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\(\dfrac{4}{5}\)(\(\dfrac{-3}{7}\)+\(\dfrac{-2}{7}\))= \(\dfrac{4}{5}\).\(\dfrac{-5}{7}\)= \(\dfrac{-20}{35}\)=\(\dfrac{-4}{7}\)

\(\dfrac{1}{4}\)x= \(\dfrac{5}{8}\)- \(\dfrac{3}{4}\)

\(\dfrac{1}{4}\)x= \(\dfrac{5}{8}\)- \(\dfrac{6}{8}\)

\(\dfrac{1}{4}\)x= \(\dfrac{-1}{8}\)

   x= \(\dfrac{-1}{8}\): \(\dfrac{1}{4}\)

   x= \(\dfrac{-1}{8}\): \(\dfrac{2}{8}\)

   x= \(\dfrac{-1}{8}\). \(\dfrac{8}{2}\)

   x= \(\dfrac{-1}{2}\)

a. x x(1,2+1,8)=45

    x x 3= 45

    x= 45:3

    x=15

b. \(\dfrac{13+x}{20}\)= \(\dfrac{15}{20}\)

=>  13+x=15

            x=15-13

            x=2

1d.<=> \(\dfrac{\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\)- \(\dfrac{x}{x\left(x+1\right)}\)= \(\dfrac{2x-1}{x\left(x+1\right)}\)

ĐK: x khác 0, x khác -1

<=> (x-1)(x+1) - x = 2x-1

<=> x²-1 -x -2x+1=0

<=> x² -3x=0

<=> x(x-3)=0

<=> x=0 (loại)  hoặc x-3=0

                                 x=3

S={3}

1c. <=> (x-2)(x+2) + (x-2)(x-9)= 0

<=> (x-2)(x+2+x-9)=0

<=> (x-2)(2x-7)=0

<=> x-2=0 hoặc 2x-7=0

<=> x=2             x=\(\dfrac{7}{2}\)

S={2;\(\dfrac{7}{2}\)}

1b. <=> \(\dfrac{2\left(2x+1\right)}{6}\) + \(\dfrac{3\left(x+1\right)}{6}\)= \(\dfrac{12}{6}\)

<=> 4x + 2 + 3x + 3= 12

<=> 7x= 7

<=> x= 1

S={1}

1a. th1: ta có x+2=3x+4 nếu x+2>0 => x > -2

x+2=3x+4

x-3x=4-2

-2x=2

x= -1 (chọn)

th2: ta có x+2= -3x+4 nếu x+2<0 => x< -2

x+2 = -3x+4

x+3x= 4-2

4x=2

x=\(\dfrac{1}{2}\)(loại)

S={-1}

2. \(\dfrac{3\left(x-2\right)}{6}\)- \(\dfrac{4}{6}\)> \(\dfrac{6x}{6}\)-\(\dfrac{6}{6}\)

<=> 3x - 6 - 4> 6x - 6

<=> 3x - 6 - 4 - 6x + 6> 0

<=> -3x -4 > 0

<=> -3x >  4

<=> x < \(\dfrac{-4}{3}\)

S= {x/x <\(\dfrac{-4}{3}\)}

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