a: =>4y+15/16=1

=>4y=1/16

hay y=1/64

b: =>10y+1023/1024=1

=>10y=1/1024

hay  y=1/10240

\(S=\dfrac{1}{2^2}\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)

=>\(S< =\dfrac{1}{4}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\right)\)

=>\(S< =\dfrac{1}{4}\cdot\left(1-\dfrac{1}{n}\right)=\dfrac{1}{4}\cdot\dfrac{n-1}{n}< =\dfrac{1}{4}\)

\(a.x+\dfrac{1}{6}=-\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{13}{24}\)

\(b.2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

\(\Leftrightarrow2-\dfrac{3}{4}+x=\dfrac{7}{12}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c.\dfrac{1}{2}x+\dfrac{1}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{5}{8}x=\dfrac{3}{4}\)

\(\Leftrightarrow x=\dfrac{6}{5}\)

\(d.75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}-\left(\dfrac{-1}{2}\right)^2\)

\(=\dfrac{75}{100}-\dfrac{3}{2}+\dfrac{1}{2}:\dfrac{5}{12}-\dfrac{1}{4}\)

\(=-\dfrac{3}{4}+\dfrac{6}{5}-\dfrac{1}{4}\)

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

a) \(x+\dfrac{1}{6}=\dfrac{-3}{8}\)

            \(x=\dfrac{-3}{8}-\dfrac{1}{6}\)

           \(x=\dfrac{-13}{24}\)

vậy x =....

b) \(2-\left(\dfrac{3}{4}-x\right)=\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=2-\dfrac{7}{12}\)

             \(\dfrac{3}{4}-x=\dfrac{17}{12}\)

                    \(x=\dfrac{3}{4}-\dfrac{17}{12}\)

                   \(x=\dfrac{-2}{3}\)

vậy x =....

a: \(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{10}-\dfrac{3}{202}=\dfrac{150}{101}\)

b: 

1. Tìm n, biết:

a) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)

(-2)^{n + 2} = (-2)⁵

n + 2 = 5

n = 5 - 2

n = 3.

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow\dfrac{2^3}{2^n}=2\)

\(\Rightarrow\) 2ⁿ . 2 = 2³

n + 1 = 3

n = 3 - 1

n = 2.

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)

2n - 1 = 3

2n = 3 + 1

2n = 4

n = 4 : 2

n = 2.

2. Tính:

a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)

\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)

\(=\left(\dfrac{1}{2}\right)^7\)

\(=\dfrac{1}{128}\)

b) 27³ : 9³

= (3³)³ : (3²)³

= 3⁹ : 3⁶

= 3³

= 27

c) 125² : 25³

= (5³)² : (5²)³

= 5⁶ : 5⁶

= 1

d) \(\dfrac{27^2.8^5}{6^6.32^3}\)

\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)

\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)

\(=\dfrac{3^6}{6^6}\)

\(=\dfrac{1}{64}.\)

B2 :

b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)

c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1

@Luân Đào

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)

\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)