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

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

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

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

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

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

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

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

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

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

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

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`

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

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

\(\dfrac{5}{11}.\left(\dfrac{-3}{7}\right)+\dfrac{5}{11}.\left(\dfrac{-5}{7}\right)+\left(\dfrac{-8}{7}\right).\dfrac{6}{11}\\ =\dfrac{5}{11}.\left(\dfrac{-3}{7}+\dfrac{-5}{7}\right)+\left(\dfrac{-8}{7}\right).\dfrac{6}{11}\\ =\dfrac{5}{11}.\dfrac{-8}{7}+\dfrac{-8}{7}.\dfrac{6}{11}\\ =\dfrac{-8}{7}.\left(\dfrac{5}{11}+\dfrac{6}{11}\right)\\ =\dfrac{-8}{7}.\dfrac{11}{11}\\ =\dfrac{-8}{7}.1=-\dfrac{8}{7}\)

Ta có: M1 đối đỉnh với M3 

⇒ M1 kề bù với M2 

⇒ \(M1+M2=180^o\) 

Mà: \(M1=3\cdot M2\)

\(\Rightarrow3\cdot M2+M2=180^o\)

\(\Rightarrow4\cdot M2=180^o\)

\(\Rightarrow M2=\dfrac{180^o}{4}=45^o\)

Mà: M4 = M2 = `45^o` 

⇒ M1 = 3.M2 = 3.45 = `135^o` 

Mà: M1 = M3 

⇒ M3 = `135^o`  

`5x^3 : x + 2x - 5x^2 = 1`

`<=> 5x^2 + 2x - 5x^2 = 1`

`<=> 2x = 1`.

`<=> x = 1/2`

Không dùng dấu ⇔ nhé em!