a) \(x:\left(-\frac{1}{3}\right)^3=-\frac{1}{3}\)

\(\Rightarrow x=\left(-\frac{1}{3}\right).\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow x=\left(-\frac{1}{3}\right)^4\)

\(\Rightarrow x=\frac{1}{81}\)

Vậy \(x=\frac{1}{81}.\)

b) \(\frac{3}{4}:\frac{41}{99}=x:\frac{75}{90}\)

\(\Rightarrow\frac{297}{164}=x:\frac{75}{90}\)

\(\Rightarrow x=\frac{297}{164}.\frac{75}{90}\)

\(\Rightarrow x=\frac{495}{328}\)

Vậy \(x=\frac{495}{328}.\)

c) \(x+\left|-\frac{1}{2}\right|=3\frac{1}{3}-4\frac{1}{2}\)

\(\Rightarrow x+\frac{1}{2}=\frac{10}{3}-\frac{9}{2}\)

\(\Rightarrow x+\frac{1}{2}=-\frac{7}{6}\)

\(\Rightarrow x=\left(-\frac{7}{6}\right)-\frac{1}{2}\)

\(\Rightarrow x=-\frac{5}{3}\)

Vậy \(x=-\frac{5}{3}.\)

Chúc bạn học tốt!

=>x-100=0

=>x=100

ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)

\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)

ta gọi B là biểu thức thứ2

\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)

\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)

\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)

\(\Rightarrow x=1\)

mk nghĩ vậy bạn ạ, mk mong nó đúng

a, 2/1/3:1/3=7/9:x                                    b, x:1/3=12/99:15/90

   7           = 7/9 : x                                      x :1/3= 8/11

  x          = 7/9:7                                           x    = 8/11 * 1/3

  x          = 1/9                                               x =   8/33

c, 0,15:x=3/1/3:2,25                       d, 3/4:0,75=x:75/90

    0,15:x =40/27                                 1         =x:75/90

    x        = 0,15:40/27                         x = 1*75/90

    x         = 81/800                              x = 75/90

e, x/-15=-60/x                              f, -2/x=-x/8

=>x*x=-15*(-60)                           => (-2)*8=x*-x

=>x²=900                                   => -16 = -x²

=>x²=30²    hoặc x²=(-30)²          => 16=x²

=> x=30 hoặc x= -30                 => x²=4² hoặc x²=(-4)²

                                                => x=4 hoặc x=-4

a)\(2\frac{1}{3}:\frac{1}{3}=\frac{7}{9}:x\)

\(\frac{7}{3}\times3=\frac{7}{9}:x\)

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

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

\(x=\frac{7}{9}\times\frac{1}{7}\)

\(x=\frac{1}{9}\)

b)\(x:\frac{1}{3}=\frac{12}{99}:\frac{15}{90}\)

\(x:\frac{1}{3}=\frac{12}{99}\times\frac{90}{15}\)

\(x:\frac{1}{3}=\frac{8}{11}\)

\(x=\frac{8}{11}\times\frac{1}{3}\)

\(x=\frac{8}{33}\)

x=-100

\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)

\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)

\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)

\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)

a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)

\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)

\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)

b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)

\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)

\(\simeq40.39\)