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) 

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

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 

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

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

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

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

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)

\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);                                                   

b)

\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);

c)

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

d)

\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).

a) \(\dfrac{13}{6}\)b) \(\dfrac{7}{18}\)c) \(5\) d) \(\dfrac{2}{3}\)

1.

a)x\(=\)\(\dfrac{7}{6}\)\(-\dfrac{2}{3}\)

  x\(=\dfrac{1}{2}\)

b)x\(=\dfrac{9}{2}:\dfrac{3}{4}\)

   x\(=6\)

LỘN LỚP GÙI CÁI ĐẤY KIẾN THỨC LỚP 4 MÀ CỤ

Bạn ơi, bạn viết lại đề đi. Khó nhìn quá

Bài 5: 

a. 1 - 2y + y²

= (1 - y)²

b. (x + 1)² - 25

= (x + 1)² - 5²

= (x + 1 - 5)(x + 1 + 5)

= (x - 4)(x + 6)

c. 1 - 4x²

= 1² - (2x)²

= (1 - 2x)(1 + 2x)

d. 8 - 27x³

= 2³ - (3x)³

= (2 - 3x)(4 + 6x + 9x²)

e. (đề hơi khó hiểu ''x3'' !?)

g. x³ + 8y³

= (x + 2y)(x² - 2xy + y²)

trong sách giáo khoa hả

hnay mik cg hc 

Bài 3: 

a: x+3/8=14/9x10/35

\(\Leftrightarrow x+\dfrac{3}{8}=\dfrac{4}{9}\)

=>x=4/9-3/8=32/72-27/72=5/72

b: \(x-\dfrac{2}{7}=\dfrac{21}{10}\cdot\dfrac{15}{28}=\dfrac{21}{28}\cdot\dfrac{15}{10}=\dfrac{3}{4}\cdot\dfrac{3}{2}=\dfrac{9}{8}\)

\(\Leftrightarrow x=\dfrac{9}{8}+\dfrac{2}{7}=\dfrac{63}{56}+\dfrac{16}{56}=\dfrac{79}{56}\)

c: \(\dfrac{11}{7}-x=\dfrac{8}{3}\cdot\dfrac{9}{32}=\dfrac{1}{4}\cdot3=\dfrac{3}{4}\)

nên \(x=\dfrac{11}{7}-\dfrac{3}{4}=\dfrac{44}{28}-\dfrac{21}{28}=\dfrac{23}{28}\)

d: \(x+\dfrac{2}{5}=\dfrac{12}{35}\cdot\dfrac{14}{9}=\dfrac{4}{3}\cdot\dfrac{2}{5}=\dfrac{8}{15}\)

nên \(x=\dfrac{8}{15}-\dfrac{2}{5}=\dfrac{8}{15}-\dfrac{10}{15}=-\dfrac{2}{15}\)

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

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