Ta có :

$$\begin{gathered} \mathrm{B}=\frac{2.3.5.2^{14}.3^{29}-5.2^5.7^5.2^{12}}{2.3^3.2^{14}.3^{14}.3^{14}-2^2.3.2^{15}.7^5} \\ \mathrm{B}=\frac{2^{15}.3^{30}.5-2^{17}.5.7^5}{2^{15}.2^{31}-2^{17}.3.7^5} \\ \mathrm{B}=\frac{5\left(2^{15}.3^{30}-2^{17}.7^5\right)}{3\left(2^{15}.3^{30}-2^{17}.7^5\right)}=\frac{5}{3} \end{gathered}$$

$$\begin{gathered} {\left(\mathrm{x}+5\right)}^3=-125 \\ \Rightarrow {\left(\mathrm{x}+5\right)}^3={\left(-5\right)}^3 \\ \Rightarrow \mathrm{x}+5=-5 \\ \Rightarrow \mathrm{x}=-10 \end{gathered}$$

Ta có: $3^4.27=3^4.3^3=3^{4+3}=3^7$

Ta có: ${\mathrm{a}}^{\mathrm{n}}:{\mathrm{a}}^2={\mathrm{a}}^{\mathrm{n}-2}$.

$$\begin{gathered} {\left(\mathrm{x}+2\right)}^2=64\Rightarrow {\left(\mathrm{x}+2\right)}^2=8^2 \\ \Rightarrow \mathrm{x}+2=8 \mathrm{hoặc} \mathrm{x}+2=-8 \\ \Rightarrow \mathrm{x}=6 \mathrm{hoặc} \mathrm{x}=-10 \end{gathered}$$

Suy ra $S=\left\{6; -10\right\}$.

Do đó tổng các phần tử là: 6 + (-10) = -4.

Ta có 3².3  = 3³.

Do đó ta có:

$3^3\le 3^{\mathrm{n}}\le 3^5\iff 3\le \mathrm{n}\le 5\Rightarrow \mathrm{n}\in \left\{3;4;5\right\}$

Vậy có tất cả 3 giá trị của n thỏa mãn điều kiện.

Xét

$$\begin{gathered} 3.\mathrm{A}=3^{2021}-3^{2020}+3^{2019}-3^{2018}+...+3^3-3^2+3 \\ \Rightarrow 3\mathrm{A}+\mathrm{A}=3^{2021}+1 \\ \Rightarrow \mathrm{A}=\frac{3^{2021}+1}{4} \end{gathered}$$

Ta có: (-2x + 1)³ = -0,001

(-2x + 1)³ = (-0,1)³

-2x + 1 = -0,1

-2x = -1,1

x = 0,55.

Vậy có 1 giá trị của x thỏa mãn.

Xét :

$$\begin{gathered} \frac{1}{2^2}.\mathrm{S}=\frac{1}{2^4}-\frac{1}{2^6}+\frac{1}{2^8}-...+\frac{1}{2^{2016}}-\frac{1}{2^{2018}} \\ \Rightarrow \frac{1}{2^2}.\mathrm{S}+\mathrm{S}=\frac{1}{2^2}-\frac{1}{2^{2020}} \\ \Rightarrow \frac{5}{4}\mathrm{S}<\frac{1}{4} \\ \Rightarrow \mathrm{S}<\frac{1}{5} \mathrm{hay} \mathrm{S}<0,2 \end{gathered}$$

$$\begin{gathered} \mathrm{A}=\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3} \\ \mathrm{A}=\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)} \\ \mathrm{A}=\frac{5.\left(-6\right)}{9}=\frac{-10}{3} \end{gathered}$$

Do đó A, B, D đúng và C sai.

(7x + 3)⁴ = 625

(7x + 3)⁴ = 5⁴ = (-5)⁴

7x + 3 = 5 hoặc 7x + 3 = -5

$\mathrm{x}=\frac{2}{7} \mathrm{hoặc} \mathrm{x}=-\frac{8}{7}$

$$\begin{gathered} {\left(0,5\right)}^5:{\left(-0,5\right)}^3-{\left(\frac{17}{2}\right)}^7:{\left(\frac{17}{2}\right)}^6 \\ ={\left(-0,5\right)}^2-\frac{17}{2} \\ =\frac{1}{4}-\frac{17}{2} \\ =-\frac{33}{4} \end{gathered}$$

Ta có:

$$\begin{gathered} \frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.{10}^2} \\ =\frac{1}{9^1}-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{{10}^2} \\ =1-\frac{1}{{10}^2}=\frac{99}{100}<1 \end{gathered}$$

Vậy $\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.{10}^2}<1$.

Do đó cả A và B đều đúng.

Ta có:

$$\begin{gathered} \mathrm{P}=\left(\frac{5}{12}.\frac{6}{20}\right)+{\left(\frac{1}{6}\right)}^2 \\ \mathrm{P}=\frac{1}{8}+\frac{1}{36}=\frac{11}{17} \\ \mathrm{Q}=\frac{4^2.2^3}{2^6}=\frac{2^4.2^3}{2^6} \\ \mathrm{Q}=\frac{2^7}{2^6}=2 \end{gathered}$$

Khi đó: $\mathrm{P}-\mathrm{Q}=\frac{11}{72}-2=\frac{-133}{72}$

a) $7^6+7^5-7^4⋮55$

Ta có:

$$\begin{gathered} 7^6+7^5-7^4 \\ =7^4.\left(7^2+7-1\right) \\ =7^4.\left(49+7-1\right) \\ =7^4.55⋮55 \end{gathered}$$

Vậy $7^6+7^5-7^4⋮55$

b) ${81}^7-{27}^9+3^{29}⋮33$

Ta có:

$$\begin{gathered} {81}^7-{27}^9+3^{29} \\ ={\left(3^4\right)}^7-{\left(3^3\right)}^9+3^{29} \\ =3^{28}-3^{27}+3^{29} \\ =3^{26}.\left(3^2-2+3^3\right) \\ =3^{26}.33⋮33 \end{gathered}$$

Vậy ${81}^7-{27}^9+3^{29}⋮33$

c) $8^{12}-2^{33}-2^{30}⋮55$

Ta có:

$$\begin{gathered} 8^{12}-2^{33}-2^{30} \\ =8^{12}-2^{33}-2^{30} \\ ={\left(2^3\right)}^{12}-2^{33}-2^{30} \\ =2^{36}-2^{33}-2^{30} \\ =2^{30}.\left(2^6-2^3-1\right) \\ =2^{30}.55⋮55 \end{gathered}$$

Vậy $8^{12}-2^{33}-2^{30}⋮55$

Do đó cả 3 phát biểu đều đúng.