Ta có: 

A=\(\frac{1}{1.101}+\frac{1}{2.102}+...+\frac{1}{25.125}\)

=\(\frac{1}{100}\left(\frac{100}{1.101}+\frac{100}{2.102}+...+\frac{100}{25.125}\right)\)

=\(\frac{1}{100}\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{25}-\frac{1}{125}\right)\)

=\(\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)

B=\(\frac{1}{1.26}+\frac{1}{2.27}+...+\frac{1}{100.125}\)

=\(\frac{1}{25}\left(\frac{25}{1.26}+\frac{25}{2.27}+...+\frac{25}{100.125}\right)\)

=\(\frac{1}{25}\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+...+\frac{1}{100}-\frac{1}{125}\right)\)

=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{125}\right)\right]\)

=\(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)+\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)

= \(\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]\)

=> \(\frac{A}{B}\)=\(\frac{\frac{1}{100}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}{\frac{1}{25}\left[\left(1+\frac{1}{2}+...+\frac{1}{25}\right)-\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{125}\right)\right]}\)=\(\frac{1}{\frac{100}{\frac{1}{25}}}\)=\(\frac{1}{100}\cdot25=\frac{25}{100}=\frac{1}{4}\)

A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{125}\)

giup minh nhé

\(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{102}+...+\frac{1}{125}\)

( Nếu muốn dùng thử lại : )

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a, \(\left(\frac{1}{2}\right)^x+\left(\frac{1}{2}\right)^{x+4}=17\)

\(\Rightarrow\frac{1}{2^x}+\frac{1}{2^x}\cdot\frac{1}{16}=17\)

\(\Rightarrow\frac{1}{2^x}\left(1+\frac{1}{16}\right)=17\)

\(\Rightarrow\frac{1}{2^x}\cdot\frac{17}{16}=17\)

\(\Rightarrow\frac{1}{2^x}=17:\frac{17}{16}=\frac{1}{16}=\frac{1}{2^4}\)

=> x = 4

b, Ta có: \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....;\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+...+x+\frac{1}{99.100}=100x\)

\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow99x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=100x\)

\(\Rightarrow100x-99x=1-\frac{1}{100}\)

\(\Rightarrow x=\frac{99}{100}\)

a) \(2^x+2^{x+1}2^{x+2}=112\)

    \(2^x.\left(1+2+4\right)=112\)

     \(2^x=112:7=16\)

Mà \(2^4=16\)

\(\Rightarrow2^x=2^4\)

Vậy x = 4

b) \(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...\left|x+\frac{1}{99.100}\right|=100x\)

Vì \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;....\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow\left(x+x+...x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow100x+\left(1-\frac{1}{100}\right)=100x\)

\(\Rightarrow\frac{99}{100}=x\)

a) 2^x+2^x+1+2^x+2=112

  2^x(1+2+2²)=112

2^x.7=12

2^x=16

x=4

\(\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+\left|x+\frac{1}{3.4}\right|+.............+\left|x+\frac{1}{99.100}\right|=100x^{\left(1\right)}\)

Ta có \(\left|x+\frac{1}{1.2}\right|\ge0;\left|x+\frac{1}{2.3}\right|\ge0;\left|x+\frac{1}{3.4}\right|\ge0;...........;\left|x+\frac{1}{99.100}\right|\ge0\)

\(\Rightarrow VT\ge0\Rightarrow VP\ge0\Rightarrow100x\ge0\Rightarrow x\ge0\)

Với \(x\ge0\).Từ (1) \(\Rightarrow x+\frac{1}{1.2}+x+\frac{1}{2.3}+x+\frac{1}{3.4}+.................+x+\frac{1}{99.100}=100x\)

\(\Rightarrow\left(x+x+x+.......+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+............+\frac{1}{99.100}\right)=100x\)

\(\Rightarrow99x+\left(1-\frac{1}{100}\right)=100x\)

\(\Rightarrow99x+\frac{99}{100}=100x\)

\(\Rightarrow x=\frac{99}{100}\left(TMĐK:x\ge0\right)\)

Vì GTTĐ luôn lớn hơn hoặc bằng 0 với mọi x

\(\Rightarrow\left|x+\frac{1}{1\cdot2}\right|+\left|x+\frac{1}{2\cdot3}\right|+...+\left|x+\frac{1}{99\cdot100}\right|\ge0\)

\(\Rightarrow100x\ge0\)

\(\Rightarrow x\ge0\)

Từ điều kiện trên ta có :

\(x+\frac{1}{1\cdot2}+x+\frac{1}{2\cdot3}+...+x+\frac{1}{99\cdot100}=100x\)

\(50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(50x=1-\frac{1}{100}\)

\(50x=\frac{99}{100}\)

\(x=\frac{99}{5000}\)

Do \(\left|a\right|\ge0\forall a\) nên:

\(A=\left|x+\frac{1}{1.2}\right|+\left|x+\frac{1}{2.3}\right|+...+\left|x+\frac{1}{99.100}\right|\ge0\forall x\)

\(\Leftrightarrow100x\ge0\) hay \(x\ge0\)

Do vậy ta có: \(A=\left(x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)=100x\) ( 50 chữ số x)

\(\Leftrightarrow A=50x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)=100x\)

\(\Leftrightarrow50x+\left(1-\frac{1}{100}\right)=100x\Leftrightarrow50x+\frac{99}{100}=100x\)

\(\Leftrightarrow50x=\frac{99}{100}\Leftrightarrow x=\frac{99}{100.50}=\frac{99}{5000}\)

tưởng là có 99x lận mà

#)Giải :

a) x + 2x + 3x + ... + 100x = - 213

=> 100x + ( 2 + 3 + 4 + ... + 100 ) = - 213 

=> 100x + 5049 = - 213 

<=> 100x = - 5262

<=> x = - 52,62

#)Giải :

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{3}+\frac{1}{6}\)

\(\Rightarrow\frac{1}{2}x+\frac{1}{4}x=\frac{1}{2}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{4}\right)x=\frac{1}{2}\)

\(\Rightarrow\frac{3}{4}x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{2}{3}\)

a) x + 2x + 3x + ... +100x = -213

=>  x . (1 + 2 + 3 +... + 100) = - 213

=> x . 5050 = -213

=> x           = - 213 : 5050

=> x           = -213/5050

b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}x-\frac{1}{6}\)

=> \(\frac{1}{2}x-\frac{1}{4}x=\frac{1}{3}-\frac{1}{6}\)

=> \(x.\left(\frac{1}{2}-\frac{1}{4}\right)=\frac{1}{6}\)

=> \(x.\frac{1}{4}=\frac{1}{6}\)

=> \(x=\frac{1}{6}:\frac{1}{4}\)

=> \(x=\frac{2}{3}\)

c) 3(x-2) + 2(x-1) = 10

=> 3x - 6 + 2x - 2 = 10

=> 3x + 2x - 6 - 2 = 10

=> 5x - 8 = 10

=> 5x = 10 + 8

=> 5x = 18

=> x = 18:5

=> x = 3,6

d) \(\frac{x+1}{3}=\frac{x-2}{4}\)

=> \(4\left(x+1\right)=3\left(x-2\right)\)

=>\(4x+4=3x-6\)

=> \(4x-3x=-4-6\)

=> \(x=-10\)