a) \( - 2{x^2} + 6{x^2} = ( - 2 + 6).{x^2} = 4{x^2}\);                  

b) \(4{x^3} - 8{x^3} = (4 - 8).{x^3} =  - 4{x^3}\);

c) \(3{x^4}( - 6{x^2}) = 3.( - 6).{x^4}.{x^2} =  - 18{x^{4 + 2}} =  - 18{x^6}\);                 

d) \(( - 24{x^6}):( - 4{x^3}) = ( - 24: - 4).({x^6}:{x^3}) = 6{x^{6 - 3}} = 6{x^3}\). 

a: \(\dfrac{1}{8}\cdot\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2\cdot3}=\dfrac{1}{6}\)

b: \(=\dfrac{8-3}{12}\cdot\dfrac{6}{5}=\dfrac{6}{12}=\dfrac{1}{2}\)

c: \(=\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{3}{4}\)

d: =63/21+15/21-7/21=71/21

a \(\dfrac{1}{8}\times\dfrac{4}{5}\times\dfrac{10}{6}=\dfrac{1}{6}\)

b \(\left(\dfrac{8}{12}-\dfrac{3}{12}\right)\times\dfrac{6}{5}=\dfrac{5}{12}\times\dfrac{6}{5}=\dfrac{1}{2}\)

c \(\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{24}{35}\times\dfrac{35}{32}=\dfrac{3}{4}\)

d \(\dfrac{63}{21}+\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{71}{21}\)

a,`9/5+2/5xx4/6=9/5+4/15=27/15+4/15=31/15`

b,`3/8xx2-6/7xx1/3=3/4-2/7=21/18-8/28=13/28`

a) 9/5+ 4/15

= 27/15+4/15

= 31/15

b) 3/4- 2/7

= 13/28

\(a,=\dfrac{9}{5}+\dfrac{4}{15}=\dfrac{27}{15}+\dfrac{4}{15}=\dfrac{31}{15}\\ b,=\dfrac{3}{4}-\dfrac{2}{7}=\dfrac{13}{28}\)

1/4

17/2

1/4

17/2

em hong có bít∞❤

a: \(=\dfrac{18+6}{27+6}=\dfrac{24}{33}=\dfrac{8}{11}\)

b: \(=\dfrac{21\cdot7\cdot11}{3\cdot11\cdot4}=\dfrac{147}{12}=\dfrac{49}{4}\)

a) \(({x^2} + 2x + 3) + (3{x^2} - 5x + 1) = ({x^2} + 3{x^2}) + (2x - 5x) + (3 + 1) = 4{x^2} - 3x + 4\);        

b) \(\begin{array}{l}(4{x^3} - 2{x^2} - 6) - ({x^3} - 7{x^2} + x - 5) = 4{x^3} - 2{x^2} - 6 - {x^3} + 7{x^2} - x + 5\\ = (4{x^3} - {x^3}) + ( - 2{x^2} + 7{x^2}) - x + ( - 6 + 5) = 3{x^3} + 5{x^2} - x - 1\end{array}\);

c) \(\begin{array}{l} - 3{x^2}(6{x^2} - 8x + 1) =  - 3{x^2}.6{x^2} -  - 3{x^2}.8x +  - 3{x^2}.1\\ =  - 18{x^{2 + 2}} + 24{x^{2 + 1}} - 3{x^2} =  - 18{x^4} + 24{x^3} - 3{x^2}\end{array}\);               

d) \(\begin{array}{l}(4{x^2} + 2x + 1)(2x - 1) = (4{x^2} + 2x + 1).2x - (4{x^2} + 2x + 1).1 = 4{x^2}.2x + 2x.2x + 1.2x - 4{x^2} - 2x - 1\\ = 8{x^{2 + 1}} + 4{x^{1 + 1}} + 2x - 4{x^2} - 2x - 1 = 8{x^3} + 4{x^2} + 2x - 4{x^2} - 2x - 1 = 8{x^3} - 1\end{array}\);

e) \(\begin{array}{l}({x^6} - 2{x^4} + {x^2}):( - 2{x^2}) = {x^6}:( - 2{x^2}) - 2{x^4}:( - 2{x^2}) + {x^2}:( - 2{x^2})\\ =  - \dfrac{1}{2}{x^{6 - 2}} + {x^{4 - 2}} - \dfrac{1}{2}{x^{2 - 2}} =  - \dfrac{1}{2}{x^4} + {x^2} - \dfrac{1}{2}.\end{array}\);  

g) 

 \(({x^5} - {x^4} - 2{x^3}):({x^2} + x)=x^3-2x^2\)

a) 4 + 5/7 = .... 28/7 + 5/7=.33/7..............

    5/9 x 6/7 = ..10/21....

    3 - 7/5 =  15/5 - 7/5...8/5........

    3/5 x 4/8 = ..3/10....

b) 2 - 1/4 = .8/4 - 1/4...7/4...

    4 : 5/9 = ..4 x 9/5 = 36/5....

    2/9 x 3/5 = .....2/15.....

    3/8 : 4 = ......3/8 x 1/4 = 3/2.....

a) \(4+\dfrac{5}{7}=\dfrac{4}{1}+\dfrac{5}{7}=\dfrac{28}{7}+\dfrac{5}{7}=\dfrac{33}{7}\)

   \(\dfrac{5}{9}\times\dfrac{6}{7}=\dfrac{10}{21}\)

    \(3-\dfrac{7}{5}=\dfrac{3}{1}-\dfrac{7}{5}=\dfrac{15}{5}-\dfrac{7}{5}=\dfrac{8}{5}\)

    \(\dfrac{3}{5}\times\dfrac{4}{8}=\dfrac{12}{40}=\dfrac{3}{10}\)

b) \(2-\dfrac{1}{4}=\dfrac{2}{1}-\dfrac{1}{4}=\dfrac{8}{4}-\dfrac{1}{4}=\dfrac{7}{4}\)

    \(4:\dfrac{5}{9}=\dfrac{4}{1}:\dfrac{5}{9}=\dfrac{4}{1}\times\dfrac{9}{5}=\dfrac{36}{5}\)

    \(\dfrac{2}{9}\times\dfrac{3}{5}=\dfrac{6}{45}=\dfrac{2}{15}\)

     \(\dfrac{3}{8}:4=\dfrac{3}{8}:\dfrac{4}{1}=\dfrac{3}{8}\times\dfrac{1}{4}=\dfrac{3}{2}\)

3/4 x 3/2 = 9/8

3/2 - 1/3 = 7/6

3/4x3/2=9/8

3/2-1/3=7/6

a, = 3/4 x 3/2 = 9/8

b, 3/2 - 1/3 = 7/6