1/3x4 + 1/4x5 + 1/5x6 +...+ 1/97x 98 + 1/98x99 + x =1

=> 1/3-1/4+1/4-1/5+1/5-1/6+....+1/97-1/98 + 1/98-1/99 +x = 1

=> 1/3 - 1/99 +x=1

=>  32/99+x=1

=> x= 1-32/99

=> x = 67/99

67/99

\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)

Ta có: 1+2+3+...+98=98.99:2=4851

Đặt B=1.2+2.3+3.4+...+98.99  => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)

=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100

=> B=33.98.100. Thay vào A được:

\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)

A=1/2

\(\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)

=\(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{99}-\frac{1}{100}\)

=\(\frac{1}{4}-\frac{1}{100}\)

=\(\frac{6}{25}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}=\dfrac{21-2}{42}=\dfrac{19}{42}\)

Lời giải:
Gọi biểu thức số 1 là A và số 2 là B

\(A=\frac{3-2}{2\times 3}+\frac{4-3}{3\times 4}+\frac{5-4}{4\times 5}+\frac{6-5}{5\times 6}+\frac{7-6}{6\times 7}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{2}-\frac{1}{7}=\frac{5}{14}\)

B tương tự A:
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{20}-\frac{1}{21}\)

\(=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\)

=1x2x3+2x3x3+...+98x99x3

=1x2x3+2x3x(4-1)+...+98x99x(100-97)

=1x2x3+2x3x4-1x2x3+...+98x99x100-97x98x99

=98x99x100

=970200

B=1/3*4-1/4*5-....-1/99*100

-(1/3*4+1/4*5+...+1/99*100)

=-(1/3-1/4+1/5-1/5+...+1/99-1/100)

=-(1/3-1/100)

=-(100/300-3/300)=-97/300

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+....+\frac{1}{99\times100}\)

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\frac{1}{1}-\frac{1}{100}\)

\(\frac{100-1}{100}\)

\(\frac{99}{100}\)

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{99\times100}\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\frac{1}{1}-\frac{1}{100}\)
\(\frac{100-1}{100}\)
\(\frac{99}{100}\)

 

99/100

=1+1/3-1/4+1/4-1/5+1/5-1/6

=4/3-1/6=7/6

Đợi nhé

Bài nào khó lắm thì mới hỏi thôi chứ bài này dễ mà bạn tự vận động não đi

\(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{11\cdot12}=x\)

\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{12}=\dfrac{4}{12}-\dfrac{1}{12}=\dfrac{1}{4}\)

dễ lắm ko cần hỏi đâu ,tự vận động não đi