{x^2+9-3x+9}{x\left(x-3\right)\left(x+3\right)}\)

=\(\dfrac{x^2-3x+18}{3\left(x-3\right)\left(x+3\right)}\)

c) \(\dfrac{6}{7}-\dfrac{3}{4}=\dfrac{3}{28}\)

d) \(\dfrac{35}{28}-\dfrac{48}{64}=\dfrac{1}{2}\)

 ( x + 2,7 ) : 4,9 = 25,3

         ( x + 2,7 ) = 25,3 x 4,9

            x + 2,7 = 123,97

                     x = 123,97 - 2,7

                     x = 121,27

(x+2,7):4,9=25,3

x+2,7        =25,3x4,9

x+2,7        =123,97

x               =123,97-2,7

x               =121,27

(x + 2,7) : 4,9 = 25,3

x + 27 = 25,3 x 4,9

x + 27 = 123,97

x         = 123,97 - 27

x         = 96,97

Câu 8: B