a: \(\dfrac{8}{18}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{5}{3}=\dfrac{4}{9}+\dfrac{15}{9}=\dfrac{4+15}{9}=\dfrac{19}{9}\)

b: \(\dfrac{8}{24}+\dfrac{4}{48}=\dfrac{1}{3}+\dfrac{1}{12}=\dfrac{4}{12}+\dfrac{1}{12}=\dfrac{4+1}{12}=\dfrac{5}{12}\)

c: \(\dfrac{20}{15}-\dfrac{4}{45}=\dfrac{4}{3}-\dfrac{4}{45}=\dfrac{60}{45}-\dfrac{4}{45}=\dfrac{60-4}{45}=\dfrac{56}{45}\)

d: \(\dfrac{40}{32}-\dfrac{1}{2}=\dfrac{5}{4}-\dfrac{1}{2}=\dfrac{5-2}{4}=\dfrac{3}{4}\)

`\sqrt{[27(x-1)^2]/12} +3/2 - (x - 2)\sqrt{[50x^2]/[8(x-2)^2]}`   `(1 < x < 2)`

`=\sqrt{[3(x-1)]^2 .3}/\sqrt{2^2 .3} + 3/2 - (x - 2) \sqrt{(5x)^2 . 2}/\sqrt{[2(x - 2)]^2 . 2}`

`=[3\sqrt{3}|x-1|]/[2\sqrt{3}]+3/2-(x-2)[5\sqrt{2}|x|]/[2\sqrt{2}|x-2|]`

`=[3(x-1)]/2+3/2-[5x(x-2)]/[2(2-x)]`   (Vì `1 < x < 2`)

`=3/2x - 3/2 + 3/2 + 5/2x`

`=4x`

Câu 1: \(\sqrt{8}\) − \(\sqrt{18}\) + \(2\sqrt{32}\) = \(\sqrt{4\text{×}2}\) −  \(\sqrt{\text{9×2}}\) + 2\(\sqrt{\text{16×2}}\)

                                           =2\(\sqrt{2}\) − 3\(\sqrt{2}\) + 2×4\(\sqrt{2}\) 

                                           =(2− 3+ 8)\(\sqrt{2}\)

                                           =7\(\sqrt{2}\)

Câu 2: Mik ko chắc làm đúng hay ko  nên ko làm

b: \(=\dfrac{\sqrt{x}-4+\sqrt{x}+4}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\cdot\dfrac{\sqrt{x}+4}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-4}\)

a) \(\dfrac{25}{60}+6\times\dfrac{1}{8}=\dfrac{25}{60}+\dfrac{3}{4}=\dfrac{25}{69}+\dfrac{45}{60}=\dfrac{25+45}{60}=\dfrac{70}{60}=\dfrac{7}{6}\)

b) \(\dfrac{15}{9}\times\left(6:\dfrac{30}{9}\right)=\dfrac{15}{9}\times\left(6\times\dfrac{30}{9}\right)=\dfrac{15}{9}\times20=\dfrac{100}{3}\)

\(B=\left(ab+bc+ca\right)\left(\dfrac{ab+bc+ca}{abc}\right)-abc\left(\dfrac{a^2b^2+b^2c^2+c^2a^2}{a^2b^2c^2}\right)\)

\(=\dfrac{\left(ab+bc+ca\right)^2-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=\dfrac{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)-\left(a^2b^2+b^2c^2+c^2a^2\right)}{abc}\)

\(=2\left(a+b+c\right)\)

B = \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{19}{20}\)

= \(\dfrac{1}{20}\)

\(A=\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)\cdot...\left(1+\dfrac{1}{2499}\right)\)

\(=\dfrac{4}{3}\cdot\dfrac{9}{8}\cdot...\cdot\dfrac{2500}{2499}\)

\(=\dfrac{2\cdot2}{1\cdot3}\cdot\dfrac{3\cdot3}{2\cdot4}\cdot...\cdot\dfrac{50\cdot50}{49\cdot51}\)

\(=\dfrac{2\cdot3\cdot4\cdot...\cdot50}{1\cdot2\cdot3\cdot...\cdot49}\cdot\dfrac{2\cdot3\cdot...\cdot50}{3\cdot4\cdot...\cdot51}\)

\(=\dfrac{50}{1}\cdot\dfrac{2}{51}=\dfrac{100}{51}\)

b: \(C=75\left(2-128+128\right)=75\cdot2=150\)

e: \(E=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{69\cdot74}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{69}-\dfrac{1}{74}\right)\)

\(=\dfrac{1}{5}\cdot\dfrac{70}{74}=\dfrac{14}{74}=\dfrac{7}{37}\)