x=\(-\frac{125}{27}\)

\(x=-\frac{125}{27}\)

x = âm 125 / 27

k đii

\(\left(-\frac{5}{3}\right)^3=\left(-\frac{5}{3}\right).\left(-\frac{5}{3}\right).\left(-\frac{5}{3}\right)=-\frac{125}{27}\)

                           Đáp số: \(-\frac{125}{27}\)

Chúc bạn học tốt nha  Linh Trần Thị Thùy

Ta có : 

\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)

\(\Leftrightarrow\)\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)

\(\Leftrightarrow\)\(\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)=32-35+3\)

\(\Leftrightarrow\)\(\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=-3+3\)

\(\Leftrightarrow\)\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)

Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)

Nên \(x+2005=0\)

\(\Rightarrow\)\(x=-2005\)

Vậy \(x=-2005\)

Chúc bạn học tốt ~ 

Ta có: \(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)

\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)

\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=-3\)

\(\Rightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1=-3+3\)

\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}+\frac{x+3+2002}{2002}=0\)

\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=0\)

\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)

Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)

Nên x + 2005 = 0

=> x                = -2005

Vậy x = -2005

\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)

\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1+32=32\) 

\(\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=32-32\) 

\((x+2005).(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002})=0\) 

Mà  \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)

\(\Rightarrow x+2005=0\) 

\(\Rightarrow x=-2005\)

Vậy x=-2005

a) (1,5 . 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125

=> (2,85 - x - 0,5) : 0,25 = 60

=> (2,85 - 0,5) - x = 60 . 0,25

=> 2,35 - x = 15

=> x = 2,35 - 15

=> x = -12,65

Vậy x = -12,65

b) \(1-\left(5\frac{2}{9}+x-7\frac{7}{18}\right)\div2\frac{1}{6}=0\)

\(\Rightarrow\left(5\frac{2}{9}-7\frac{7}{18}+x\right)\div2\frac{1}{6}=1-0\)

\(\Rightarrow\left(\frac{47}{9}-\frac{133}{18}+x\right)\div2\frac{1}{6}=1\)

\(\Rightarrow\frac{-13}{6}+x=2\frac{1}{6}\)

\(\Rightarrow x=2\frac{1}{6}-\frac{-13}{6}\)

\(\Rightarrow x=\frac{13}{6}+\frac{13}{6}\)

\(\Rightarrow x=\frac{26}{6}\)

\(\Rightarrow x=\frac{13}{3}\)

Vậy \(x=\frac{13}{3}\)

c) \(35\left(2\frac{1}{5}-x\right)=32\)

\(\Rightarrow2\frac{1}{5}-x=32\div35\)

\(\Rightarrow\frac{11}{5}-x=\frac{32}{35}\)

\(\Rightarrow x=\frac{11}{5}-\frac{32}{35}\)

\(\Rightarrow x=\frac{9}{7}\)

Vậy \(x=\frac{9}{7}\)

d) \(\frac{4}{3}+\left(x\div2\frac{2}{3}-0,5\right).1\frac{35}{55}=0,6\)

\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{3}{5}-\frac{4}{3}\)

\(\Rightarrow\left(x\div\frac{8}{3}-\frac{1}{2}\right).\frac{18}{11}=\frac{-11}{15}\)

\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-11}{15}\div\frac{18}{11}\)

\(\Rightarrow x\div\frac{8}{3}-\frac{1}{2}=\frac{-121}{270}\)

\(\Rightarrow x\div\frac{8}{3}=\frac{-121}{270}+\frac{1}{2}\)

\(\Rightarrow x\div\frac{8}{3}=\frac{7}{135}\)

\(\Rightarrow x=\frac{7}{135}.\frac{8}{3}\)

\(\Rightarrow x=\frac{56}{405}\)

Vậy \(x=\frac{56}{405}\)

e) \(1\frac{1}{3}.2\frac{2}{4}\div\frac{5}{6}.1\frac{1}{11}=11-5\div x\)

\(\Rightarrow\frac{4}{3}.\frac{5}{2}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)

\(\Rightarrow\frac{10}{3}\div\frac{5}{6}.\frac{12}{11}=11-5\div x\)

\(\Rightarrow4.\frac{12}{11}=11-5\div x\)

\(\Rightarrow11-5\div x=\frac{48}{11}\)

\(\Rightarrow5\div x=11-\frac{48}{11}\)

\(\Rightarrow5\div x=\frac{73}{11}\)

\(\Rightarrow x=5\div\frac{73}{11}\)

\(\Rightarrow x=\frac{55}{73}\)

Vậy \(x=\frac{55}{73}\)

a) (1,5 * 1,9 - x - 0,5) : 0,25 = 7,5 : 0,125

(2,85 - x - 0,5) : 0,25 = 60

(2,85 - x - 0,5) = 60 x 0,25

(2,85 - x - 0,5) = 15

2,35 - x = 15

x = 2,35 - 15

x = -12,65

b) 1 - \(\left(5\frac{2}{9}+x-7\frac{7}{18}\right):2\frac{1}{6}=0\)

1 - \(\left(\frac{47}{9}+x-\frac{133}{18}\right):\frac{13}{6}=0\)

1 - \(\left(\frac{47}{9}+x-\frac{133}{18}\right)=0\)

\(\frac{47}{9}+x-\frac{133}{18}=1\)

\(x-\frac{13}{6}=1\)

\(x=1+\frac{13}{6}\)

\(x=\frac{19}{6}\)

loại trừ bớt đi rùi tìm x 

\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{x.\left(x+2\right)}=\frac{24}{35}\)

\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{x.\left(x+2\right)}\right)=\frac{24}{35}\)

\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)

\(\frac{3}{10}-\frac{3}{2x+4}=\frac{24}{35}\)

\(\frac{3}{2x+4}=\frac{-27}{70}\)

tự làm nốt

\(\frac{3}{35}+\frac{3}{63}+\frac{3}{99}+....+\frac{3}{x\left(x+2\right)}=\frac{24}{35}\)

\(\Leftrightarrow3\left(\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+....+\frac{1}{x\left(x+2\right)}\right)=\frac{24}{25}\)

\(\Leftrightarrow3\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{x\left(x+2\right)}\right)=\frac{24}{35}\)

\(\Leftrightarrow3\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{24}{35}\)

\(\Leftrightarrow3\left(\frac{1}{5}-\frac{1}{x+2}\right)=\frac{24}{35}\)

\(\Leftrightarrow\frac{3}{5}-\frac{3}{x+2}=\frac{24}{35}\)

\(\Leftrightarrow\frac{3}{x+2}=\frac{3}{5}-\frac{24}{35}\)

\(\Leftrightarrow\frac{3}{x+2}=\frac{21-24}{35}\)

\(\Leftrightarrow\frac{3}{x+2}=\frac{3}{-35}\)

\(\Rightarrow x+2=-35\Leftrightarrow x=-35-2\Leftrightarrow x=-37\)

Vậy \(x=-37\)