Cách 1:

a)  \( - 2020 = 2.\left( { - 1010} \right)\) nên \(\left( { - 2020} \right):2 =  - 1010\)

b) \(64 = \left( { - 8} \right).\left( { - 8} \right) \Rightarrow 64:\left( { - 8} \right) =  - 8\)

c) \( - 90 = \left( { - 45} \right).2 \Rightarrow \left( { - 90} \right):\left( { - 45} \right) = 2\)

d) \( - 2121 = 3.\left( { - 707} \right) \Rightarrow \left( { - 2121} \right):3 =  - 707\)

Cách 2:

a) \( - 2020 :2=-(2020:2)=  - 1010\)

b)  \(64:(-8)=-(64:8) =  - 8\)

c) \( -90:(-45)=90:45= 2\)

d) \( - 2121 :3=-(2121:3)=  - 707\)

\(9,\left(2x-5\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-5-x-1\right)\left(2x-5+x+1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(3x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy \(S=\left\{6;\dfrac{4}{3}\right\}\)

\(10,\left(x+3\right)^2-x^2=45\)

\(\Leftrightarrow x^2+6x+9-x^2-45=0\\ \Leftrightarrow6x=36\\ \Leftrightarrow x=6\)

Vậy \(S=\left\{6\right\}\)

\(11,\left(5x-4\right)^2-49x^2=0\\ \Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\\ \Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\\ \Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(S=\left\{-2;\dfrac{1}{3}\right\}\)

\(12,16\left(x-1\right)^2-25=0\\ \Leftrightarrow4^2\left(x-1\right)^2-5^2=0\\ \Leftrightarrow\left[4\left(x-1\right)\right]^2-5^2=0\\ \Leftrightarrow\left(4x-4\right)^2-5^2=0\\ \Leftrightarrow\left(4x-4-5\right)\left(4x-4+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-9=0\\4x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{4};\dfrac{9}{4}\right\}\)

\(\left(\frac{5}{12}-\frac{5}{7}-\frac{22}{45}+\frac{7}{12}-\frac{23}{45}\right).\left|x\right|-9=-2\)

\(\left(\frac{12}{12}-\frac{5}{7}-\frac{45}{45}\right).\left|x\right|=-2+9\)

\(\left(1-\frac{5}{7}-1\right).\left|x\right|=7\)

\(\frac{-5}{7}.\left|x\right|=7\)

\(\left|x\right|=7:\left(\frac{-5}{7}\right)\)

\(\left|x\right|=\frac{-49}{5}\)

\(\Rightarrow x\in\varnothing\) vì trị tuyệt đối của 1 số luôn dương 

Ta có \(\dfrac{a}{b}=\dfrac{36}{45}=\dfrac{4}{5}\Rightarrow a=4k,b=5k\) 

BCNN (a,b) =300 mà \(\left(4,5\right)=1\Rightarrow k=300:\left(4.5\right)=15\)

Vậy \(a=4.15=60;b=5.15=75\)

75