1)Tính

a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)

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

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

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

b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

2) tìm x

\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)

\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)

\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)

\(\dfrac{4}{5}x=0\)

\(x=0:\dfrac{4}{5}\)

\(x=0\)

b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)

\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)

\(\dfrac{2}{5}x=\dfrac{31}{10}\)

\(x=\dfrac{31}{10}:\dfrac{2}{5}\)

\(x=\dfrac{31}{4}\)

Giải

S=1+2+2²+2³+....................+2⁶²+2⁶³

2S=2(1+2+2²+2³+.................+2⁶²+2⁶³)

2S=2+2²+2³+2⁴+............................+2⁶³+2⁶⁴)

2S-S=(2+2²+2³+2⁴+...................+2⁶³+2⁶⁴)-(1+2+2²+2³+.....................+2⁶²+2⁶³)

S=2⁶⁴-1

Tính:

\(3.27-19+2010\)⁰

\(=3.3.9-18+1-1\)

\(=3.3.9-18\)

\(=9.9-18\)

\(=9.9-9.2\)

\(=9(9-2)\)

\(=9.7\)

\(=63\)

45^x.45¹²³=45¹²⁶

45^x=45¹²⁶:45¹²³

45⁵=45³

\(\Leftrightarrow\)x=3

Gọi d là UCLN(21n+4;14n+3)

\(\Leftrightarrow21n+4⋮d\Rightarrow2\left(21n+4\right)⋮d\Rightarrow42n+8⋮d\)

\(14n+3⋮d\Rightarrow3\left(14n+3\right)⋮d\Rightarrow42n+9⋮d\)

Vì:

\(42n+8;42n+9\)\(⋮d\)

Nên:

\(\left(42n+9\right)-\left(42n+8\right)⋮d\)

\(42n+9-42n-8⋮d\)

\(1⋮d\)

\(\Rightarrow\dfrac{21n+4}{14n+3}\)tối giản với mọi n

Đặt

S=1+2+2²+2³+........+2²⁰¹⁴

2S=2(1+2+2²+2³+........+2²⁰¹⁴)

2S=2+2²+2³+2⁴+..............+2²⁰¹⁵

2S-S=(2+2²+2³+2⁴+.............+2²⁰¹⁵)-(1+2+2²+2³+.........+2²⁰¹⁴)

S=2²⁰¹⁵-1

Ta thấy: S là tử B

thay S vào B ta có:

B=\(\dfrac{2^{2015}-1}{1-2^{2015}}=-1\)

Ta có: 286:a dư 48=>(286-48)\(⋮\)a\(\Rightarrow\)238\(⋮\)a(a>48) (1)

Ta có:969:a dư 17=>(969-17)\(⋮\)a\(\Rightarrow\)952\(⋮\)a(a>17) (2)

Từ(1) và (2) ta có:

a là ƯC(238;952)

Mà 952\(⋮\)238

nên 238 sẽ là UCLN(952;238)

mà Ư(238)={1;2;7;14;17;34;119;238}

Mà a>48

nên a=119;238