1

3/2

1

\(\frac{4x5x6}{12x15x18}\)

\(\frac{9x8x5}{6x4x15}=\frac{9x2}{6x3}=\frac{18}{18}=1\)

\(\frac{9\cdot8\cdot5}{6\cdot48\cdot15}\)

\(\Rightarrow\frac{3\cdot2\cdot1}{2\cdot1\cdot3}\)

\(\Leftrightarrow1\)

Bằng 1 nhé

Các bạn ủng hộ cho tớ đê !!!!!!!!!!!!!!!!!!!!1

9 x 8 x 5 = 9 x 5 x 8

= 40 x 8 

= 320

6 x 4 x 15 = 4 x 15 x 6

= 60 x 6

= 360

ghi cách làm đi

= \(\frac{3x3x4x4x5}{3x2x4x5x3}=\frac{4x5}{2}=\frac{20}{2}=10\)

\(\frac{9x8x5}{6x4x15}\)

\(\frac{3x3x2x4x5}{3x2x4x5x3}\)

1

\(\frac{9\cdot8\cdot5}{6\cdot4\cdot15}\)

\(=\frac{3\cdot2\cdot1}{2\cdot1\cdot3}\)

\(=1\)

nếu thấy đúng thì k cho tớ nha

a) 1/3

b) 1

a, \(\frac{2x3x4x5}{3x4x5x6}\)= \(\frac{2}{6}\)=\(\frac{1}{3}\)

b, \(\frac{9x8x5}{6x4x15}\)=\(\frac{3x3x4x2x5}{2x3x4x3x5}\)=1

a, 1 x 1x1x1 / 1x1x1x3 

= 1 / 3

b, 9x2x1 / 6x1x3

= 18 / 18 

= 1

\(9.8.5=\frac{360}{1}\)                         \(6.4.15=\frac{360}{1}\)

1

tick minh nha ai tick minh người ấy có tiền cả năm

khong phai dau ra phan so co ma

a)

\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)

b)

\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).

c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)

1/2-1/94 ban tu tinh nhe

d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)

1-1/94 ban tu tinh nhe 

tk nha

a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)

\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)

\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)

c,d dễ bn tách ra rồi trừ đi

\(A.\)\(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}+\frac{-1}{n+1}=\frac{n+1}{n\left(n+1\right)}+\frac{-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)

\(B.\)\(\frac{1}{n}-\frac{1}{n+3}=\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)

\(C.\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)

\(D.\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}=\frac{1}{1}-\frac{1}{94}=\frac{93}{94}\)