1) \(x^2-4=x^2-2^2=\left(x-2\right)\left(x+2\right)\)

2) \(1-4x^2=1-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)

3) \(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)

4) \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)

5) \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)

6) \(9x^2-36=\left(3x\right)^2-6^2=\left(3x-6\right)\left(3x+6\right)\)

7) \(\left(3x\right)^2-y^2=\left(3x+y\right)\left(3x-y\right)\)

8) \(x^2-\left(2y\right)^2=\left(x+2y\right)\left(x-2y\right)\)

9) \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)

a:Ta có: 22>21

=>-22<-21

=> \(-\frac{22}{35}<-\frac{21}{35}=-\frac35\) (1)

ta có: \(\frac{-3}{5}=\frac{-3\cdot35,4}{5\cdot35,4}=\frac{-106,2}{177}\)

mà \(-\frac{106.2}{177}<-\frac{103}{177}\left(-106,2<-103\right)\)

nên \(-\frac35<-\frac{103}{177}\) (2)

Từ (1) và (2) suy ra \(-\frac{22}{35}<-\frac{103}{177}\)

b:Ta có: 38<65

=>\(\frac{13}{38}>\frac{13}{65}\)

=> \(\frac{-13}{38}<-\frac{13}{65}=-\frac15\) (3)

Ta có: \(75<88\)

=>\(\frac{15}{75}>\frac{15}{88}\)

=>\(-\frac{15}{75}<-\frac{15}{88}\)

=>\(-\frac15<-\frac{15}{88}\) (4)

Từ (3),(4) suy ra \(-\frac{13}{38}<-\frac{15}{88}\)

c: Ta có: 66<67

=>\(\frac{22}{66}>\frac{22}{67}\)

=>\(-\frac{22}{66}<-\frac{22}{67}\)

=>\(-\frac13<-\frac{22}{67}\)

=>\(-\frac{22}{67}>\frac{-1}{3}\left(5\right)\)

Ta có: 151<153

=>\(\frac{51}{151}>\frac{51}{153}\)

=>\(\frac{51}{151}>\frac13\)

=>\(-\frac{51}{151}<-\frac13\) (6)

Từ (5),(6) suy ra \(-\frac{51}{151}<-\frac{22}{67}\)

d: Ta có: 17<17,5

=>-17>-17,5

=>\(\frac{-17}{35}>\frac{-17,5}{35}=-\frac12\) (7)

Ta có: 85<86

=>\(\frac{43}{85}>\frac{43}{86}\)

=>\(-\frac{43}{85}<\frac{-43}{86}=-\frac12\) (8)

Từ (7),(8) suy ra \(-\frac{43}{85}<\frac{-17}{35}\)

\(3x-\dfrac{3}{5}=\dfrac{-7}{10}\\ 3x=\dfrac{-7}{10}+\dfrac{3}{5}\\ 3x=\dfrac{-7}{10}+\dfrac{6}{10}\\ 3x=\dfrac{-1}{10}\\ x=\dfrac{-1}{30}\)

___________________

\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\\ \dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\\ \dfrac{1}{3}:x=\dfrac{9}{15}-\dfrac{10}{15}\\ \dfrac{1}{3}:x=\dfrac{-1}{15}\\ x=\dfrac{1}{3}:\dfrac{-1}{15}\\ x=-5\)

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

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

\(3x\)        \(=-\dfrac{1}{10}\)

 \(x\)         \(=-\dfrac{1}{10}:3\)

 \(x\)         \(=-\dfrac{1}{30}\)

Vậy \(x=-\dfrac{1}{30}\)

\(\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{3}{5}\)

        \(\dfrac{1}{3}:x=\dfrac{3}{5}-\dfrac{2}{3}\)

        \(\dfrac{1}{3}:x=-\dfrac{1}{15}\)

              \(x=\dfrac{1}{3}:-\dfrac{1}{15}\)

              \(x=-5\)

Vậy \(x=-5\)

\(\left(\dfrac{2}{7}-\dfrac{9}{4}\right)-\left(-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(\dfrac{2}{4}-\dfrac{9}{7}\right)\)

\(=\dfrac{2}{7}-\dfrac{9}{4}+\dfrac{3}{7}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{9}{7}\)

\(=\left(\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\right)-\left(\dfrac{9}{4}+\dfrac{5}{4}+\dfrac{2}{4}\right)\)

\(=2-4\)

\(=-2\)

-2

\(\dfrac{4}{7}+\dfrac{3}{7}x=\dfrac{1}{2}\\ \dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{4}{7}\\ \dfrac{3}{7}x=-\dfrac{1}{14}\\ x=-\dfrac{1}{14}:\dfrac{3}{7}\\ x=-\dfrac{1}{6}\)

Vậy....

\(\dfrac{17}{6}-\left(x+\dfrac{7}{6}\right)=\dfrac{7}{4}\\x+\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}\\ x+\dfrac{7}{6}=\dfrac{13}{12}\\ x=\dfrac{13}{12}-\dfrac{7}{6}\\ x=-\dfrac{1}{12} \)

Vậy....

\(\dfrac{4}{7}+\dfrac{3}{7}x=\dfrac{1}{2}\\ \Rightarrow\dfrac{3}{7}x=\dfrac{1}{2}-\dfrac{4}{7}\\ \Rightarrow\dfrac{3}{7}x=-\dfrac{1}{14}\\ \Rightarrow x=-\dfrac{1}{14}:\dfrac{3}{7}\\ \Rightarrow x=-\dfrac{1}{6}\)

___________

\(\dfrac{17}{6}-\left(x+\dfrac{7}{6}\right)=\dfrac{7}{4}\\ \Rightarrow\dfrac{17}{6}-x-\dfrac{7}{6}=\dfrac{7}{4}\\ \Rightarrow\dfrac{5}{3}-x=\dfrac{7}{4}\\ \Rightarrow x=\dfrac{5}{3}-\dfrac{7}{4}\\ \Rightarrow x=-\dfrac{1}{12}\)

\(\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\left(\dfrac{5}{3}-\dfrac{3}{2}\right)+\left(\dfrac{7}{3}-\dfrac{5}{2}\right)\\ =\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{3}{2}+\dfrac{7}{3}-\dfrac{5}{2}\\ =\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)+\left(-\dfrac{1}{3}-\dfrac{5}{3}+\dfrac{7}{3}\right)\\ =-\dfrac{1}{2}+\dfrac{1}{3}\\ =-\dfrac{1}{6}\)

a)

\(\dfrac{-7}{20}\) = \(\dfrac{4}{5}\) + \(\left(\dfrac{-23}{20}\right)\) 

b)

\(\dfrac{-7}{20}\) = \(\dfrac{-1}{5}\) + \(\left(\dfrac{-3}{20}\right)\)

Do x và y là hai đại lượng tỉ lệ nghịch nên:

\(a=xy=2\cdot\left(-15\right)=-30\)