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}`

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

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

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

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

a: \(-0,7< \dfrac{-13}{19}< -0,6\)

\(\dfrac{19}{-23}< -0,8\)

mà -0,8<-0,7

nên \(\dfrac{19}{-23}< -\dfrac{13}{19}\)

b: \(\dfrac{1}{83}:\dfrac{6}{331}=\dfrac{1}{83}\cdot\dfrac{331}{6}=\dfrac{331}{498}< 1\)

=>\(\dfrac{1}{83}< \dfrac{6}{331}\)

=>\(\dfrac{1}{83}+1< \dfrac{6}{331}+1\)

=>\(\dfrac{84}{83}< \dfrac{337}{331}\)

=>\(\dfrac{84}{-83}>\dfrac{-337}{331}\)