sorry mình nhầm

ta có:

M=\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).\(\frac{4^2}{4.5}\)

=\(\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}\)

=\(\frac{1}{5}\)

vậy M=\(\frac{1}{5}\)

\(M=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)

ta có:

\(\frac{1^2}{1.2}\).\(\frac{2^2}{2.3}\).\(\frac{3^2}{3.4}\).

hình như là 3² chứ k f 3³

\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)

\(B=\frac{\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)}{\left(1\cdot2\right)\left(2\cdot3\right)\left(3\cdot4\right)\left(4\cdot5\right)}\)

\(B=\frac{\left(1\cdot2\cdot3\cdot4\right)\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\left(2\cdot3\cdot4\cdot5\right)}\)

\(=\frac{1}{5}\)

\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)

\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)

\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1^2\cdot2^2\cdot3^2\cdot4^2\cdot5}=\frac{1}{5}\)

\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)

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

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

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

\(A=9.\frac{99}{100}\)

\(A=\frac{891}{100}\)

kết quả là 891/100 nha

A=909/100

đáp án là thế nha!!

nếu đúng thì tích mk cái

Ta có:

\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}=\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}=\frac{1}{5}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}...\frac{100^2}{100.101}\)

\(=\frac{1.1.2.2.3.3...100.100}{1.2.2.3.3.4.4...100.101}\)

\(=\frac{\left(1.2.3...100\right)\left(1.2.3...100\right)}{\left(1.2.3..100\right)\left(2.3.4...101\right)}=\frac{1}{101}\)

\(\text{Đặt C = 1.2 + 2.3 + 3.4 + ..... +98.99 }\)

\(\text{ Và A = 1.98 + 2.97 + 3.96 + .... + 98.1 }\)

\(\text{Khi đó : }A=1+\left(1+2\right)+....+\left(1+2+...+98\right)\)

                      \(=\frac{1.2}{2}+\frac{2.3}{2}+\frac{3.4}{2}+....+\frac{98.99}{2}\)

                      \(=\frac{1.2+2.3+3.4+....+98.99}{2}=\frac{C}{2}\)

\(\Rightarrow B=\frac{B}{\frac{2}{B}}=\frac{1}{2}\)

\(\frac{1}{1.2}\)\(+\frac{1}{2.3}+\)\(\frac{1}{3.4}\)\(+\)\(.............+\)\(\frac{1}{2017.2018}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{2017}-\frac{1}{2018}\)

\(=\frac{1}{1}-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{2017.2018}\)

 \(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+......+\frac{2018-2017}{2017.2018}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2017}-\frac{1}{2018}\)

\(=1-\frac{1}{2018}\)

\(=\frac{2017}{2018}\)

1/1.2+1/2.3+1/3.4+...+1/2017.2018

=1-1/2+1/2-1/3+1/3-1/4+...+1/2017-1/2018

=1-1/2018

=2017/2018

\(M=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}=\frac{1}{5}\)

\(N=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)

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

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

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

N=1/2x(1/3-1/101)

N=1/2x98/101

N=49/101

nhầm 98/303 mới đúng

Đặt : S=1.2+2.3+3.4+........+98.99 
3S=1.2.3+2.3.3+3.4.3+........+98.99.3 
3S=1.2.3+2.3.(4 -1) +3.4.(5 -2)+........+98.99.(100 -97) 
3S=1.2.3+2.3.4 -1.2.3 +3.4.5 -2.3.4 +........+98.99.100 -97.98.99 
3S=98.99.100 
Vậy : S=\(\frac{98.99.100}{3}\)

Thay S vào dãy 1.2+2.3+3.4+........+98.99  ta được :

\(A=\frac{\frac{98.99.100}{3}.y}{26950}\)=\(\frac{48}{7}\)

đến đây chắc tự tính được ( quy đồng 48/7 lên có mẫu bằng 26950 rồi sẽ có bài tìm y .... tự làm )

tử của VP=S= 1.2 + 2.3 + 3.4 + ...+ 99.100

3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3

3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)

3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100

3S = 99.100.101  3S = 3.33.100.101 

 S=33.100.101= 333300

thay S vào tử của VP ta được \(\frac{333300y}{26950}=\frac{12.6.2}{7.3}\)

\(\frac{609y}{49}=\frac{48}{7}\)

\(\frac{609y}{49}=\frac{336}{49}\)

=>609y=336

=>y=\(\frac{16}{29}\)

tử của VP=S= 1.2 + 2.3 + 3.4 + ...+ 99.100

3S = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3

3S= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)

3S= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100

3S = 99.100.101  3S = 3.33.100.101 

 S=33.100.101= 333300