\(Q=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)

\(Q=\left(\frac{1}{2}\right).\left(\frac{2}{3}\right).\left(\frac{3}{4}\right)...\left(\frac{99}{100}\right)\)

\(Q=\frac{1}{100}\)

\(P=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)...\left(1+\frac{1}{99.101}\right)\)

\(P=\left(\frac{1.3}{1.3}+\frac{1}{1.3}\right)\left(\frac{2.4}{2.4}+\frac{1}{2.4}\right)\left(\frac{3.5}{3.5}+\frac{1}{3.5}\right)...\left(\frac{99.101}{99.101}+\frac{1}{99.101}\right)\)

\(P=\left(\frac{4}{1.3}\right)\left(\frac{9}{2.4}\right)\left(\frac{16}{3.5}\right)...\left(\frac{10000}{99.101}\right)\)

\(P=\left(\frac{2^2}{1.3}\right)\left(\frac{3^2}{2.4}\right)\left(\frac{4^2}{3.5}\right)...\left(\frac{100^2}{99.101}\right)\)

Bạn tự tách ra rồi bạn sẽ ra kết quả như ở dưới

\(P=\frac{201}{100}\)

Xinloi, t ghi thiếu đề

\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)

\(\left|x+\frac{1}{1.3}\right|+\left|x+\frac{1}{3.5}\right|+...+\left|x+\frac{1}{97.99}\right|=50x\)

Vì \(\left|x+\frac{1}{1.3}\right|\ge0\forall x\)

     \(\left|x+\frac{1}{3.5}\right|\ge0\forall x\)

        ................

       \(\left|x+\frac{1}{97.99}\right|\ge0\forall x\)

(VT: Vế trái; VP: Vế phải)

\(\Rightarrow VT\ge0\Rightarrow VP=50x\ge0\)mà \(50>0\)

\(\Rightarrow x>0\)

\(\Rightarrow x+\frac{1}{1.3}>0\forall x\)

        ..............

      \(x+\frac{1}{97.99}>0\forall x\)(1)

(1) \(\Leftrightarrow x+\frac{1}{1.3}+x+\frac{1}{3.5}+...+x+\frac{1}{97.99}=50x\)

\(\Leftrightarrow49x+\left(\frac{1}{1.3}+...+\frac{1}{97.99}\right)=50x\)

\(\Leftrightarrow50x-49x=\frac{1}{2}\left(\frac{2}{1.3}+...+\frac{2}{97.99}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\left(1-\frac{1}{99}\right)\)

\(\Leftrightarrow x=\frac{1}{2}\cdot\frac{98}{99}=\frac{49}{99}\)

Vậy....

P/s: Làm bừa :) Ko chắc đúng nhé

a, Đề sai hả bạn ??

b, \(\dfrac{\left(1,16-x\right).5,25}{\left(10\dfrac{5}{9}-7\dfrac{1}{4}\right).2\dfrac{2}{17}}=75\%\)

\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{95}{9}-\dfrac{29}{4}\right).\dfrac{36}{17}}=\dfrac{75}{100}\)

\(\dfrac{\left(1,16-x\right).5,25}{\left(\dfrac{380}{36}-\dfrac{261}{36}\right).\dfrac{36}{17}}=\dfrac{3}{4}\)

\(\dfrac{\left(1,16-x\right).5,25}{\dfrac{119}{36}.\dfrac{36}{17}}=\dfrac{3}{4}\)

\(\dfrac{\left(1,16-x\right).5,25}{7}=\dfrac{3}{4}\)

=> \(\left[\left(1,16-x\right).5,25\right].4=3.7\)

\(\left[\left(1,16-x\right).5,25\right].4=21\)

( 1,16 - x ) . 5,25 = 21/4

1,16 - x = 21/4 : 5,25

1,16 - x = 1

x = 1,16 - 1

x = 0,16

Vậy x = 0,16

c, \(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{21}\right).420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(\dfrac{1}{2}.\dfrac{20}{21}.420-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(200-\left[0,4.\left(7,5-2,5x\right)\right]:0,25=212\)

\(0,4.\left(7,5-2,5x\right):0,25=200-212\)

\(0,4.\left(7,5-2,5x\right):0,25=-12\)

0,4 . ( 7,5 - 2,5x ) = -12 . 0,25

0,4 . ( 7,5 - 2,5x ) = -3

7,5 - 2,5x = -3 :0,4

7,5 - 2,5x = -7,5

2,5x = 7,5-(-7,5)

2,5x = 15

x = 6

Vậy x = 6

Vậy x = 51

Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{\left(2x-1\right)\left(2x+1\right)}\)

\(2A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{\left(2x-1\right)\left(2x+1\right)}\) 

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{\left(2x-1\right)}-\frac{1}{\left(2x+1\right)}\)

\(2A=1-\frac{1}{2x+1}=\frac{2x}{2x+1}\)

\(A=\frac{x}{2x+1}\) 

Mà \(A=\frac{49}{99}\) \(\Leftrightarrow\frac{x}{2x+1}=\frac{49}{99}\Leftrightarrow x=49\)

x=49

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x-1\right).\left(2x+1\right)}=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x-1\right)\left(2x-1\right)}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x-1}+\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{2x+1}\right)=\frac{49}{99}\)

\(\Rightarrow\frac{x}{2x+1}=\frac{49}{99}\)

\(\Rightarrow99x=49\left(2x+1\right)\)

\(\Rightarrow99x=98x+49\)

\(\Rightarrow x=49\)

Vậy : \(x=49\)

3. a) \(đk:x\ne1;x\ne-2\)

Ta có: \(A=\frac{3x-3+2}{x-1}=\frac{3\left(x-1\right)+2}{x-1}=3+\frac{2}{x-1}\)

Để A là số nguyên thì x là số nguyên và x-1 là ước của 2 . Ta có bảng:

Lại có: \(B=\frac{2x^2+4x-3x-6+5}{x+2}=\frac{2x\left(x+2\right)-3\left(x+2\right)+5}{x+2}=2x-3+\frac{5}{x+2}\)

Để B là số nguyên thì x là số nguyên và x+2 là ước của 5. Ta có bảng:

b) Để A và B cùng nguyên thì \(x\in\left\{-1;3\right\}\)

mo sach ra ma tim

Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1) 
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1) 
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1) 
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1) 
 A = n/(2n + 1)=49/99

Tự tính nha !

Đặt A =\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2x-1\right)\left(2x+1\right)}=\frac{49}{99}\)

\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2x-1}-\frac{1}{2x+1}=\frac{98}{99}\)

\(\Rightarrow2A=1-\frac{1}{2x+1}=\frac{98}{99}\)

\(\Rightarrow2A=\frac{2x+1-1}{2x+1}=\frac{98}{99}\)

\(\Rightarrow\frac{2x}{2x+1}=\frac{98}{99}\)

\(\Rightarrow2x=98\)

\(\Rightarrow x=98:2=49\)