a) A = (1.1)/(1.2) x (2.2)/(2.3) x ... (99.99)/(99.100) = 1/2 . 2/3 . 3/4. ..99/100 = 1/100 

Tk cho em nha

bạn hãy rút gọn vế phải: x/200=1/2.2/3.3/4......98/99.99/100

  Rồi sẽ có cái phương trình:x/200=1/100

từ đó suy ra:x/200=2/200 =>x=2

:)))))

\(\dfrac{x}{200}=\dfrac{1^2}{1.2}.\dfrac{2^2}{2.3}.\dfrac{3^2}{3.4}...\dfrac{99^2}{99.100}\)

\(\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}...\dfrac{99}{100}\)

\(\Leftrightarrow\dfrac{x}{200}=\dfrac{1}{100}\)

\(\Leftrightarrow x=2\)

dễ ghê chiều làm cho giờ off

Đặt \(A=1\cdot2+2\cdot3+\cdots+99\cdot100\)

\(=1\left(1+1\right)+2\left(2+1\right)+\cdots+99\left(99+1\right)\)

\(=\left(1^2+2^2+\cdots+99^2\right)+\left(1+2+\cdots+99\right)\)

\(=\frac{99\left(99+1\right)\left(2\cdot99+1\right)}{6}+\frac{99\cdot\left(99+1\right)}{2}\)

\(=\frac{99\cdot100\cdot199}{6}+\frac{99\cdot100}{2}=\frac{99\cdot100\cdot199+3\cdot99\cdot100}{6}\)

\(=\frac{99\cdot100\cdot202}{6}=33\cdot100\cdot101=333300\)

Ta có: \(x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)\)

\(=\left(x^2+0\right)+\left(x^2+1\right)+\cdots+\left(x^2+99\right)\)

\(=\left(x^2+x^2+\cdots+x^2\right)+\left(0+1+\cdots+99\right)=100x^2+\frac{99\cdot100}{2}=100x^2+99\cdot50=100x^2+4950\)

Ta có: \(\frac{1\cdot2+2\cdot3+\cdots+99\cdot100}{x^2+\left(x^2+1\right)+\left(x^2+2\right)+\cdots+\left(x^2+99\right)}=50\frac{116}{131}\)

=>\(\frac{333300}{100x^2+4950}=50+\frac{116}{131}\)

=>\(\frac{6666}{2x^2+99}=\frac{6666}{131}\)

=>\(2x^2+99=131\)

=>\(2x^2=32\)

=>\(x^2=16\)

=>\(\left[\begin{array}{l}x=4\\ x=-4\end{array}\right.\)

Mình gõ câu a bị lỗi nha , thực chất câu a là

a) Tìm các số tự nhiên x, y biết : 2xy + x + 2y = 13

a)Bạn làm nha vì bài này dễ rồi

b)+)Ta có:A=1.2+2.3+3.4+..................+99.100

=>3A=1.2.3+2.3.3+3.4.3+.................+99.100.3

=>3A=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)

=>3A=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101

=>3A=99.100.101

=>A=\(\frac{99.100.101}{3}=333300\)

+)Ta lại có:B=1²+2²+3²+..................+99²

=>B=1.1+2.2+3.3+............+99.99

=>B=1.(2-1)+2.(3-1)+3.(4-1)+..........+99.(100-1)

=>B=1.2-1+2.3-2+3.4-3+........................+99.100-99

=>B=(1.2+2.3+3.4+............+99.100)-(1+2+3+..............+99)

Đặt N=1.2+2.3+3.4+....................+99.100

=>3N=1.2.3+2.3.3+3.4.3+.................+99.100.3

=>3N=1.2.3+2.3.(4-1)+3.4.(5-2)+................+99.100.(101-98)

=>3N=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...................-98.99.100+99.100.101

=>3N=99.100.101

=>N=\(\frac{99.100.101}{3}=333300\)

Đặt M=1+2+3+..............+99(có 99 số hạng)

=>M=\(\frac{\left(1+99\right).99}{2}=4950\)

+)Ta thấy A-B=333300-(333300-4950)

=>A-B=333300-333300+4950

=>A-B=4950\(⋮\)50

Vậy A-B\(⋮\)50

Chúc bn học tốt

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)