Gọi phép tính là: \(A\)

Ta có:

\(A=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+...+\dfrac{2}{9\times10}\\ A\div2=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\\ A\div2=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\\ A\div2=\dfrac{9}{10}\\ A=\dfrac{18}{10}=\dfrac{9}{5}\)

\(=2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}\right)\)

\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=2\cdot\dfrac{9}{10}=\dfrac{9}{5}\)

Đặt phép tính  là A

\(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+\dfrac{2}{9.10}\)

\(2A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(2A=1-\dfrac{1}{10}\)

\(2A=\dfrac{9}{10}\)

\(A=\dfrac{9}{5}\)

... tìm số dư khi chia hết???

nếu nó chia hết thì số dư bằng 0 rồi

bạn nếu cách làm đi

a)\(\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.5\right)^{10}=1^{10}=1\)

b)\(5^2.3^5.\left(\frac{3}{5}\right)^2=\left(\frac{3}{5}.5\right)^2.3^5=3^2.3^5=3^7\)

c)\(\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{2.3}:\left(\frac{1}{8}\right)^2=\left(\frac{1}{8}\right)^{6+2}=\left(\frac{1}{8}\right)^8\)

\(a.\left(\frac{1}{5}\right)^{10}.5^{20}=\left(\frac{1}{5}\right)^{10}.5^{10.2}=\left(\frac{1}{5}\right)^{10}.\left(5^2\right)^{10}=\left(\frac{1}{5}\right)^{10}.25^{10}=\left(\frac{1}{5}.25\right)^{10}=5^{10}.\)

\(b.5^2.3^5.\left(\frac{3}{5}\right)^2=\left[5^2.\left(\frac{3}{5}\right)^2\right].3^5=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)\(c.\left(\frac{1}{16}\right)^3:\left(\frac{1}{8}\right)^2=\left[\left(\frac{1}{4}\right)^2\right]^3:\left[\left(\frac{1}{2}\right)^3\right]^2=\left(\frac{1}{4}\right)^6:\left(\frac{1}{2}\right)^6=\left(\frac{1}{4}:\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^6\)

Rùi sao má mỗi thế thôi à

a)

Ta có : \(32^{13}=\left(2^5\right)^{13}=2^{65}\)

            \(64^{10}=\left(2^6\right)^{10}=2^{60}\)

Mà \(2^{65}>2^{60}\Rightarrow.....\)

b)

A = 2 + 2.2 + 2.2.2 + ... + 2.2.2.2....2

A = \(2+2^2+2^3+...+2^{100}\)

2A = \(2^2+2^3+2^4+...+2^{101}\)

\(2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)

\(A=2^{101}-2\)

1.

a) Ta có : 32¹³ = ( 2⁵ ) ¹³ = 2⁶⁵

               64¹⁰ = ( 2⁶ ) ¹⁰ = 2⁶⁰ 

Vì 2⁶⁵ > 2⁶⁰ nên 32¹³ > 64¹⁰

b) A = 2 + 2.2 + 2.2.2 + 2.2.2.2 + ... + 2.2.2.2.2...2 ( 100 số 2 )

A = 2 . ( 1 + 2 + 2.2 + 2.2.2 + ... + 2.2.2.2...2 )

A = 2. ( 1 + 2 + 2² + 2³ + ... + 2⁹⁹ )

gọi B là biểu thức trong ngoặc

Lại có : B = 1 + 2 + 2² + 2³ + ... + 2⁹⁹

         2B = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰

         2B - B = ( 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰ ) - ( 1 + 2 + 2² + 2³ + ... + 2⁹⁹ )

         B = 2¹⁰⁰ - 1

\(\Rightarrow\)A = 2 . ( 2¹⁰⁰ - 1 )

\(\Rightarrow\)A = 2¹⁰¹ - 2

a)Ta có: 64¹⁰=(32x2)¹⁰=32^2x10=32²⁰

mà  32²⁰ > 32¹³

nên 32²⁰   <  64¹⁰

b)  A = 2+2.2+2.2.2+2.2.2.2+...+2.2.2.2.2...2(100 số 2)

   2 A =  2.[ 2+2.2+2.2.2+2.2.2.2+...+2.2.2.2.2...2(100 số 2)]

   2 A  =  2.2+2.2.2+2.2.2.2+2.2.2.2.2+...+2.2.2.2.2...2.2(101 số 2)

2  A - A  = 2.2.2.2.2...2.2(101 số 2) - 2

       A = 2¹⁰¹ -  2