A=\(\frac{1}{1.2}+\frac{1}{2.3}+.............+\frac{1}{80.81}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.........+\frac{1}{80}-\frac{1}{81}\)

\(=1-\frac{1}{81}\)

\(=\frac{80}{81}\)

A = 1/2 + 1/6 + 1/12 + ... + 1/6480

A = 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/80.81 

A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/80 - 1/81

A = 1/1 - 1/81

A = 80/81

Vậy A = 80/81

\(\text{Ta có: }A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{80.81}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{80}-\frac{1}{81}\)

\(\Rightarrow A=1-\frac{1}{81}\)

\(\Rightarrow A=\frac{80}{81}\)

\(A=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{80\times81}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{80}-\frac{1}{81}\)

\(A=\frac{1}{1}-\frac{1}{81}\)

\(A=\frac{80}{81}\)

 Vậy A = \(\frac{80}{81}\)

\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{80.81}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{80}-\frac{1}{81}\)

\(\Rightarrow A=1-\frac{1}{81}\)

\(\Rightarrow A=\frac{80}{81}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{80.81}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)

\(=1-\frac{1}{81}\)

\(=\frac{80}{81}\)

A = 1/2 + 1/6 + 1/12 + .. + 1/6480

A= 1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/80 x 81

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/80 - 1/81

A = 1 - 1/81

A = 80/81

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(A=\frac{1}{1}\times2+\frac{1}{2}\times3+\frac{1}{3}\times4+...+\frac{1}{80}\times81\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{80}-\frac{1}{81}\)

\(A=1-\frac{1}{81}\)

\(A=\frac{80}{81}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6084}\)

\(\Rightarrow A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)

\(\Rightarrow A=1-\frac{1}{81}=\frac{80}{81}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{80}-\frac{1}{81}=1-\frac{1}{81}=\frac{80}{81}\)

\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)

\(A=1-\frac{1}{81}\)

\(A=\frac{80}{81}\)

Cái này là toán lớp 6 nha bn

Ủng hộ mk nha ^_-

A=\(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{80x81}\)

A=\(1-\frac{1}{81}=\frac{80}{81}\)

\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(Z=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(Z=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)

\(Z=\frac{1}{1}-\frac{1}{81}=\frac{80}{81}\)

\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{6480}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{80.81}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+.....+\frac{1}{80}-\frac{1}{81}\)

\(=1-\frac{1}{81}\)

\(=\frac{80}{81}\)

\(Z=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{6480}\)

\(Z=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{80.81}\)

\(Z=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{80}-\frac{1}{81}\)

\(Z=1-\frac{1}{81}=\frac{80}{81}\)

Ủng hộ mk nha !!! *_*

Bạn đợi tí nha Trần Phương Thùy

Mình đang trả lời !

\(a,\text{ }3^{x+4}-3^x=6480\)

\(3^x\left(3^4-1\right)=6480\)

\(3^x\cdot80=6480\)

\(3^x=6480\text{ : }80\)

\(3^x=81\)

\(3^x=3^4\)

\(\Rightarrow\text{ }x=4\)

\(b,\text{ }x^5=4x^3\)

\(x^5-4x^3=0\)

\(x^3\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3=0\\x^2-4=0\end{cases}}\)                    \(\Rightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)               \(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm2\end{cases}}\)

                   \(\Rightarrow\text{ }x\in\left\{0\text{ ; }-2\text{ ; }2\right\}\)

a) ( 1 - 1/2 ) x ( 1 - 1/3 ) x ( 1 - 1/4 ) x ( 1 - 1/5 ) x ( 1 - 1/6 )

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6 

= \(\frac{1.2.3.4.5}{2.3.4.5.6}\)

= 1/6

\(\frac{1}{6}\)

1/6

đáp số:1/6