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

8 x 1 = 8       8 x 3 = 24       8 x 5 = 40       8 x 7 = 56       8 x 9 = 72

8 x 2 = 16       8 x 4 = 32       8 x 6 = 48       8 x 8 = 64       8 x 10 = 80

8 x 1 = 8

8 x 5 = 40

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 1 = 8

8 x 5 = 40

8 x 0 = 0

8

64

16

32

48

72

24

56 80

Học sinh tự nhẩm và ghi như sau:

8 x 3 = 24

8 x 2 = 16

8 x 4 = 32

8 x 1 = 8

8 x 5 = 40

8 x 6 = 48

8 x 7 = 56

0 x 8 = 0

8 x 8 = 64

8 x 10 = 80

8 x 9 = 72

8 x 0 = 0

1.24

2.16

3.32

4.8

5.40

6.48

7.56

8.0

9.64

10.80

11.72

12.0

1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)

\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)

=0

2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)

\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)

\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)

3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)

\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)

\(=\dfrac{52546}{15}\)

4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)

\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)

\(=\dfrac{4}{7}\)

1) Tính biểu thức:

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

=1/8x(3/2-1/4)+3/8x5/4

=1/8x5/4+3/8x5/4

=5/4x(1/8+3/8)

=5/8

\(C=\dfrac{1}{8}\times\dfrac{3}{2}-\dfrac{1}{8}\times\dfrac{1}{4}+\dfrac{3}{8}\times\dfrac{5}{8}\)

\(C=\dfrac{1}{8}\times\left(\dfrac{3}{2}-\dfrac{1}{4}\right)+\dfrac{3}{8}\times\dfrac{5}{4}\)

\(C=\dfrac{1}{8}\times\dfrac{5}{4}+\dfrac{3}{8}\times\dfrac{5}{4}\)

\(C=\left(\dfrac{1}{8}+\dfrac{3}{8}\right)\times\dfrac{5}{4}\)

\(C=\dfrac{1}{2}\times\dfrac{5}{4}\)

\(C=\dfrac{5}{8}\)

( 4/5 x 3/8 + 4/5 x 5/8 - 4/5 x 7/8 ) : 1/2

= 4/5 x ( 3/8 + 5/8 - 7/8 ) : 1/2

= 4/5 x 1/8 : 1/2

= 1/10 : 1/2

=1/5

\(ĐKXĐ:x\ne1;-1;2;-2\)

\(\frac{\left(x+4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x-4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{\left(x-8\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+8\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{8}{3}\)

\(\Leftrightarrow\frac{x^2+x+4x+4+x^2-x-4x+4}{x^2-1}=\frac{x^2-2x-8x+16+x^2+2x+8x+16}{x^2-4}-\frac{8}{3}\)

\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{2x^2+32}{x^2-4}-\frac{8}{3}\)

\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{3\left(2x^2+32\right)}{3\left(x^2-4\right)}-\frac{8\left(x^2-4\right)}{3\left(x^2-4\right)}\)

\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{9x^2+96-8x^2+32}{3\left(x^2-4\right)}\)

\(\Leftrightarrow\frac{2x^2+8}{x^2-1}=\frac{x^2+128}{3\left(x^2-4\right)}\)

\(\Leftrightarrow3\left(x^2-4\right)\left(2x^2+8\right)=\left(x^2+128\right)\left(x^2-1\right)\)

\(\Leftrightarrow9x^4+24x^2-24x^2-96=x^4-x^2+128x^2-128\)

\(\Leftrightarrow9x^4+24x^2-24x^2-x^4+x^2+128x^2=-128+96\)

\(\Leftrightarrow8x^4+129x^2=-32\)

\(\Leftrightarrow8x^4+129x^2+32=0\)

\(\Leftrightarrow x=\frac{1}{2}\left(tmđkxđ\right)\)

bạn sai r bạn ơi cái chỗ chuyển vế dòng tương đương số 8 : x^4 - x^2 + 128x^2 - 128 đáng ra sau khi chuyển px là -x^4 +x^2 - 128x^2 + 128 chứ sao lại là x^4 + x^2 + 128x^2 +128

Dài thế trời!!!

1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14

2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84

3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9

4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1

5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5

6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3