\(\dfrac{4}{7}:\dfrac{4}{3}+5:\dfrac{4}{3}=\left(\dfrac{4}{7}+5\right):\dfrac{4}{3}=\dfrac{39}{7}:\dfrac{4}{3}=\dfrac{39}{7}\times\dfrac{3}{4}=\dfrac{117}{28}\)

\(\dfrac{4}{7}:\dfrac{4}{3}+5:\dfrac{4}{3}=\dfrac{3}{7}+\dfrac{15}{4}=\dfrac{12}{28}+\dfrac{105}{28}=\dfrac{117}{28}\)

\(\dfrac{3}{4}\times\dfrac{2}{5}-\dfrac{3}{4}\times\dfrac{2}{7}=\dfrac{3}{4}\times\left(\dfrac{2}{5}-\dfrac{2}{7}\right)=\dfrac{3}{4}\times\dfrac{4}{35}=\dfrac{12}{140}=\dfrac{3}{35}\)

\(\dfrac{3}{4}\times\dfrac{2}{5}-\dfrac{3}{4}\times\dfrac{2}{7}=\dfrac{3}{10}-\dfrac{3}{14}=\dfrac{42}{140}-\dfrac{30}{140}=\dfrac{12}{140}=\dfrac{3}{35}\)

\(\left(\dfrac{2}{3}-\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{2}{21}\times\dfrac{2}{5}=\dfrac{4}{105}\)

\(\left(\dfrac{2}{3}-\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{2}{3}\times\dfrac{2}{5}-\dfrac{4}{7}\times\dfrac{2}{5}=\dfrac{4}{15}-\dfrac{8}{35}=\dfrac{140}{525}-\dfrac{120}{525}=\dfrac{20}{525}=\dfrac{4}{105}\)

#YVA

4/7:4/3+5:4/3

= 3/7 + 15/4

=4 5/28

4/7:4/3+5:4/3

= (4/7+5) x 3/4

=4 5/28

3/4x2/5-3/4x2/7

=3/10 - 3/14

=3/35

3/4x2/5-3/4x2/7

=3/4x(2/5-2/7)

=3/35

(2/3-4/7)x2/5

=2/21x2/5

=4/105

(2/3-4/7)x2/5

=2/3x2/5-4/7x2/5

=4/105

kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

a) C1 : 2/5 x 3/7 + 2/7 x 4/7

         = 2/5 x (3/7 + 4/7)

        = 2/5 x 1

        = 2/5

C2 : 2/5 x 3/7 + 2/5 x 4/7

= 6/35 + 8/35

= 2/5 

b) (2/3 - 4/7) x 5/5

= 2/21 x 5/5

= 2/21

C2 : (2/3 - 4/7) x 5/5

= 2/3 x 1 - 4/7 x 1

= 2/3 - 4/7

= 2/21

c) 3/4 x 2/5 - 3/4 x 2/7

= 3/4 x (2/5 - 2/7)

= 3/4 x 4/35

= 3/35

C2 : 3/4 x 2/5 - 3/4 x 2/7

= 3/10 - 3/14

= 3/35

a, 

Cách 1

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{6}{35}+\frac{8}{35}\)

\(=\frac{2}{5}\)

Cách 2 :

\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)

\(=\frac{2}{5}\times\left(\frac{3}{7}+\frac{4}{7}\right)\)

\(=\frac{2}{5}\times1\)

\(=\frac{2}{5}\)

b,

Cách 1 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{21}\times1\)

\(=\frac{2}{21}\)

Cách 2 :

\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)

\(=\frac{2}{3}\times\frac{5}{5}-\frac{4}{7}\times\frac{5}{5}\)

\(=\frac{2}{3}-\frac{4}{7}\)

\(=\frac{2}{21}\)

c, 

Cách 1 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{10}-\frac{3}{14}\)

\(=\frac{3}{35}\)

Cách 2 :

\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)

\(=\frac{3}{4}\times\left(\frac{2}{5}-\frac{2}{7}\right)\)

\(=\frac{3}{4}\times\frac{4}{35}\)

\(=\frac{3}{35}\)

Mk nhanh nhất đó

Đúng 100%

Tk mk mk tk lại

Cảm ơn bạn nhiều

Thank you very  much

( ^ _ ^ )

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)​

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)

1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)

hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)

2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)

hay \(x\in\left\{1;5\right\}\)

3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)

hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)

4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)

hay \(x\in\left\{-4;3;-3\right\}\)

5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)

6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)

\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)

\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)

hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)

1.

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)

\(\Leftrightarrow x+3=5x-2\)

\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)

2.

\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)

\(\Leftrightarrow x^2+x+1=x^2-2x+16\)

\(\Leftrightarrow3x=15\Leftrightarrow x=5\)

3.

\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)

7.

\(\Leftrightarrow x^2+2x-15=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)

\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)

9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)

\(\Leftrightarrow x+2=3-2x\)

\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)

a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x-5\right)=-4\)

\(\Leftrightarrow x^2+5x+6-x^2+7x-10=-4\)

\(\Leftrightarrow12x=0\)

hay x=0

b: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)

\(\Leftrightarrow x^3+1-x^3+9x=8\)

\(\Leftrightarrow9x=7\)

hay \(x=\dfrac{7}{9}\)

c: Ta có: \(4x^2-9=\left(3x+1\right)\left(2x-3\right)\)

\(\Leftrightarrow\left(3x+1\right)\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x+1-2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)

Bài 3 là hỗn số hả em?