8 x 1 = 8        8 x 2 = 16       8 x 3 = 24       8 x 4 = 32

1 x 8 = 8        2 x 8 = 16       3 x 8 = 24       4 x 8 = 32

8 x 5 = 40       8 x 6 = 48       8 x 7 = 56       8 x 8 = 64

5 x 8 = 40       6 x 8 = 48       7 x 8 = 56       8 x 9 = 72

\(P\left(x\right)=\sqrt[3]{\sqrt{x+8}.\left[x^3\left(x+8\right)+12x\right]+6x^2\left(x+8\right)+8}\)

Đặt:  \(\sqrt{x+8}=a>0\) =>  \(x+8=a^2\)

Khi đó ta có:

\(P\left(x\right)=\sqrt[3]{a\left(x^3a^2+12x\right)+6x^2a^2+8}\)

\(=\sqrt[3]{x^3a^3+12xa+6x^2a^2+2}\)

\(=\sqrt[3]{\left(ax+2\right)^3}\)

\(=ax+2\)

\(=x\sqrt{x+8}+2\)

a) 7 × 3 = 7 + 7 + 7 = 21

    7 × 3 = 21

b) 8 × 4 = 8 + 8 + 8 + 8 = 32

    8 × 4 = 32

c) 6 × 5 = 6 + 6 + 6 + 6 + 6 =30

    6 × 5 = 30.

mình đang cần gấp câu trả lời ,bạn nào giải được nhanh k luôn ,hứa đấy

bạn đi học thêm à ?

1) -5+(-3+12-x)=7-(2-8)

=>-5+(9-x)=7-(-6)

=>-5+9-x=13

=>x=9-5-13=-9

2) 8-(7-x+3)=-|-7+2|-(-9)

=>8-(10-x)=-|-5|+9

=>8-10+x=-5+9

=>x=-5+9-8+10=6

3)-(12-5+9-x)=3-|5-8+2|

=>-(16-x)=3-|-1|

=>-16+x=3-1

=>x=3-1+16=18

4)-(-3-x+8)-(-3+5)=|5-8|

=>-(5-x)-2=|-3|

=>-5+x-2=3

=>x=3+5+2=10

5)-|5-8|-(4-x+5)=4+|3-9|

=>-|-3|-(9-x)=4+|-6|

=>-3-9+x=4+6

=>x=4+6+3+9=22

\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)

\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)

\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)

Vậy x = 5/19

\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

Vậy x = 1/2 hoặc x = -6

\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Vậy x = 7 hoặc x = -1

b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)

\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)

\(\Leftrightarrow x+3=11\)

hay x=8

c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)

\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)

\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)

Suy ra: x=5

d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)

\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)

\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)

\(\Leftrightarrow8^x=8^{20}\)

Suy ra: x=20

Học sinh nhẩm và ghi:

8 x 5 = 40

8 x 4 = 32

8 x 6 = 48

8 x 3 = 24

40 : 8 = 5

32 : 8 = 4

48 : 8 = 6

24 : 8 = 3

40 : 5 = 8

32 : 4 = 8

48 : 6 = 8

24 : 3 = 8

(chú ý: mối quan hệ giữa nhân và chia)

Tính nhẩm:

8 x 5 = 40

8 x 4 = 32

8 x 6 = 48

8 x 3 = 24

40 : 8 = 5

32 : 8 = 4

48 : 8 = 6

24 : 8 = 3

40 : 5 = 8

32 : 4 = 8

48 : 6 = 8

24 : 3 = 8

k cho mik với

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)

8 x 0 =0

8 x 8 =64

8 x 2 =16

8 x 4 =32

8 x 6 =48

8 x 9 =72

8 x 3 =24

8 x 7 =56

8 x 10 =80

8 x 0 =0

8 x 8 =64

8 x 2 =16

8 x 4 =32

8 x 6 =48

8 x 9 =72

8 x 3 =24

8 x 7 =56

8 x 10 =80

8 x 0 =0

8 x 8 =64

8 x 2 =16

8 x 4 =32

8 x 6 =48

8 x 9 =72

8 x 3 =24

8 x 7 =56

8 x 10 =80

a, \(\left(\frac{4}{9}.\frac{3}{7}\right).\frac{7}{4}=\frac{4}{9}\left(\frac{3}{7}.\frac{7}{4}\right)=\frac{4}{9}.\frac{3}{4}=\frac{3}{9}=\frac{1}{3}\)

b, \(\left(\frac{6}{5}.\frac{4}{5}\right).\frac{25}{16}=\frac{6}{5}\left(\frac{4}{5}.\frac{25}{16}\right)=\frac{6}{5}.\frac{5}{4}=\frac{6}{4}=\frac{3}{2}\)

c, \(\left(\frac{7}{8}.\frac{16}{9}\right).\frac{3}{14}=\frac{7}{8}\left(\frac{16}{9}.\frac{3}{14}\right)=\frac{7}{8}.\frac{8}{21}=\frac{7}{21}=\frac{1}{3}\)

a, $\left(\frac{4}{9}.\frac{3}{7}\right).\frac{7}{4}=\frac{4}{9}\left(\frac{3}{7}.\frac{7}{4}\right)=\frac{4}{9}.\frac{3}{4}=\frac{3}{9}=\frac{1}{3}$

b, $(\frac{}{}$