ta có:1/8^100

       -1/4^200=(-1/4^2)^100=1/16^100

=>1/8^100 >1/16^100

=>1/8^100 >-1/4^200

bằng nhau

ta có:\(\left(-\frac{1}{8}\right)^{180}=\left(\frac{1}{8}\right)^{180}=\left(\frac{1}{4}\right)^{2^{180}}=\left(\frac{1}{4}\right)^{360}\) 

ta có :\(\left(-\frac{1}{4}\right)^{200}=\left(\frac{1}{4}\right)^{200}\)

=>(1/4)^360<(1/4)^200

Vậy : (-1/8)^180 < ( -1/4)^200

Ta có: \(\left(\frac{-1}{8}\right)^{180}=\frac{-1^{180}}{8^{180}}=\frac{1}{8^{180}}\)

\(\left(\frac{-1}{4}\right)^{200}=\frac{-1^{200}}{4^{200}}=\frac{1}{4^{200}}\)

Suy ra để so sánh \(\left(\frac{-1}{8}\right)^{180}\)và\(\left(\frac{-1}{4}\right)^{200}\)thì ta chỉ cần so sánh 8¹⁸⁰ và 4²⁰⁰

Ta có: 8¹⁸⁰=(2³)¹⁸⁰=2^3.180=2⁵⁴⁰

             4²⁰⁰=(2²)²⁰⁰=2^2.200=2⁴⁰⁰

       Ta thấy 2=2 nhưng 540>400 suy ra 8¹⁸⁰>4²⁰⁰ suy ra \(\frac{1}{8^{180}}< \frac{1}{4^{200}}\)suy ra \(\left(\frac{-1}{8}\right)^{180}< \left(\frac{-1}{4}\right)^{200}\)

\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)

\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)

M=-(\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{1-100^2}{100^2}\))

=-(\(\frac{1.3}{2.2}.\frac{2.4}{3.3}\frac{3.5}{4.4}...\frac{99.100}{100.100}\))

=-(\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...100}{2.3.4..100}\))

=-(\(\frac{1}{100}.\frac{1}{2}\))

=\(\frac{-1}{200}\)

thanks you very very much

Ta có : \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-99}{100^2}=-\frac{3.8.15...9999}{\left(2.3.4...100\right)\left(2.3.4...100\right)}=-\frac{\left(1.2.3...99\right)\left(3.4.5...101\right)}{\left(2.3.4...100\right)\left(2.3.4...100\right)}\)

\(=-\frac{101}{100.2}=-\frac{101}{200}< -\frac{100}{200}=-\frac{1}{2}\)

đúng đó bạn

\(x=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{4}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{10}\right)\)

\(=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}.\frac{9}{10}=\frac{63}{256}< \frac{63}{210}=0,3\)

\(x=\sqrt{0,1}>\sqrt{0,09}=0,3\)

=> y<x

Ta có :

\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{99}{100}=\frac{3.8.15.....99}{4.9.16.....100}=\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)\(=\frac{1.2.3...9}{2.3...10}.\frac{3.4...11}{2.3...10}=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}< \frac{11}{19}\)

ta có M = (1- 1/4) (1- 1/9)... ( 1- 1/100)

             = 3/2^2.8/3^2 ... 99/10^2

             = 1.3/2^2 . 2.4/3^2 ... 9.11/10^ 2

             = 1.2.3...9/ 2.3.4...10 . 3.4.5... 11/ 2.3.4... 10

             = 1/10 . 11/2 = 11/20 < 11/19

              Vậy M < 11/19