1: \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)

\(=3+2\sqrt{2}+\sqrt{5}-2=2\sqrt{2}+\sqrt{5}+1\)

2: \(\dfrac{1}{\sqrt{3}+\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}+\dfrac{2\left(1+\sqrt{7}\right)}{-6}\)

\(=\dfrac{\sqrt{7}-\sqrt{3}}{4}-\dfrac{1+\sqrt{7}}{3}\)

\(=\dfrac{3\left(\sqrt{7}-\sqrt{3}\right)-4\left(\sqrt{7}+1\right)}{12}=\dfrac{-\sqrt{7}-3\sqrt{3}-4}{12}\)

3:

\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{2-\sqrt{a}}=-\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)}{\sqrt{a}-2}=-\sqrt{a}\)

4:

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{xy}\)

1) \(\dfrac{1}{3-2\sqrt{2}}+\dfrac{1}{\sqrt{5}+2}\)

\(=\dfrac{3+2\sqrt{2}}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\)

\(=\dfrac{3+2\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}+\dfrac{\sqrt{5}-2}{\left(\sqrt{5}\right)^2-2^2}\)

\(=\dfrac{3+2\sqrt{2}}{1}+\dfrac{\sqrt{5}-2}{1}\)

\(=3+2\sqrt{2}+\sqrt{5}-2\)

\(=2\sqrt{2}+\sqrt{5}+1\)

2) \(\dfrac{1}{\sqrt{3}-\sqrt{7}}+\dfrac{2}{1-\sqrt{7}}\)

\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{\left(1-\sqrt{7}\right)\left(1+\sqrt{7}\right)}\)

\(=\dfrac{\sqrt{3}+\sqrt{7}}{\left(\sqrt{3}\right)^2-\left(\sqrt{7}\right)^2}+\dfrac{2\cdot\left(1+\sqrt{7}\right)}{1^2-\left(\sqrt{7}\right)^2}\)

\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{2\cdot\left(1+\sqrt{7}\right)}{6}\)

\(=\dfrac{-\sqrt{3}-\sqrt{7}}{4}-\dfrac{1+\sqrt{7}}{3}\)

\(=\dfrac{-3\sqrt{3}-3\sqrt{7}}{12}-\dfrac{4+4\sqrt{7}}{12}\)

\(=\dfrac{-3\sqrt{3}-3\sqrt{7}-4-4\sqrt{7}}{12}\)

\(=\dfrac{-3\sqrt{3}-7\sqrt{7}-4}{12}\)

3) \(\dfrac{a-2\sqrt{a}}{2-\sqrt{a}}\)

\(=-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\)

\(=-\dfrac{\sqrt{a}\cdot\left(\sqrt{a}-2\right)}{\sqrt{a}-2}\)

\(=-\sqrt{a}\)

4) \(\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{x}\cdot\sqrt{xy}+\sqrt{y}\cdot\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(=\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)

\(=\sqrt{xy}\)

a) \(x=2\)

b) \(x=\frac{5}{36}\)

a) \(1-\left(\frac{2x}{3}+2\right)=-1\cdot\frac{1}{3}\)

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

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

\(-\frac{2}{3}x=\frac{2}{3}\)

\(x=-1\)

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b) \(\frac{2}{5}x-1\cdot\frac{1}{2}x+x=\frac{1}{3}\)

\(\left(\frac{2}{5}-\frac{1}{2}+1\right)x=\frac{1}{3}\)

\(\frac{9}{10}x=\frac{1}{3}\)

\(x=\frac{1}{3}:\frac{9}{10}\)

\(x=\frac{10}{27}\)

a, \(x\) - \(\dfrac{5}{7}\) = \(\dfrac{1}{9}\) 

    \(x\)        = \(\dfrac{1}{9}\) + \(\dfrac{5}{7}\)

    \(x\)        = \(\dfrac{52}{63}\)

b, \(\dfrac{2x}{5}\) = \(\dfrac{6}{2x+1}\)

     2\(x\).(2\(x\) + 1) = 30

     4\(x^2\)+ 2\(x\)  - 30 = 0

  4\(x^2\) + 12\(x\) - 10\(x\) - 30 = 0

   (4\(x^2\) + 12\(x\)) - (10\(x\) + 30) =0

   4\(x\).(\(x\) + 3) - 10.(\(x\) +3) = 0

        2 (\(x\) + 3).(2\(x\) -  5) = 0

            \(\left[{}\begin{matrix}x+3=0\\2x-5=0\end{matrix}\right.\)

             \(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-3; \(\dfrac{5}{2}\)}

c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\) = \(\dfrac{85}{x}\)

     \(\dfrac{85}{x}\)         = \(\dfrac{85}{24}\)

       \(x\)           = 24

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)..\left(1+\frac{1}{2019}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)

\(=\frac{2020}{2}=1010\)

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)

\(=\frac{2020}{2}\)

\(=1010\)

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)

\(=\frac{3}{2}.\frac{4}{3}....\frac{2020}{2019}\)

\(=\frac{2020}{2}\)

\(=1010\)

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x.\left(x+1\right)}=\dfrac{499}{500}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{499}{500}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{499}{500}\)

\(\Leftrightarrow x=499\)

giúp e vs

?????????????????

bài 1

a. = -10/3

b. = 14/5

c = 1/9

p = 1/9

bài 2.

a. x= 271/75

b. x = 1/3

bài 2

a. x - 10/3 = 7/15 . 3/5

    x - 10/3 = 7/25

           x   = 7/25 + 10/3

           x = 271/75

b. 8/23 . 46/24 - x = 1/3

               2/3 - x = 1/3

                      x = 2/3 - 1/3

                      x = 1/3

a. -5/2 : 3/4 = -5/2 . 4/3 = -20/6 = -10/3

b. (-3 1/7) : (-1 6/49) = -22/7 : (-55/49) = -22/7 . -49/55 = 1078/385 = 14/5

c. -1. 1/8 . 4/51 . (-11 1/3) = -1/8 . 4/51 . (-34/3) = -4/408 . (-34/13) = -1/102 . (-34/13) = 34/1326 = 1/39

d. (-1/6).(-15/19).38/45 = 15/114.38/45 = 5/38 . 38/45 = 5/45 = 1/9