\(\frac{1999\cdot2001-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot\left(2000+1\right)-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot2000+1999-1}{1998+1999.2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot2000+1998}{1998+1999.2000}\cdot\frac{7}{5}=1\cdot\frac{7}{5}=\frac{7}{5}\)

thank you very much

1999.2001-1/1998+1999.2000 x 7/5

=1999.2001-(1999-1998)/1998+1999.2000 x7/5

=1999.2001-1999+1998/1998+1999.2000 x7/5

=1999(2001-1)+1998/1998+1999.2000x7/5

=1999.2000+1998/1998+1999.2000x7/5

=1x7/5=7/5

\(\frac{\left(16-8:5\right)x177}{199x2001}\frac{\left(16-16\right)x177}{199x2001}=\frac{0x177}{199x2001}=\frac{0}{199x2001}=0\)

\(M=1+\frac{1}{199}+1+\frac{2}{198}+1+....+\frac{198}{2}+1=\frac{200}{200}+\frac{200}{199}+\frac{200}{198}+....+\frac{200}{2}\)

  \(=200.\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)=200 T

\(S=\frac{T}{200T}=\frac{1}{200}\)

\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)

\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

 \(=\frac{1}{2}-\frac{1}{9}\)

            * LÀM NỐT *

                              #Louis

1/2.1/3+1/3.1/4+1/4.1/5+...+1/8.1/9

=1/2.3=1/3.4+1/4.5+...+1/8.9\

=1/2-1/3+1/3-1/4=1/4-1/5+...+1/8.1/9

=1/2-1/9

=9/18-2/18

=7/18

HỌC TỐT NHA BẠN

1) Có nhận xét sau:

\(\frac{1}{a\sqrt{a+1}+\left(a+1\right)\sqrt{a}}=\frac{1}{\sqrt{a^2+a}\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a^2+a}}\)

\(=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}.\)Do đó biểu thức có giá trị bằng: \(\frac{1}{1}-\frac{1}{\sqrt{2}}+..-\frac{1}{\sqrt{1999}}=1-\frac{1}{\sqrt{1999}}.\)

2) Có nhận xét sau:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\sqrt{a+1}-\sqrt{a}.\) Thay vào biểu thức ta được biểu thức

có giá trị bằng: \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{1999}-\sqrt{1998}=\sqrt{1999}-1.\)

\(\frac{1999.2001-1}{1998+1999.2000}=\frac{1999.2001-\left(1999-1998\right)}{1998+1999.2000}=\frac{1999.2001-1999+1998}{1998+1999.2000}=\frac{1999.\left(20001-1\right)+1998}{1998+1999.2000}=\frac{1999.2000+1998}{1998+1999.2000}=1\)=> đáp án là 7/5

có bị thiếu dấu ngoặc không vậy

1999 x 2001 - 1 =3999998 ; 1998+1999 x 2000 =3999998 

Suy ra : \(\frac{3999998}{3999998}\)=1 Suy ra 1x7/5 = 7/5

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