a)x=1;2;-2(bạn nên tự giải)

b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x

=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)

=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x

=>x = 1/7680

c)=>4^x - 2^x = 6^x - 3^x

=>2^x (2^x-1)= 3^x(2^x-1)

=> 2^x = 3^x

=>x = 0

\(a.\)

\(-\dfrac{2}{3}\cdot\dfrac{?}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{?}{4}=\dfrac{1}{2}:-\dfrac{2}{3}=\dfrac{1}{2}\cdot-\dfrac{3}{2}=-\dfrac{3}{4}\)

\(\Leftrightarrow?=-3\)

\(b.\)

\(\dfrac{?}{3}\cdot\dfrac{5}{8}=-\dfrac{5}{12}\)

\(\Leftrightarrow\dfrac{?}{3}=\dfrac{-5}{12}:\dfrac{5}{8}=\dfrac{-5}{12}\cdot\dfrac{8}{5}=-\dfrac{2}{3}\)

\(\Leftrightarrow?=-2\)

\(c.\)

\(\dfrac{5}{6}\cdot\dfrac{3}{?}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{3}{?}=\dfrac{1}{4}:\dfrac{5}{6}=\dfrac{1}{4}\cdot\dfrac{6}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow?=10\)

Mk gọi ? = x nha

a) \(\dfrac{-2}{3}.\dfrac{x}{4}=\dfrac{1}{2}\) 

           \(\dfrac{x}{4}=\dfrac{1}{2}:\dfrac{-2}{3}\) 

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

⇒x=-3

b)\(\dfrac{x}{3}.\dfrac{5}{8}=\dfrac{-5}{12}\)  

       \(\dfrac{x}{3}=\dfrac{-5}{12}:\dfrac{5}{8}\) 

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

⇒x=-2

c)\(\dfrac{5}{6}.\dfrac{3}{x}=\dfrac{1}{4}\) 

        \(\dfrac{3}{x}=\dfrac{1}{4}:\dfrac{5}{6}\) 

        \(\dfrac{3}{x}=\dfrac{3}{10}\) 

⇒x=10

a: \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}-\dfrac{8}{13}\right)+\left(\dfrac{-18}{35}-\dfrac{17}{35}\right)\)

=1-1-1

=-1

b: \(=\dfrac{-3}{8}\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+\dfrac{-5}{8}=\dfrac{-3}{8}-\dfrac{5}{8}=-1\)

c: \(=\dfrac{4}{4}\cdot\dfrac{5}{15}\cdot\dfrac{11}{11}=\dfrac{1}{3}\)

a) \(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(-\dfrac{5}{13}+-\dfrac{8}{13}\right)+\left(-\dfrac{18}{35}+-\dfrac{17}{35}\right)=1+-1+-1=-1\)

b) \(=-\dfrac{3}{8}\cdot\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}-\dfrac{10}{16}=-1\)

c) \(=\left(-\dfrac{4}{11}\cdot-\dfrac{11}{4}\right)\cdot\dfrac{5}{15}=1\cdot\dfrac{1}{3}=\dfrac{1}{3}\)

a)\(=\left(-\dfrac{5}{13}+\dfrac{-8}{13}\right)+\left(-\dfrac{18}{35}-\dfrac{17}{35}\right)+\left(\dfrac{3}{14}+\dfrac{14}{17}\right)=-1-1+1=-1\)

b)\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)-\dfrac{10}{16}=-\dfrac{3}{8}.1-\dfrac{10}{16}=-\dfrac{6}{16}-\dfrac{10}{16}=-\dfrac{16}{16}=-1\)

c)\(\dfrac{-4.5.11}{11.5.3.-4}=\dfrac{1}{3}\)

\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)

\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)

\(\Leftrightarrow11\left(x-2\right)=-20x+40\)

\(\Leftrightarrow11x-22=-20x+40\)

\(\Leftrightarrow31x=62\)

\(\Rightarrow x=2\)

\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60

=>11x-22=-20x-20+60

=>31x=62

=>x=2

[3/7.4/15+1/3.(9^15)]^0.1/3.6^8/12^4

= 1.1/3.(2.3)^8/(3.4)^4

= 1/3.2^8.3^8/3^4.4^4

= 1/3.2^8.3^8/3^4.2^8

= 1/3.3^8/3^4

= 1/3.3^4=27

(dấu . là nhân nha)

=1.1/3x6^8/6^8

=1/3.1

=1/3

gmae là dễ

\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)

\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)

\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)

\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)

Ta có : M . N = \(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{100}{101}\) 

= \(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{100}{101}\) 

= \(\dfrac{1}{101}\) 

Vậy M . N = \(\dfrac{1}{101}\)