Lời giải:

$A=9+2.3^2+2.3^3+2.3^4+...+2.3^{2023}$

$A-9=2(3^2+3^3+3^4+...+3^{2023})$

$3(A-9)=2(3^3+3^4+3^5+...+3^{2024})$

$\Rightarrow 3(A-9)-(A-9)=2(3^{2024}-3^2)$

$2(A-9)=2.3^{2024}-18$

$\Rightarrow 2A-18=2.3^{2024}-18$

$\Rightarrow A=3^{2024}\vdots 3^{2023}$ (đpcm)

13/5

13/5

\(3-\dfrac{2}{5}=\dfrac{3}{1}-\dfrac{2}{5}=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

\(\left(\dfrac{3}{4}+\dfrac{2}{3}\right):\dfrac{17}{4}-\dfrac{3}{4}\)

\(=\left(\dfrac{9}{12}+\dfrac{8}{12}\right):\dfrac{17}{4}-\dfrac{3}{4}\)

\(=\dfrac{17}{12}:\dfrac{17}{4}-\dfrac{3}{4}\)

\(=\dfrac{17.4}{17.12}-\dfrac{3}{4}\)

\(=\dfrac{1}{3}-\dfrac{3}{4}\)

\(=\dfrac{4}{12}-\dfrac{9}{12}\)

\(=-\dfrac{5}{12}\)

de qua ma

x = 2/9 : 2/3 = 1/3

\(\frac{2}{9}:x=\frac{2}{3}\)

\(\Rightarrow x=\frac{2}{9}:\frac{2}{3}\)

\(\Rightarrow x=\frac{1}{3}\)

\(\text{vậy x = }\frac{1}{3}\)

\(3\dfrac{2}{3}=\dfrac{3\times3+2}{3}=\dfrac{11}{3}\)

\(\dfrac{11}{3}\)

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\left(x\ne-3;x\ne1\right)\)

suy ra: \(4\left(x-1\right)=2\left(x+3\right)\\ < =>4x-4=2x+6\\ < =>4x-2x=6+4\\ < =>2x=10\\ < =>x=5\left(tm\right)\)

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\text{ĐKXĐ}:x\ne-3;1\)

\(\Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}MTC:\left(x+3\right)\left(x-1\right)\)

\(\Rightarrow4x-4=2x+6\)

\(\Leftrightarrow4x-4-2x-6=0\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{5\right\}\)

ĐKXĐ: `x+3\ne0=>x\ne-3` ; `x-1\ne0=>x\ne1`

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\\ \Leftrightarrow4\left(x-1\right)=2\left(x+3\right)\\ \Leftrightarrow4x-4=2x+6\\ \Leftrightarrow4x-2x=6+4\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\left(tm\right)\)