A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)

A=\(\frac{1.2.3.4...2015}{2.3.4...2016}=\frac{1}{2016}\)

Hok tốt

A = \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2015}\right).\left(1-\frac{1}{2016}\right)\)

= \(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2014}{2015}.\frac{2015}{2016}\)

= \(\frac{1}{2016}\)

Vậy ...

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)......\left(1-\frac{1}{2015}\right)\left(1-\frac{1}{2016}\right)\)

\(\Rightarrow A=\frac{1.2.3.4....2015}{2.3.4....2016}\)

\(\Rightarrow A=\frac{1}{2016}\)

#hok tốt

M<1 => \(\frac{x-3}{x+2}\)<1

       <=> \(\frac{x-3}{x+2}\)- 1 < 0

       <=> \(\frac{x-3}{x+2}\)-\(\frac{x+2}{x+2}\)< 0

       <=> \(\frac{x-3-x-2}{x+2}\)< 0

       <=>              -5         < 0

=> Vô nghiệm

Ta có:\(\frac{-157}{623}=\frac{-331441}{132699}\)

           \(\frac{-47}{213}=\frac{-29281}{132699}\)

  Vì \(-331441< -29281\)nên\(\frac{-331441}{132699}< \frac{-29281}{132699}\)

                                                     hay\(\frac{-157}{623}< \frac{-47}{213}\)

              Vậy \(\frac{-157}{623}< \frac{-47}{213}\)

\(\frac{-157}{623}< \frac{-47}{213}\)

B=2/1.3 + 2/3.5 + 2/5.7 +...+ 2/299.301

B=1-1/3+1/3-1/5+1/5-1/7+...+1/299-1/301=1-1/301=300/301

\(Ta có: \frac{2}{3}=\frac{1}{1}-\frac{1}{3}\);

\(\frac{2}{15}=\frac{1}{3}-\frac{1}{5}\);

\(\frac{2}{35}=\frac{1}{5}-\frac{1}{7}\) ; ... ; \(\frac{2}{89999}=\frac{1}{299}-\frac{1}{301}\).

=> B= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{299}-\frac{1}{301}\)

=> B=\(\frac{1}{1}-\frac{1}{301}\)

=> B=\(\frac{300}{301}\)

Ix+9I x 2=10

<=> Ix+9I=5

<=> x+9=5 hoặc x+9=-5

<=> x=-4 hoặc x=-14

\(\left|x+9\right|.2=10\)

\(\Leftrightarrow\left|x+9\right|=10:2=5\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=5\\x+9=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5-9=-4\\x=-5-9=-14\end{cases}}}\)

Vậy: \(\orbr{\begin{cases}x=-4\\x=-14\end{cases}}\)

Mình làm phần đẩu thôi.

|x + 9 | .2  = 10

|x + 9 | = 10 : 2

|x + 9 | = 5

=> x + 9 = \(\pm5\)

<=> x = 4 hoặc -4

    A = 8 - 8² + 8 + 8³ + 8⁴+ ......+ 8⁹⁹

        A = ( 8 - 8 ) + ( 8² + 8³ + 8⁴ +......+ 8⁹⁹ )

        A = 8² + 8³ + 8⁴ +......+ 8⁹⁹

       8A = 8³ + 8⁴ + 8⁵ +.......+ 8¹⁰⁰ 

  8A - A = ( 8³ + 8⁴ +...+ 8¹⁰⁰ ) - ( 8² + 8³ + ...+ 8⁹⁹ )

       7A = 8¹⁰⁰ - 8²

    => A = \(\frac{8^{100}-8^2}{7}\)

VẬY, \(A=\frac{8^{100}-8^2}{7}\)