Tìm số tự nhiên x:  \(2^{x-1}+5.2^{x-2}=224\Leftrightarrow2.2^{x-2}+5.2^{x-2}=224\)

\(\Leftrightarrow2^{x-2}.\left(5+2\right)=224\Leftrightarrow2^{x-2}.7=224\)

\(\Rightarrow2^{x-2}=32\Leftrightarrow2^{x-2}=2^5\)\(\Rightarrow x-2=5\Leftrightarrow x=7\)

Vậy x=7

Tìm x biết: \(\frac{3}{7}=\frac{2x+1}{3x+5}\)

\(\Rightarrow3\left(3x+5\right)=7\left(2x+1\right)\Leftrightarrow9x+15=14x+7\)

\(\Leftrightarrow14x+7-\left(9x+15\right)=0\Rightarrow5x+\left(-8\right)=0\)

\(\Leftrightarrow5x=8\Rightarrow x=\frac{8}{5}\)

Vậy x=8/5

phân số thỏa mãn là 35/8 nhé bạn

Ta có: \(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)

Ta lại có: \(\frac{8}{14}< \frac{x}{14}< \frac{x}{10}< \frac{10}{14}\)

\(\Rightarrow\frac{8}{14}< \frac{9}{14}< \frac{9}{10}< \frac{10}{14}\)

Vậy giá trị của x là 9.

Ta có:

\(\frac{4}{7}< \frac{x}{10}< \frac{5}{7}\)

\(\Rightarrow\frac{40}{70}< \frac{7x}{70}< \frac{50}{70}\)

\(\Rightarrow40< 7x< 50\)

Vì \(7x⋮7\)

\(\Rightarrow7x\in\left\{42;49\right\}\)

\(\Rightarrow x\in\left\{6;7\right\}\)

Vậy: x = 6 hoặc x = 7

 x = 6 hoặc 7

12 + x / 43 - x có tổng là 12 + 43 = 55

Tổng số phần bằng nhau :

 2 + 3 = 5 ( phần )

 Giá trị 1 phần :

 55 : 5 = 11

x là :

  ( 11 x 2 ) - 12 = 10 

b ) x/5 < 3/7 <=> 7x/35  < 15/35

=> 7x = 14

       x = 2

c ) 1 < 11/x < 2

     11/5 > 2 

=> 5 < x < 11

x { 6 ; 7 ; 8 ; 9 ; 10 }

Sao nhiều quá vại??

mk lm k nổi đâu

Dài quá nhìn lòi bảng họng lun ak

Bài : 4 

a/ \(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+....+\frac{1}{24\cdot25}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+....+\frac{1}{24}-\frac{1}{25}\)

\(=\frac{1}{5}-\frac{1}{25}\)

\(=\frac{4}{25}\)

b/ \(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{99\cdot101}\)

\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\)

\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)

\(=\frac{1}{1}-\frac{1}{101}\)

\(=\frac{100}{101}\)

c/ \(\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+\frac{5^2}{11\cdot16}+\frac{5^2}{16\cdot21}+\frac{5^2}{21\cdot26}+\frac{5^2}{26\cdot31}\)

\(=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\frac{25}{11\cdot16}+\frac{25}{16\cdot21}+\frac{25}{21\cdot26}+\frac{25}{26\cdot31}\)

\(=\frac{6-1}{1\cdot6}+\frac{11-6}{6\cdot11}+....+\frac{31-26}{26\cdot31}\)

\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+....+\frac{1}{26}-\frac{1}{31}\right)\)

\(=\frac{25}{5}\cdot\left(\frac{1}{1}-\frac{1}{31}\right)\)

\(=\frac{25}{5}\cdot\frac{30}{31}\)

\(=\frac{150}{31}\)

d/ \(\frac{3}{1\cdot3}+\frac{3}{3\cdot5}+\frac{3}{5\cdot7}+....+\frac{3}{49\cdot51}\)

\(=\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+....+\frac{51-49}{49\cdot51}\)

\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{51}\right)\)

\(=\frac{3}{2}\cdot\left(\frac{1}{1}-\frac{1}{51}\right)\)

\(=\frac{3}{2}\cdot\frac{50}{51}\)

\(=\frac{25}{17}\)

e/ \(\frac{1}{7}+\frac{1}{91}+\frac{1}{247}+\frac{1}{475}+\frac{1}{775}+\frac{1}{1147}\)

\(=\frac{1}{1\cdot7}+\frac{1}{7\cdot13}+\frac{1}{13\cdot19}+\frac{1}{19\cdot25}+\frac{1}{25\cdot31}+\frac{1}{31\cdot37}\)

\(=\frac{7-1}{1\cdot7}+\frac{13-7}{7\cdot13}+....+\frac{37-31}{31\cdot37}\)

\(=\frac{1}{6}\cdot\left(1-\frac{1}{7}+\frac{1}{7}-\frac{1}{13}+....+\frac{1}{31}-\frac{1}{37}\right)\)

\(=\frac{1}{6}\cdot\left(1-\frac{1}{37}\right)\)

\(=\frac{1}{6}\cdot\frac{36}{37}\)

\(=\frac{6}{37}\)

Bài 5 : 

b/ \(\left(\frac{2}{11\cdot13}+\frac{2}{13\cdot15}+...+\frac{2}{19\cdot21}\right)-x+4+\frac{221}{231}=\frac{7}{3}\)

\(\left(\frac{13-11}{11\cdot13}+\frac{15-13}{13\cdot15}+...+\frac{21-19}{19\cdot21}\right)-x+4=\frac{7}{3}-\frac{221}{231}\)

\(\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{19}-\frac{1}{21}\right)-x+4=\frac{106}{77}\)

\(\left(\frac{1}{11}-\frac{1}{21}\right)-x=\frac{106}{77}-4\)

\(\frac{10}{231}-x=-\frac{202}{77}\)

\(x=\frac{10}{231}-\left(-\frac{202}{77}\right)\)

\(x=\frac{8}{3}\)

Ta quy đ

Ta có : 

\(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{n^2-1}{n^2}\)

\(S=\frac{4-1}{4}+\frac{9-1}{9}+\frac{16-1}{16}+...+\frac{n^2-1}{n^2}\)

\(S=\frac{2^2-1}{2^2}+\frac{3^2-1}{3^2}+\frac{4^2-1}{4^2}+...+\frac{n^2-1}{n^2}\)

\(S=\frac{2^2}{2^2}-\frac{1}{2^2}+\frac{3^2}{3^2}-\frac{1}{3^2}+\frac{4^2}{4^2}-\frac{1}{4^2}+...+\frac{n^2}{n^2}-\frac{1}{n^2}\)

\(S=1-\frac{1}{2^2}+1-\frac{1}{3^2}+1-\frac{1}{4^2}+...+1-\frac{1}{n^2}\)

\(S=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)\)

Vì từ \(2\) đến \(n\) có \(n-2+1=n-1\) số \(1\) nên : 
\(S=n-1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)< n-1\) \(\left(1\right)\)

Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\) ta lại có : 

\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(A< 1-\frac{1}{n}< 1\)

\(\Rightarrow\)\(S=n-1-A>n-1-1=n-2\) 

\(\Rightarrow\)\(S>n-2\) \(\left(2\right)\)

Từ (1) và (2) suy ra : 

\(n-2< S< n-1\)

Vì \(n>3\) nên \(S\) không là số tự nhiên 

Vậy \(S\) không là số tự nhiên 

Chúc bạn học tốt ~