a: \(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-2\dfrac{2}{5}\)

=>\(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-\dfrac{12}{5}\)

=>\(\dfrac{1}{3}-5x=\dfrac{32}{15}:\dfrac{-12}{5}=\dfrac{32}{15}\cdot\dfrac{-5}{12}=\dfrac{-160}{180}=\dfrac{-8}{9}\)

=>\(5x=\dfrac{1}{3}+\dfrac{8}{9}=\dfrac{3}{9}+\dfrac{8}{9}=\dfrac{11}{9}\)

=>\(x=\dfrac{11}{9}:5=\dfrac{11}{45}\)

b: \(\left(2x-\dfrac{1}{3}\right)^5=\dfrac{32}{243}\)

=>\(\left(2x-\dfrac{1}{3}\right)^5=\left(\dfrac{2}{3}\right)^5\)

=>\(2x-\dfrac{1}{3}=\dfrac{2}{3}\)

=>\(2x=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

=>\(x=\dfrac{1}{2}\)

c: \(\left(\dfrac{5}{6}x+3\right)^2=\dfrac{4}{9}\)

=>\(\left[{}\begin{matrix}\dfrac{5}{6}x+3=\dfrac{2}{3}\\\dfrac{5}{6}x+3=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{6}x=\dfrac{2}{3}-3=-\dfrac{7}{3}\\\dfrac{5}{6}x=-\dfrac{2}{3}-3=-\dfrac{11}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-\dfrac{7}{3}:\dfrac{5}{6}=-\dfrac{7}{3}\cdot\dfrac{6}{5}=\dfrac{-14}{5}\\x=-\dfrac{11}{3}:\dfrac{5}{6}=-\dfrac{11}{3}\cdot\dfrac{6}{5}=\dfrac{-22}{5}\end{matrix}\right.\)

a) 

\(\dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-2\dfrac{2}{5}\\ \dfrac{32}{15}:\left(\dfrac{1}{3}-5x\right)=-\dfrac{8}{5}\\ \dfrac{1}{3}-5x=\dfrac{32}{15}:-\dfrac{8}{5}\\ \dfrac{1}{3}-5x=\dfrac{-4}{3}\\ 5x=\dfrac{1}{3}+\dfrac{4}{3}\\ 5x=\dfrac{5}{3}\\ x=\dfrac{5}{3}:5\\ x=\dfrac{1}{3}\) 

b) \(\left(2x-\dfrac{1}{3}\right)^5=\dfrac{32}{243}\) 

\(\left(2x-\dfrac{1}{3}\right)^5=\left(\dfrac{2}{3}\right)^5\)

\(2x-\dfrac{1}{3}=\dfrac{2}{3}\)

\(2x=\dfrac{2}{3}+\dfrac{1}{3}\)

\(2x=1\\ x=\dfrac{1}{2}\) 

c) \(\left(\dfrac{5}{6}x+3\right)^2=\dfrac{4}{9}\)

\(\left(\dfrac{5}{6}x+3\right)^2=\left(\dfrac{2}{3}\right)^2\)

TH1: 

\(\dfrac{5}{6}x+3=\dfrac{2}{3}\\ \dfrac{5}{6}x=\dfrac{2}{3}-3\\ \dfrac{5}{6}x=-\dfrac{7}{3}\\ x=\dfrac{-7}{3}:\dfrac{5}{6}=-\dfrac{14}{5}\)

TH2: 

\(\dfrac{5}{6}x+3=-\dfrac{2}{3}\\ \dfrac{5}{6}x=-\dfrac{2}{3}-3\\ \dfrac{5}{6}x=-\dfrac{11}{3}\\ x=-\dfrac{11}{3}:\dfrac{5}{6}\\ x=-\dfrac{22}{5}\)

d: \(\dfrac{13}{27}< \dfrac{13}{26}=\dfrac{1}{2}\)

\(\dfrac{1}{2}=\dfrac{20,5}{41}< \dfrac{27}{41}\)

Do đó: \(\dfrac{13}{27}< \dfrac{27}{41}\)

c: a+1>a-1

=>\(\dfrac{1}{a+1}< \dfrac{1}{a-1}\)

a: \(\dfrac{14}{25}=0,56;\dfrac{5}{7}=0,\left(714285\right)\)

mà 0,56<0,(714285)

nên \(\dfrac{14}{25}< \dfrac{5}{7}\)

a) 

\(\dfrac{14}{25}< \dfrac{14}{21}=\dfrac{2}{3}\)

\(\dfrac{15}{21}>\dfrac{14}{21}\) hay \(\dfrac{5}{7}>\dfrac{2}{3}\)

\(\dfrac{14}{25}< \dfrac{5}{7}\)

c) \(a+1>a-1\)

\(\dfrac{1}{a+1}< \dfrac{1}{a-1}\)

đ) \(\dfrac{1119}{1999}=1-\dfrac{880}{1999};\dfrac{1999}{2000}=1-\dfrac{1}{2000}\)

Mà: \(\dfrac{880}{1999}>\dfrac{1}{2000}\) (vì 1999 < 2000 và 880 > 1)  

\(1-\dfrac{880}{1999}< 1-\dfrac{1}{2000}\)

\(\dfrac{1119}{1999}< \dfrac{1999}{2000}\)

d) Ta có: 

\(\dfrac{13}{27}< \dfrac{13,5}{27}=\dfrac{1}{2}\)

\(\dfrac{27}{41}>\dfrac{20,5}{41}=\dfrac{1}{2}\)

\(\dfrac{13}{27}< \dfrac{27}{41}\)

\(2\left(x+1\dfrac{1}{3}\right)=\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}\\ 2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{6}\\ x+\dfrac{4}{3}=\dfrac{1}{6}:2\\ x+\dfrac{4}{3}=\dfrac{1}{12}\\ x=\dfrac{1}{12}-\dfrac{4}{3}\\ x=-\dfrac{15}{12}=\dfrac{-5}{4}\)

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

=>\(2\left(x+\dfrac{4}{3}\right)=\dfrac{1}{4}\cdot\dfrac{2}{3}=\dfrac{1}{6}\)

=>\(x+\dfrac{4}{3}=\dfrac{1}{12}\)

=>\(x=\dfrac{1}{12}-\dfrac{4}{3}=\dfrac{1}{12}-\dfrac{16}{12}=-\dfrac{15}{12}=-\dfrac{5}{4}\)

Số vịt ban đầu là:

20456-12650=7806(con)

Số gà ban đầu là:

12650-7806=4844(con)

\(\left(-\dfrac{3}{5}\right)^2-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\)

=>\(\dfrac{9}{25}-\left(x-\dfrac{1}{3}\right)=\dfrac{4}{25}\)

=>\(x-\dfrac{1}{3}=\dfrac{9}{25}-\dfrac{4}{25}=\dfrac{5}{25}=\dfrac{1}{5}\)

=>\(x=\dfrac{1}{5}+\dfrac{1}{3}=\dfrac{8}{15}\)

a: Quy luật là số sau bằng số trước cộng thêm 3 đơn vị

b: B={2;5;8;11;14;17;20;23;26;29}

a: Quy luật là số trước cộng thêm 3 đơn vị thì ra số sau.

b: B={2;5;8;11;14;17;20;23;26;29}

ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AB^2+AB^2=10^2\)

=>\(2\cdot AB^2=100\)

=>\(AB^2=50\)

=>\(AB=\sqrt{50}=5\sqrt{2}\left(cm\right)\)

`x- \left(\frac54-\frac75 \right)=\frac{9}{20}`

`\Rightarrow x-\frac{-3}{20}=\frac{9}{20}`

`\Rightarrow x=\frac{9}{20}+\frac{-3}{20}`

`\Rightarrow x=\frac{3}{10}`

\(x-\left(\dfrac{5}{4}-\dfrac{7}{5}\right)=\dfrac{9}{20}\)

=>\(x-\dfrac{25-28}{20}=\dfrac{9}{20}\)

=>\(x+\dfrac{3}{20}=\dfrac{9}{20}\)

=>\(x=\dfrac{9}{20}-\dfrac{3}{20}=\dfrac{6}{20}=\dfrac{3}{10}\)