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\(\dfrac{2}{5}< \dfrac{3}{7}\)
\(\dfrac{2}{7}< \dfrac{4}{9}\)
\(\dfrac{5}{8}< \dfrac{8}{5}\)
\(\dfrac{11}{18}< \dfrac{5}{6}\)
\(\dfrac{7}{8}< \dfrac{11}{12}\)
a) \(\dfrac{x+9}{x^2-9}\)-\(\dfrac{3}{x^2+3x}\) = \(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}\)-\(\dfrac{3}{x\left(x+3\right)}\)
= \(\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}\)
= \(\dfrac{x+3}{x\left(x-3\right)}\)
\(2x-1-x^2\\ =x+x-1-x^2\\ =\left(x-x^2\right)+\left(x-1\right)\\ =-x\left(x-1\right)+\left(x-1\right)\\ =\left(x-1\right)\left(1-x\right)\)
\(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\)
= \(\dfrac{x+9}{\left(x-3\right).\left(x+3\right)}-\dfrac{3}{x.\left(x+3\right)}\)
=\(\dfrac{\left(x+9\right).x}{\left(x-3\right).\left(x+3\right).x}-\dfrac{3.\left(x-3\right)}{x.\left(x+3\right).\left(x-3\right)}\)
=\(\dfrac{x^2+9x}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3x-9}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2+9-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
=\(\dfrac{x^2-3x+18}{3\left(x-3\right)\left(x+3\right)}\)
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
\(a,\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28}\\ b,\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{3.8}{5.8}+\dfrac{3.5}{5.8}=\dfrac{6}{5}\\ c,=\dfrac{4.3}{9.3}+\dfrac{10}{27}=\dfrac{12+10}{27}=\dfrac{22}{27}\\ d,=\dfrac{2.3}{3.3}+\dfrac{7}{9}=\dfrac{6+7}{9}=\dfrac{13}{9}\\ e,=\dfrac{5.5}{12.5}+\dfrac{7.4}{15.4}=\dfrac{25+28}{60}=\dfrac{53}{60}\)
\(\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28};\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{24}{40}+\dfrac{15}{40};\dfrac{4}{9}+\dfrac{10}{27}=\dfrac{12}{27}+\dfrac{10}{27}=\dfrac{22}{27};\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{18}{27}+\dfrac{21}{27}=\dfrac{39}{27};\dfrac{5}{12}+\dfrac{7}{15}=\dfrac{75}{180}+\dfrac{84}{180}=\dfrac{159}{180}\)