5/8+3/5x5/12=

Ta có:

1/5×5 < 1/4×5

1/6×6 < 1/5×6

1/7×7 < 1/6×7

.........

1/100×100 < 1/99×100

=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100

                                      = 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100

                                    = 1/4-1/100 < 1/4  

=> 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4  (1)

Lại có:

1/5×5 > 1/6×7

1/6×6 > 1/7×8

1/7×7 > 1/8×9

........

1/100×100 > 1/101×102

=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8  +.....+1/100×101

                                   = 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101

                                   = 1/5 - 1/101 > 1/5 - 1/30 = 1/6

=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)

Từ (1) và (2)

=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4

Đặt \(A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)

Có \(\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)(1)

Lại có :\(\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)\)

Từ (1) và (2) \(\RightarrowĐCCM\)

\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+....+\frac{1}{99.100}\)

\(A< \frac{1}{4}-\frac{1}{100}\)

\(A< \frac{6}{25}< \frac{1}{4}\)

các bạn giải mau lên

`5/39xx(7 4/5xx1 2/3+8 1/3xx7 4/5)`

`=5/39xx(39/5xx5/3+25/3xx39/5)`

`=5/39xx39/5(5/3+25/3)`

`=1xx30/3`

`=10`

10

a) \(\dfrac{2}{5}+\dfrac{4}{5}\times\dfrac{5}{2}\)

\(=\dfrac{2}{5}+\dfrac{4\times5}{5\times2}\)

\(=\dfrac{2}{5}+\dfrac{4}{2}\)

\(=\dfrac{2}{5}+2\)

\(=\dfrac{2}{5}+\dfrac{10}{5}\)

\(=\dfrac{12}{5}\)

b) \(\dfrac{2008}{2009}-\dfrac{2009}{2008}+\dfrac{1}{2009}+\dfrac{2007}{2008}\)

\(=\left(1-\dfrac{1}{2009}\right)-\left(1+\dfrac{1}{2008}\right)+\dfrac{1}{2009}+\left(1-\dfrac{1}{2008}\right)\)

\(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=\left(1-1+1\right)-\left(\dfrac{1}{2009}-\dfrac{1}{2009}\right)-\left(\dfrac{1}{2008}+\dfrac{1}{2008}\right)\)

\(=1-\dfrac{2}{2008}\)

\(=\dfrac{2008}{2008}-\dfrac{2}{2008}\)

\(=\dfrac{2006}{2008}\)

\(=\dfrac{1003}{1004}\)

a: =2/5+4/2

=2/5+2

=12/5

b: \(=1-\dfrac{1}{2009}-1-\dfrac{1}{2008}+\dfrac{1}{2009}+1-\dfrac{1}{2008}\)

\(=1-\dfrac{2}{2008}=1-\dfrac{1}{1004}=\dfrac{1003}{1004}\)

\(S=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+...+\frac{1}{20.20}\)

Ta có:  \(\frac{1}{2}-\frac{1}{3}>\frac{1}{3.3}>\frac{1}{3}-\frac{1}{4}\)

           \(\frac{1}{3}-\frac{1}{4}>\frac{1}{4.4}>\frac{1}{4}-\frac{1}{5}\)

          \(\frac{1}{4}-\frac{1}{5}>\frac{1}{5.5}>\frac{1}{5}-\frac{1}{6}\)

           ...................................

           \(\frac{1}{19}-\frac{1}{20}>\frac{1}{20.20}>\frac{1}{20}-\frac{1}{21}\)

Cộng theo vế ta được:

\(\frac{1}{2}-\frac{1}{20}>S>\frac{1}{3}-\frac{1}{21}\)\(\Rightarrow\)\(\frac{1}{2}>S>\frac{1}{4}\)

tui ko viet dc kiểu p/số nên viết cái gạch ngang giữa kết quả nha

\(A=\frac{8+\frac{8}{17}+\frac{8}{37}}{4+\frac{4}{17}+\frac{4}{37}}\)

\(A=\frac{8\cdot\left(1+\frac{1}{17}+\frac{1}{37}\right)}{4\cdot\left(1+\frac{1}{17}+\frac{1}{37}\right)}\)

\(A=\frac{8}{4}\)

\(A=2\)

ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac

\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)

\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)

\(A=\frac{4-1}{36+12}\)

\(A=\frac{3}{48}=\frac{1}{16}\)