\(a.\dfrac{1}{a+1}+\dfrac{2}{1-a}+\dfrac{5a-1}{a^2-1}\)

\(=\dfrac{1}{a+1}-\dfrac{2}{a-1}+\dfrac{5a-1}{a^2-1}\)

\(=\dfrac{a-1}{\left(a+1\right)\left(a-1\right)}-\dfrac{2\left(a+1\right)}{\left(a+1\right)\left(a-1\right)}+\dfrac{5a-1}{\left(a+1\right)\left(a-1\right)}\)

\(=\dfrac{a-1-2a-2+5a-1}{\left(a+1\right)\left(a-1\right)}\)

\(=\dfrac{4a-4}{\left(a+1\right)\left(a-1\right)}\)

\(=\dfrac{4}{a+1}\)

a.\(\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}=\dfrac{3x-x+6}{2x\left(x+3\right)}=\dfrac{2x+6}{2x\left(x+3\right)}=\dfrac{1}{x}\)

\(a,\dfrac{3}{9}=\dfrac{1}{3}\\ \dfrac{16 }{24}=\dfrac{2}{3}\\ b,\dfrac{17}{51}=\dfrac{1}{3}=\dfrac{5}{15}\\ \dfrac{18}{90}=\dfrac{1}{5}=\dfrac{3}{15}\\ c,\dfrac{5}{10}=\dfrac{1}{2}=\dfrac{3}{6}\\ \dfrac{15}{45}=\dfrac{1}{3}=\dfrac{2}{6}\\ d,\dfrac{42}{63}=\dfrac{2}{3}=\dfrac{6}{9}\\ \dfrac{64}{72}=\dfrac{8}{9}\)

ngu ngu ngu ngu ngu ngu

2: ĐKXĐ: x>=0

\(\sqrt{3x}-2\sqrt{12x}+\dfrac{1}{3}\cdot\sqrt{27x}=-4\)

=>\(\sqrt{3x}-2\cdot2\sqrt{3x}+\dfrac{1}{3}\cdot3\sqrt{3x}=-4\)

=>\(\sqrt{3x}-4\sqrt{3x}+\sqrt{3x}=-4\)

=>\(-2\sqrt{3x}=-4\)

=>\(\sqrt{3x}=2\)

=>3x=4

=>\(x=\dfrac{4}{3}\left(nhận\right)\)

3: 

ĐKXĐ: x>=0

\(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18}=0\)

=>\(3\sqrt{2x}+5\cdot2\sqrt{2x}-20-3\sqrt{2}=0\)

=>\(13\sqrt{2x}=20+3\sqrt{2}\)

=>\(\sqrt{2x}=\dfrac{20+3\sqrt{2}}{13}\)

=>\(2x=\dfrac{418+120\sqrt{2}}{169}\)

=>\(x=\dfrac{209+60\sqrt{2}}{169}\left(nhận\right)\)

4: ĐKXĐ: x>=-1

\(\sqrt{16x+16}-\sqrt{9x+9}=1\)

=>\(4\sqrt{x+1}-3\sqrt{x+1}=1\)

=>\(\sqrt{x+1}=1\)

=>x+1=1

=>x=0(nhận)

5: ĐKXĐ: x<=1/3

\(\sqrt{4\left(1-3x\right)}+\sqrt{9\left(1-3x\right)}=10\)

=>\(2\sqrt{1-3x}+3\sqrt{1-3x}=10\)

=>\(5\sqrt{1-3x}=10\)

=>\(\sqrt{1-3x}=2\)

=>1-3x=4

=>3x=1-4=-3

=>x=-3/3=-1(nhận)

6: ĐKXĐ: x>=3

\(\dfrac{2}{3}\sqrt{x-3}+\dfrac{1}{6}\sqrt{x-3}-\sqrt{x-3}=-\dfrac{2}{3}\)

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

=>\(\sqrt{x-3}\cdot\dfrac{-1}{6}=-\dfrac{2}{3}\)

=>\(\sqrt{x-3}=\dfrac{2}{3}:\dfrac{1}{6}=\dfrac{2}{3}\cdot6=\dfrac{12}{3}=4\)

=>x-3=16

=>x=19(nhận)

18x -19 = 21 + 8x 

18 x - 8x = 21 + 19 

10 x = 40

x =4 

g) 18 -4x=-20-6x

-4x + 6x = -20 - 18

2x = -38

x= -19

k, -10 \(x\) - 27 = -7\(x\) + 33

   -27 - 33 = -7\(x\) + 10\(x\)

3\(x\)             = -60

   \(x\)            - 20

m, -17\(x\) - 24 = -9\(x\) - 40

     -24 + 40 = -9\(x\) + 17\(x\)

     8\(x\)          = 16

       \(x\)           = 2

h)-15x -24 = 7x + 32 

-15x - 7x = 32 + 24 

-22x = 56 

x = \(-\dfrac{14}{5}\)

i)15x -3(4x -6)=-12+36

15x - 12x + 18 = 24

3x + 18 = 24 

3x = 24 - 18

3x = 6 

x=2

18 - 4\(x\) = -20 - 6\(x\) 

-4\(x\) + 6\(x\) = - 20  - 18

      2\(x\)    = - 38

         \(x\)  = - 19

h, -15 \(\times\) 24 = -7\(x\) + 32

     7\(x\) = 360 + 32

      7\(x\) = 392

         \(x\) =  392:7

          \(x\) = 56

i, 15\(x\) -3.(4\(x\) - 6) = -12 + 36

    15\(x\) - 12\(x\) + 18 =  24

      3\(x\) = 24 - 18

       3\(x\) = 6

         \(x\) = 2

k, -10\(x\) - 27 = -7\(x\) + 33

    -27 - 33 = -7\(x\) + 10\(x\)

     3\(x\) = -60

        \(x\) = -20

m, -17\(x\) - 24 = - 9\(x\) - 40

     - 24 + 40 =  -9\(x\) + 17\(x\)

          8\(x\) = 16

             \(x\) = 2

n, -23 \(\times\) 25 = -18\(x\) + 75

     -575 = -18\(x\) + 75

       18\(x\) = 575 + 75

        18\(x\) = 650

              \(x\) = \(\dfrac{325}{9}\)

a ⇒A=\(4\sqrt{4\times3}+3\sqrt{25\times3}-5\sqrt{16\times3}=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}=3\sqrt{3}\)

b ĐKXĐ x≥2 ⇔\(\sqrt{x-2}+3\sqrt{x-2}=16\Leftrightarrow4\sqrt{x-2}=16\Leftrightarrow\sqrt{x-2}=4\Rightarrow x-2=16\Leftrightarrow x=18\)

a.   \(A=4\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)

          \(=8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)

          \(=3\sqrt{3}\)

b.    \(\sqrt{x-2}-\sqrt{9x-18}=16\)

   \(\Leftrightarrow\sqrt{x-2}-\sqrt{9\left(x-2\right)}=16\)

   \(\Leftrightarrow\sqrt{x-2}-3\sqrt{x-2}=16\)

   \(\Leftrightarrow-2\sqrt{x-2}=16\)

   \(\Leftrightarrow\sqrt{x-2}=-8\)  ( Vô lý )

   Vậy PT vô nghiệm

a, 4\(\sqrt{12}+3\sqrt{75}-5\sqrt{48}\)

=\(8\sqrt{3}+15\sqrt{3}-20\sqrt{3}\)

= (8+15-20)\(\sqrt{3}\) = 3\(\sqrt{3}\)