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

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

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

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

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

\(\dfrac{7}{5}-\left(\dfrac{2}{5}-x\right)=-\dfrac{3}{10}\)

=>\(\dfrac{7}{5}-\dfrac{2}{5}+x=-\dfrac{3}{10}\)

=>\(x+1=-\dfrac{3}{10}\)

=>\(x=-\dfrac{3}{10}-1=-\dfrac{13}{10}\)

\(\dfrac{7}{5}\) - (\(\dfrac{2}{5}\) - \(x\)) = \(\dfrac{-3}{10}\)

        \(\dfrac{2}{5}\) - \(x\) = \(\dfrac{7}{5}\) -  \(\dfrac{-3}{10}\) 

         \(\dfrac{2}{5}\) - \(x\) = \(\dfrac{14}{10}\) + \(\dfrac{3}{10}\)

        \(\dfrac{2}{5}\) - \(x\) = \(\dfrac{17}{10}\)

              \(x\) = \(\dfrac{2}{5}\) - \(\dfrac{17}{10}\)

              \(x\) = \(\dfrac{4}{10}\) - \(\dfrac{17}{10}\)

              \(x\) = \(\dfrac{-13}{10}\)

Vậy \(x=-\dfrac{13}{10}\)

Gọi mẫu số của các phân số cần tìm là x

(Điều kiện: \(x\ne0\))

Theo đề, ta có: \(\dfrac{-3}{5}< \dfrac{9}{x}< \dfrac{-4}{9}\)

=>\(\dfrac{-36}{60}< \dfrac{-36}{-4x}< \dfrac{-36}{81}\)

=>\(\dfrac{36}{60}>\dfrac{36}{-4x}>\dfrac{36}{81}\)

=>60<-4x<81

=>-15>x>-20,25

=>-20,25<x<-15

Vậy: Các phân số cần tìm có dạng là \(\dfrac{9}{x}\), với điều kiện là -20,25<x<-15

Gọi mẫu của các phân số cần tìm là x

Theo đề, ta có: \(-\dfrac{9}{11}< \dfrac{7}{x}< \dfrac{-9}{13}\)

=>\(\dfrac{-63}{77}< \dfrac{-63}{-9x}< \dfrac{-63}{91}\)

=>\(\dfrac{63}{77}>\dfrac{63}{-9x}>\dfrac{63}{91}\)

=>77<-9x<91

=>\(-\dfrac{77}{9}>x>-\dfrac{91}{9}\)

Vậy: Các phân số cần tìm có dạng là \(\dfrac{7}{x}\), với điều kiện là \(-\dfrac{91}{9}< x< -\dfrac{77}{9}\)

\(\left(x+\dfrac{3}{5}\right)-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x+\dfrac{3}{5}-\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{1}{2}\\ \Rightarrow x=\dfrac{10}{30}-\dfrac{18}{30}+\dfrac{15}{30}\\ \Rightarrow x=\dfrac{7}{30}\)

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

=>\(x=\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{3}{5}=\dfrac{5}{6}-\dfrac{3}{5}=\dfrac{25}{30}-\dfrac{18}{30}=\dfrac{7}{30}\)

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

=>\(x+\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{5}{3}\)

=>\(x=\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{2}{5}=\dfrac{7}{3}-\dfrac{2}{5}=\dfrac{35}{15}-\dfrac{6}{15}=\dfrac{29}{15}\)

`#3107.101107`

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^8\cdot2^2\cdot5}\\ =\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\\ =\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\\ =\dfrac{-2}{6}=-\dfrac{1}{3}\)

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^9}{2^8.3^8.2^2+6^8.20}\\ =\dfrac{2^{10}.3^8-2.6^9}{\left(2.3\right)^8.2^2+6^8.20}=\dfrac{2^8.3^8.2^2-2.6^9}{6^8.4+6^8.20}\\ =\dfrac{6^8.4-2.6.6^8}{6^8.\left(4+20\right)}=\dfrac{6^8.\left(4-2.6\right)}{6^8.24}\\ =\dfrac{4-12}{24}=\dfrac{-8}{24}=-\dfrac{1}{3}\)

`#3107.101107`

\(\dfrac{2^8\cdot2^{18}}{8^5\cdot4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5\cdot\left(2^2\right)^6}=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{2^{26}}{2^{15+12}}=\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2}\)

\(\dfrac{2^8.2^{18}}{8^5.4^6}=\dfrac{2^{8+18}}{\left(2^3\right)^5.\left(2^2\right)^6}\\ =\dfrac{2^{26}}{2^{15}.2^{12}}=\dfrac{2^{26}}{2^{15+12}}\\ =\dfrac{2^{26}}{2^{27}}=\dfrac{1}{2^1}=\dfrac{1}{2}\)