Ta có: 

  \(A=\frac{112}{13.20}+\frac{112}{20.27}+.........+\frac{112}{62.69}\)

\(\Rightarrow A=112.\left(\frac{1}{13.20}+\frac{1}{20.27}+..........+\frac{1}{62.69}\right)\)

\(\Rightarrow A=16.\left(\frac{7}{13.20}+\frac{7}{20.27}+.......+\frac{7}{62.69}\right)\)

\(\Rightarrow A=16.\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+........+\frac{1}{62}-\frac{1}{69}\right)\)

\(\Rightarrow A=16.\left(\frac{1}{13}-\frac{1}{69}\right)\)

\(\Rightarrow A=16.\frac{56}{897}\)

\(\Rightarrow A=\frac{896}{897}\)

Vậy: \(A=\frac{896}{897}\)

\(A=\frac{112}{13.20}+\frac{112}{20.27}+\frac{112}{27.34}+...+\frac{112}{62.69}\)

\(\Rightarrow A=112.\left(\frac{1}{13.20}+\frac{1}{20.27}+\frac{1}{27.34}+...+\frac{1}{62.69}\right)\)

\(\Rightarrow A=112.\frac{7}{7}.\left(\frac{1}{13.20}+\frac{1}{20.27}+\frac{1}{27.34}+...+\frac{1}{62.69}\right)\)

\(\Rightarrow A=112.\frac{1}{7}\left(\frac{7}{13.20}+\frac{7}{20.27}+\frac{7}{27.34}+...+\frac{7}{62.69}\right)\)

\(\Rightarrow A=16\left(\frac{1}{13}-\frac{1}{20}+\frac{1}{20}-\frac{1}{27}+\frac{1}{27}-\frac{1}{34}+...+\frac{1}{62}-\frac{1}{69}\right)\)

\(\Rightarrow A=16\left(\frac{1}{13}-\frac{1}{69}\right)\)

\(\Rightarrow A=16.\frac{56}{897}\)

\(\Rightarrow A=\frac{896}{897}\)

\(Can\)\(you\) \(k\) \(for\) \(me,everyone?\)

5x².3xy²=15x³y²

hệ số là 15

15 nha

Giả sử : \(\frac{a}{b}=\frac{c}{d}\) thì ad = bc 

Suy ra : ad < bc thì \(\frac{a}{b}< \frac{c}{d}\) (đpcm)

a) 

Có \(\frac{a}{b}< \frac{c}{d}\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Rightarrow ad< bc\) (vì bd > 0)

Vậy \(\frac{a}{b}< \frac{c}{d}\Rightarrow ad< bc\) (với b, d > 0)

b) 

Có ad < bc và bd > 0

\(\Rightarrow\frac{ad}{bd}< \frac{bc}{bd}\Rightarrow\frac{a}{b}< \frac{c}{d}\)

Vậy \(ad< bc\Rightarrow\frac{a}{b}< \frac{c}{d}\) (với b, d > 0)

4,7 nha bạn

=4,7

Kb vs Mình nha

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

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

\(\left(x-1\right)^{x+2}=\left(x+1\right)^{x+2}.\left(x+1\right)^2\)

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

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

         \(\Rightarrow\hept{\begin{cases}x+1=1\\x+1=-1\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=-2\end{cases}}\)

cảm ơn anh nhiều ạ