-2 nhé bạn

                  (3.5+5.7)x+(3.5+5.7)+(3.5+5.7)=0

                       9.2x + 9.2 +9.2 = 0

                       9.2x +9.1 x 1 + 9.2 x 1 = 0

                      9.2( x + 1 + 1 ) = 0

                      9.2(x+2)=0

                           x + 2 = 0 : 9.2 

                           x + 2 = 0 

                           x = 0 - 2 

                           x = -2

\(\left(3.5+5.7\right)x+\left(3.5+5.7\right)+\left(3.5+5.7\right)=0\)

\(\left(3.5+5.7\right)^3.x=0\)

\(125000.x=0\)

\(\Leftrightarrow x=0\)

\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\)

\(=\frac{1}{2}.2.\left(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\right)\)

\(=\frac{1}{2}.2.3.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{201.203}\right)\)

\(=\left(\frac{1}{2}.3\right).2.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{201.203}\right)\)

\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{201.203}\right)\)

Vì muốn chuyển 3/5.7 = 1/5 - 1 /7 thì tử số phải bằng hiệu của mẫu số nên 3/5.7= 3/5.7 chia 2/5.7 = 3/2 . 2/5.7 các phân số khác cũng tương tự như thế

nên ta có 3/5.7 +3/7.9 +...3/201.203 = 3/2. (2/5.7+2/7.9+...+2/201.203)

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)

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

\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}\)

\(=\frac{3}{7}\)

b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)

\(=\frac{1}{2}.\frac{2010}{2011}\)

\(=\frac{1005}{2011}\)

a)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+...+\left(\frac{1}{99}-\frac{1}{101}\right)\)

                                                               \(=1-\frac{1}{101}=\frac{100}{101}\)

b) \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}=\frac{2}{1.3}.\frac{5}{2}+\frac{2}{3.5}.\frac{5}{2}+\frac{2}{5.7}.\frac{5}{2}+...+\frac{2}{99.101}.\frac{5}{2}\)

                                                                \(=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

                                                                \(=\frac{5}{2}.\frac{100}{101}=\frac{250}{101}\)

a)100/101

b)250/101

A)100/101

B)250/101

a.2/1.3+2/3.5+2/5.7+................+2/99.101

1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101

1-1/101

100/101

b.5/1.3+5/3.5+5/5.7+............+5/99.101

5.2/1.3.2+5.2/3.5.2+5.2/5.7.2+........+5.2+99.101.2

5/2(2/1.3+2/3.5+2/5.7+........+2/99.101)

5/2(1-1/3+1/3-1/5+1/5-1/7+........+1/99-1/101)

5/2(1-1/101)

5/2.100/101

250/101

a.  

\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)

\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

b.

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)

\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)

mk đầu tiên nha bạn

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)

\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)

\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)

\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)

~ Hok tốt ~

\(\)

Viết thành 2 . (1/3.5 + 1/5.7 + 1/7.9 + ...+ 1/97.99

Tui hk bít nữa

Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}\)

\(=\dfrac{4}{15}\)

Câu 1:

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

= \(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(\dfrac{1}{3}-\dfrac{1}{101}\)

= \(\dfrac{98}{303}\)

Câu 2 làm tương tự ở câu 1 nhé

5.7=35 

bài này lớp 3 nha!!!

5.9

duyệt đi

5.7=5*7=35

dấu chấm là dấu nhân,lớp 6 sẽ được học