5 x 6 = 30       2 x 6 = 12        3 x 6 = 18        4 x 6 = 24

6 x 5 = 30       6 x 2 = 12        6 x 3 = 18        6 x 4 = 24

Tính nhẩm:

5 x 6 = 30       2 x 6 = 12        3 x 6 = 18        4 x 6 = 24

6 x 5 = 30       6 x 2 = 12        6 x 3 = 18        6 x 4 = 24

30 : 6 = 5       12 : 6 = 2         18 : 6 = 3         24 : 6 = 4

30 : 5 = 6       12 : 2 = 6         18 : 3 = 6         24 : 4 = 6

Học tốt <3

5 x 6 = 30        2 x 6 = 12       3 x 6 = 18      4 x 6 = 24

6 x 6 = 30        6 x 2 = 12       6 x 3 = 18      6 x 4 = 24

30 : 6 = 5         12 : 6 = 2       18 : 6 = 3       24 : 6 = 4

30 : 5 = 6         12 : 2 = 6       18 : 3 = 6       24 : 4 = 6

\(\frac{8}{12}\cdot x\cdot4-\frac{18}{12}\cdot x\cdot6=\frac{5}{12}\)

\(x\left(\frac{8}{12}\cdot4-\frac{18}{12}\cdot6\right)=\frac{5}{12}\)

\(x\left(\frac{32}{12}-\frac{108}{12}\right)=\frac{5}{12}\)

\(x\left(-\frac{19}{3}\right)=\frac{5}{12}\)

\(x=\frac{5}{12}\div\left(-\frac{19}{3}\right)=-\frac{5}{76}\)

Vậy \(x=-\frac{5}{76}\)

Chúc bạn học tốt ^^!!!

`a)`

`6 \times 1 = 6`

`6 \times 4 = 24`

`6 \times 6 = 36`

`b)`

`12 \div 6 = 2`

`18 \div 6 = 3`

`48 \div 6 = 8`

`c)`

`6 \times 5 = 30`

`30 \div 6 = 5`

`30 \div 5 = 6`

 a,  \(\dfrac{5\times11\times15\times18}{6\times15\times22\times10}\)

= \(\dfrac{5\times11\times15\times6\times3}{6\times15\times11\times2\times2\times5}\)

= \(\dfrac{5\times11\times15\times6}{5\times11\times15\times6}\) \(\times\) \(\dfrac{3}{2\times2}\)

= 1 \(\times\) \(\dfrac{3}{4}\)

= \(\dfrac{3}{4}\)

b, \(\dfrac{7}{12}\) \(\times\) \(\dfrac{3}{5}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{5}{12}\)

= \(\dfrac{3}{5}\) \(\times\) ( \(\dfrac{7}{12}+\dfrac{5}{12}\))

= \(\dfrac{3}{5}\) \(\times\) \(\dfrac{12}{12}\)

= \(\dfrac{3}{5}\) \(\times\) 1 

= \(\dfrac{3}{5}\)

1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)

\(\Leftrightarrow5x+20+12x-28=7x+2\)

\(\Leftrightarrow17x-7x=2+8=10\)

hay x=1

2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-6x+3x=3-4\)

hay \(x=\dfrac{1}{3}\)

3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)

\(\Leftrightarrow4x-12-x-2=6x-3\)

\(\Leftrightarrow3x-14-6x+3=0\)

\(\Leftrightarrow-3x=11\)

hay \(x=-\dfrac{11}{3}\)

4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)

\(\Leftrightarrow3x-6-8x-12=x+6\)

\(\Leftrightarrow-5x-x=6+18\)

hay x=-4

5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)

\(\Leftrightarrow6x-3+2x-6=-1\)

\(\Leftrightarrow8x=8\)

hay x=1

\(\frac{3}{3.6.9}+\frac{3}{6.9.12}+\frac{3}{9.12.15}+\frac{3}{12.15.18}=\frac{3}{6}\left(\frac{6}{3.6.9}+\frac{6}{6.9.12}+\frac{6}{9.12.15}+\frac{6}{12.15.18}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{6.9}+\frac{1}{6.9}-\frac{1}{9.12}+\frac{1}{9.12}-\frac{1}{12.15}+\frac{1}{12.15}-\frac{1}{15.18}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{15.18}\right)=\frac{1}{2}.\frac{14}{270}=\frac{7}{270}\)

1) 3(x - 4) - (8 - x) = 12

3x - 12 - 8 + x = 12

4x - 20 = 12

4x = 12 + 20

4x = 32

x = 8

2) 5(x - 6) - 2(x + 3) = 12

5x - 30 - 2x - 6 = 12

3x - 36 = 12

3x = 12 + 36

3x = 48

x = 16

3) 5 - | x + 7| = 4

|x + 7| = 4 - 5

|x + 7| = -1 x + 7 = 1; -1 x + 7 = 1 hoặc x + 7 = -1 x = 1 - 7 x = -1 - 7 x = -6 x = -8 => x = -6 hoặc x = -8 4) 6|x - 7| = 18 : (-3) 6|x - 7| = -6 6|x - 7| = -6 : 6 |x - 7| = -1 x - 7 = 1; -1 x - 7 = 1 hoặc x - 7 = -1 x = 1 + 7 x = -1 + 7 x = 8 x = 6 => x = 8 hoặc x = 6

1, \(3.\left(x-4\right)-\left(8-x\right)=12\)

\(\Leftrightarrow3x-12-8+x=12\)

\(\Leftrightarrow3x+x=12+12+8\)

\(\Leftrightarrow4x=32\)

\(\Leftrightarrow x=8\)

​​Vậy : \(x=8\)

2, \(5.\left(x-6\right)-2.\left(x+3\right)=12\)

\(\Leftrightarrow5x-30-2x-6=12\)

\(\Leftrightarrow5x-2x=12+30+6\)

\(\Leftrightarrow3x=48\)

\(\Leftrightarrow x=\frac{48}{3}=16\)

Vậy : \(x=16\)

3, \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=5-4=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=1\\x+7=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-8\end{matrix}\right.\)

Vậy : \(x\in\left\{6,-8\right\}\)

4, \(6.\left|x-7\right|=18:\left(-3\right)\)

\(\Leftrightarrow6.\left|x-7\right|=-6\)

\(\Leftrightarrow\left|x-7\right|=-1\)

Ta thấy : \(\left|x-7\right|\ge0\forall x\) mà \(-1< 0\)

\(\Rightarrow x\in\varnothing\)

Vậy : không có \(x\) thỏa mãn đề.