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

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

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) 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) \(3\left(2x-3\right)+5\left(x+2\right)=6x-9+5x+10=11x+1\)

b) \(3x\left(2x-8\right)+\left(6x+2\right)\left(5-x\right)=6x^2-24x+30x-6x^2+10-2x=4x+10\)

c) \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2=x^2-9-x^2+10x-25=10x-34\)

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

Tk :

a: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

b: Ta có: \(3x\left(2x-8\right)-\left(6x-2\right)\left(x-5\right)\)

\(=6x^2-24x-6x^2+30x+2x-10\)

\(=8x-10\)

c: Ta có: \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2\)

\(=x^2-9-x^2+10x-25\)

\(=10x-34\)

a)

Vậy \(({x^3} + 1):({x^2} - x + 1) = x + 1\).

b)

Vậy \((8{x^3} - 6{x^2} + 5) = ({x^2} - x + 1)(8x + 2) + ( - 6x + 3)\)

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

Lời giải:

a.

$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$

$=4x^2-12x+9+4x^2-4x-15$

$=24-8x$
b.

$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$

c.

$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$

$=6x^2-24x-6x^2-32x+10$

$=8x-10$

d.

$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$

$=x^2-9-x^2+10x-25=10x-34$

e.

$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$

$=-3x^2y+3xy^2=3xy(y-x)$

a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)

\(=4x^2-12x+9+2x-4x^2+15-6x\)

\(=-16x+24\)

b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)

\(=6x^2-24x+30x-6x^2-10+2x\)

\(=8x-10\)

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

b) \(\begin{array}{l}(5{x^3} - 4x):( - 2x) = 5{x^3}:( - 2x) - 4x:( - 2x) = (5: - 2).({x^3}:x) - (4: - 2).(x:x)\\ =  - \dfrac{5}{2}{x^{3 - 1}} - ( - 2) =  - \dfrac{5}{2}{x^2} + 2\end{array}\)

c) \(\begin{array}{l}( - 15{x^6} - 24{x^3}):( - 3{x^2}) = ( - 15{x^6}):( - 3{x^2}) + ( - 24{x^3}):( - 3{x^2})\\ = ( - 15: - 3).({x^6}:{x^2}) + ( - 24: - 3).({x^3}:{x^2})\\ = 5.{x^{6 - 2}} + 8.{x^{3 - 2}} = 5{x^4} + 8x\end{array}\)

a) 5/9+ 6/9= 11/9

b) 5/8

c) 6/5 x 3/2 = 9/5

d) 49/14- 29/14= 20/14 = 10/7

= 11/9

= 5/8

= 9/5

= 10/7

a, 5/9 + 2/3 = 11/9

b, 7/8 x 5/7 = 5/8

c, 6/5 : 2/3 = 9/5

d, 7/2 - 29/14 =10/7