5/4-yx5/6=-1/12

5/4-y      = -1/12:5/6

5/4-y      = -1/10

y            = 5/4-(-1/10)

y            = 27/20

\(\frac{5}{4}-y.\frac{5}{6}=\frac{1}{4}-\frac{1}{3}\)

\(\Rightarrow\frac{5}{4}-y.\frac{5}{6}=-\frac{1}{12}\)

\(\Rightarrow y.\frac{5}{6}=\frac{5}{4}-\left(-\frac{1}{12}\right)\)

\(\Rightarrow y.\frac{5}{6}=\frac{5}{4}+\frac{1}{12}\)

\(\Rightarrow y.\frac{5}{6}=\frac{4}{3}\)

\(\Rightarrow y=\frac{4}{3}:\frac{5}{6}\)

\(\Rightarrow y=\frac{8}{5}\)

Vậy  \(y=\frac{8}{5}.\)

mk nhầm nha bn

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2019}-\frac{1}{2020}\)

\(=1-\frac{1}{2020}>1\)

Thank you bạn dcv new ^ ^

nhầm dấu rồi bé hơn 1 chứ:v bạn sửa lại hộ mình nhà

M = \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\)

=> 5M = 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)

=> 5M - M = ( 1 + \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{49}\)) - ( \(\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{^{^{ }}50}\))

4M = 1 - \(\left(\frac{1}{5}\right)^{50}\)

=> M = \(\frac{1-\left(\frac{1}{5}\right)^{50}}{4}\)< \(\frac{1}{4}\)

a) Vì 4 < 5 nên \(\frac{3}{4}>\frac{3}{5}\)

b) Vì 4 < 5 nên \(\frac{4}{7}< \frac{5}{7}\)

c) Ta có :

\(\frac{2}{7}=\frac{2\times5}{7\times5}=\frac{10}{35}\)

\(\frac{4}{5}=\frac{4\times7}{5\times7}=\frac{28}{35}\)

Vì 10 < 28 nên \(\frac{10}{35}< \frac{28}{35}\)hay \(\frac{2}{7}< \frac{4}{5}\)

d) Ta có :

\(\frac{8}{9}=\frac{8\times10}{9\times10}=\frac{80}{90}\)

\(\frac{9}{10}=\frac{9\times9}{10\times9}=\frac{81}{90}\)

Vì 80 < 81 nên \(\frac{80}{90}< \frac{81}{90}\)hay \(\frac{8}{9}< \frac{9}{10}\)

e) Ta có :

\(\frac{10}{9}=\frac{10\times2}{9\times2}=\frac{20}{18}\)

Vì 19 < 20 nên \(\frac{19}{18}< \frac{20}{18}\)hay \(\frac{19}{18}< \frac{10}{9}\)

f) Ta có :

\(\frac{8}{3}=\frac{8\times4}{3\times4}=\frac{32}{12}\)

\(\frac{5}{6}=\frac{5\times2}{6\times2}=\frac{10}{12}\)

Vì 32 > 10 nên \(\frac{32}{12}>\frac{10}{12}\)hay \(\frac{8}{3}>\frac{5}{6}\)

bn nào dài được đầu tiên mà đúng mik sẽ k cho bn đó nhé

\(\frac{3}{4}>\frac{3}{5}\)

\(\frac{4}{7}< \frac{5}{7}\)

\(\frac{2}{7}< \frac{4}{5}\)

\(\frac{8}{9}< \frac{9}{10}\)

\(\frac{19}{18}< \frac{10}{9}\)

\(\frac{8}{3}>\frac{5}{6}\)

a) 2n^3 + 2n^2 - 2n^3 - 2n^2 + 6n = 6n chia hết 6

b) 3n - 2n^2 - ( n + 4n^2 - 1 - 4n ) - 1 

= 3n - 2n^2 - n - 4n^2 + 1 + 4n -1

= 6n - 6n^2 chia hết 6

c) m^3 + 8 - m^3 + m^2 - 9 - m^2 - 18

= - 19

Bài 1:

\(2n^2\left(n+1\right)-2n\left(n^2+n-3\right)\)

\(=2n\left(n^2+n-n^2-n+3\right)\)

\(=6n\)\(⋮\)\(6\)
Bài 2:

\(n\left(3-2n\right)-\left(n-1\right)\left(1+4n\right)-1\)

\(=3n-2n^2-\left(n+4n^2-1-4n\right)-1\)

\(=6n-6n^2=6\left(n-n^2\right)\)\(⋮\)\(6\)

Bài 3:

\(\left(m^2-2m+4\right)\left(m+2\right)-m^3+\left(m+3\right)\left(m-3\right)-m^2-18\)

\(=m^3+8-m^3+m^2-9-m^2-18\)

\(=-19\)

\(\Rightarrow\)đpcm

a,  <=> 2n[ n(n+1)-n²-n+3)

<=> 2n( n²+n-n²-n+3)

<=> 6n chia hết cho 6 với mọi n nguyên

b, <=> 3n-2n²-(n+4n²-1-4n) -1

<=> 3n-2n²-n-4n²+1+4n-n-1

<=> 6n-6n²

<=> 6(n-n²)  chiiaia hhehethet cchchocho 6

c ,<=> m³-2³-m³+m²-3²-m²-18

<=>-35 => ko phụ thuộc vào biến

\(\dfrac{3}{5}\times\dfrac{4}{7}\times1\dfrac{1}{2}\)

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

\(=\dfrac{3\times4\times3}{5\times7\times2}\)

\(=\dfrac{3\times2\times3}{5\times7}\)

\(=\dfrac{18}{35}\)