\(a,\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28}\\ b,\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{3.8}{5.8}+\dfrac{3.5}{5.8}=\dfrac{6}{5}\\ c,=\dfrac{4.3}{9.3}+\dfrac{10}{27}=\dfrac{12+10}{27}=\dfrac{22}{27}\\ d,=\dfrac{2.3}{3.3}+\dfrac{7}{9}=\dfrac{6+7}{9}=\dfrac{13}{9}\\ e,=\dfrac{5.5}{12.5}+\dfrac{7.4}{15.4}=\dfrac{25+28}{60}=\dfrac{53}{60}\)

\(\dfrac{2}{7}+\dfrac{1}{4}=\dfrac{8}{28}+\dfrac{7}{28}=\dfrac{15}{28};\dfrac{3}{5}+\dfrac{3}{8}=\dfrac{24}{40}+\dfrac{15}{40};\dfrac{4}{9}+\dfrac{10}{27}=\dfrac{12}{27}+\dfrac{10}{27}=\dfrac{22}{27};\dfrac{2}{3}+\dfrac{7}{9}=\dfrac{18}{27}+\dfrac{21}{27}=\dfrac{39}{27};\dfrac{5}{12}+\dfrac{7}{15}=\dfrac{75}{180}+\dfrac{84}{180}=\dfrac{159}{180}\)

Đi mà, các bạn giúp mik đi. Mik tick cho!

\(\frac{232}{16}-\frac{658}{456}\)

\(=\frac{19}{2}-\frac{329}{228}\)

\(=\frac{2166}{228}-\frac{329}{228}\)

\(=\frac{2166-329}{228}\)

\(=\frac{1837}{228}\)

Vậy 232/16 - 658/456 = 1837/228

\(\frac{1837}{228}\)

\(\frac{232}{16}-\frac{658}{456}=\frac{2977}{228}\)

( 1000 + x ) . 5 = 10 . x + 5

1000 . 5 + 5 . x = 10 . x + 5

10 . x - 5 . x = 1000 . 5 - 5

5 . x = 5 . ( 1000 - 1 )

5 . x = 5 . 999

x = 999

( 1000 + x ) * 5 = 10 * x + 5

=> 1000 * 5 + 5 = 10 * x + 5

=> 10* x - 5      = 1000 * 5 - 5 ( Vì chuyển vế phải đổi dấu )

=> 5  * x          = 5 * ( 1000 - 1 )

=> 5 * x           = 5 * 999

=> 5 * x           = 4995

=> x               = 4995 : 5

=> x               = 999

Chúc bn học vui^^

Ta có:

\(A=\dfrac{\sqrt{x}-x}{\sqrt{x}-1}=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)

Vậy \(A=-\sqrt{x}\)

\(A=\dfrac{-\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=-\sqrt{x}\)

\(=\sqrt{x}\)

B

CHọn B

B

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

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

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

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

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

Ta có: \(A=\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+......+\frac{5}{64.69}\right)\)

\(\Rightarrow A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+......+\frac{1}{64}-\frac{1}{69}\right)\)

\(\Rightarrow A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{69}\right)=\frac{1}{5}\times\frac{65}{276}\)

\(\Rightarrow A=\frac{13}{276}\)

Vậy \(A=\frac{13}{276}\)

\(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{64.69}\)

\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{64}-\frac{1}{69}\)

\(5A=\frac{1}{4}-\frac{1}{69}\)

\(A=\frac{65}{276}:5\)

\(A=\frac{13}{276}\)

\(A=\frac{1}{4.9}+\frac{1}{9.14}+....+\frac{1}{64.69}\)

\(A=\frac{9-4}{1}+\frac{14-9}{9.14}+....+\frac{69-64}{64.69}\)

\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+....+\frac{1}{64}-\frac{1}{64}-\frac{1}{69}\)

\(5A=\frac{1}{4}-\frac{1}{69}\)

\(5A=\frac{65}{276}\)

\(A=\frac{65}{276}\div5=\frac{13}{276}\)

\(\Rightarrow12+20x-60=45x-15\Leftrightarrow25x=-33\Leftrightarrow x=-\dfrac{33}{25}\)

12+20x−60=45x−15

25x=−33

x=−\(\dfrac{33}{25}\)

\(\dfrac{3+5x}{5}-3=\dfrac{9x-3}{4}\)

\(\Leftrightarrow\dfrac{5x-12}{5}=\dfrac{9x-3}{4}\)

\(\Leftrightarrow20x-48=45x-15\)

\(\Leftrightarrow25x=-33\)

\(\Leftrightarrow x=-\dfrac{33}{25}\)