a: \(A=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{3+x}{3-x}-\dfrac{3-x}{3+x}-\dfrac{12x^2}{x^2-9}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\left(\dfrac{-\left(x+3\right)}{x-3}+\dfrac{x-3}{x+3}-\dfrac{12x^2}{\left(x-3\right)\left(x+3\right)}\right)\)

\(=\dfrac{x+1}{x\left(3-x\right)}:\dfrac{-x^2-6x-9+x^2-6x+9-12x^2}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{-\left(x+1\right)}{x\left(x-3\right)}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-12x^2-12x}\)

\(=\dfrac{-\left(x+1\right)\cdot\left(x+3\right)}{-12x^2\left(x+1\right)}=\dfrac{x+3}{12x^2}\)

b: Ta có: |2x-1|=5

=>2x-1=5 hoặc 2x-1=-5

=>x=-2

Thay x=-2 vào A, ta được:

\(A=\dfrac{-2+3}{12\cdot\left(-2\right)^2}=\dfrac{1}{48}\)

c: Để \(A=\dfrac{2x+1}{x^2}\) thì \(\dfrac{x+3}{12x^2}=\dfrac{2x+1}{x^2}\)

=>x+3=24x+12

=>24x+12=x+3

=>23x=-9

hay x=-9/23

d: Để A<0 thì x+3<0

hay x<-3

a) \(\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)

\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)

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

\(=\dfrac{14}{5}\)

b) \(\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)

\(=\dfrac{7}{5}-\dfrac{1}{6}\)

\(=\dfrac{42}{30}-\dfrac{5}{30}\)

\(=\dfrac{37}{30}\)

\(a,\dfrac{9}{5}+\dfrac{9}{5}:\dfrac{9}{5}\)

\(=\dfrac{9}{5}+\dfrac{9}{5}\times\dfrac{5}{9}\)

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

\(=\dfrac{9}{5}+\dfrac{5}{5}\)

\(=\dfrac{14}{5}\)

\(b,\dfrac{7}{5}-\dfrac{1}{2}\times\dfrac{1}{3}\)

\(=\dfrac{7}{5}-\dfrac{1}{6}\)

\(=\dfrac{42}{30}-\dfrac{5}{30}\)

\(=\dfrac{37}{30}\)

ĐKXĐ: \(x\ne-3\)

Tại \(x=2\Rightarrow A=\dfrac{2^2+3}{3.2+9}=\dfrac{7}{15}\)

a) A xác định khi \(3x+9\ne0\Leftrightarrow x\ne-3\).

b) Với \(x=2\) thì \(A=\dfrac{2^2+3}{3\cdot2+9}=\dfrac{7}{15}\).

a) \(\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{4}{9}\times2=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{23}{9}\)

b) \(\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{2}{5}\times\dfrac{3}{2}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)

Câu 4 

\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times4\times3\times5\times4\times5}{5\times2\times4\times4\times5\times5}=\dfrac{3\times3}{5\times2}=\dfrac{9}{10}\)

Câu 3:

\(a.\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{15}{9}+\dfrac{8}{9}=\dfrac{23}{9}\)

\(b.\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)

Câu 4:

\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times3\times1}{2\times1\times5}=\dfrac{9}{10}\)

 a) 3² . 5³ + 9² = 9 . 125 + 81

= 1 125 + 81 = 1 206

b) 8³ : 4² - 5² = 512 : 16 - 25 = 32 - 25 = 7

c) 3³ . 9² - 5².9 + 18 : 6 = 27 . 81 - 25 . 9 + 3

= 2 187 - 225 + 3 = 1 962 + 3 = 1 965

 a) 3² . 5³ + 9² = 9 . 125 + 81

= 1 125 + 81 = 1 206

b) 8³ : 4² - 5² = 512 : 16 - 25 = 32 - 25 = 7

c) 3³ . 9² - 5².9 + 18 : 6 = 27 . 81 - 25 . 9 + 3

= 2 187 - 225 + 3 = 1 962 + 3 = 1 965

a) \(A=\left(-13\right)+\left(-27\right)+\left(-65\right)\)

        \(=-105\)

b) \(B=\left(-9\right)+\left(-14\right)+27\)

         \(=4\)

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

a) 32 - 6 . (8 - 2³) + 18 =  32 - 6 . (8 - 8) + 18

= 32 - 6 . 0 + 18 = 32 + 18 = 50

b) (3 . 5 - 9)³ . (1 + 2 . 3)² + 4²

= (15 - 9)³ . (1 + 6)² + 4²

= 6³ . 7² + 4² = 216 . 49 + 16 = 10 584 + 16 = 10 600

\(a,ĐK:x\ne2\\ b,A=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\\ c,x=\dfrac{2021}{1010}\Leftrightarrow A=\dfrac{3}{\dfrac{2021}{1010}-\dfrac{2020}{1010}}=\dfrac{3}{\dfrac{1}{1010}}=3030\)