1) \(5\frac{2}{7}\).\(\frac{8}{11}\)+\(5\frac{2}{7}.\frac{5}{11}+5\frac{2}{7}.\frac{-2}{11}\)

= \(5\frac{2}{7}\).[\(\frac{8}{11}+\frac{5}{11}+\frac{-2}{11}\)]

= \(5\frac{2}{7}\).1

=\(5\frac{2}{7}\)

2)

5 2/7.8/9+5 2/7.5/11-5 2/7.2/11

=5 2/7.(8/11+5/11+2/11)

=5 2/7.15/11

=240/77.

(67/111+2/33-15/117).(1/3-1/4-1/12)

=811/1221.1/12

=811/14652

Ta có \(x.a=b\)

\(x=b:a\)

Do đó \(-\frac{3}{5}.x=\frac{5}{6}\)

\(x=\frac{5}{6}:\frac{-3}{5}\)

\(x=-\frac{1}{2}\)

Tương tự 

2. \(x=\frac{6}{8}:\frac{2}{4}=\frac{3}{2}\)

Bạn tự làm câu 3

1.\(\frac{-3}{5}\).x=\(\frac{5}{6}\)

x=\(\frac{5}{6}\):\(\frac{-3}{5}\)

x=\(\frac{-25}{18}\)

2.\(\frac{2}{4}\):x=\(\frac{6}{8}\)

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

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

x=\(\frac{2}{3}\)

3.\(\frac{67}{89}\).x=\(\frac{12}{13}\)

x=\(\frac{12}{13}\):\(\frac{67}{89}\)

x=\(\frac{1068}{871}\)

K MIK NHÉ

đáp án là 0 nha em

\(C=\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right)x\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)

\(\Rightarrow C=\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right)x\left(\frac{1}{12}-\frac{1}{12}\right)\)

\(\Rightarrow C=\left(\frac{67}{11}+\frac{2}{33}-\frac{15}{117}\right)x0\)

\(\Rightarrow C=0\)

\( a)5\left( {x - 3} \right) - 4 = 2\left( {x - 1} \right) + 7\\ \Leftrightarrow 5x - 15 - 4 = 2x - 2 + 7\\ \Leftrightarrow 5x - 19 = 2x + 5\\ \Leftrightarrow 5x - 2x = 5 + 19\\ \Leftrightarrow 3x = 24\\ \Leftrightarrow x = 8\\ b)\dfrac{{8x - 3}}{4} - \dfrac{{3x - 2}}{2} = \dfrac{{2x - 1}}{2} + \dfrac{{x + 3}}{4}\\ \Leftrightarrow 8x - 3 - \left( {3x - 2} \right).2 = \left( {2x - 1} \right).2 + x + 3\\ \Leftrightarrow 8x - 3 - 6x + 4 = 4x - 2 + x + 3\\ \Leftrightarrow 2x + 1 = 5x + 1\\ \Leftrightarrow 2x - 5x = 0\\ \Leftrightarrow - 3x = 0\\ \Leftrightarrow x = 0 \)

\( c)\dfrac{{2\left( {x + 5} \right)}}{3} + \dfrac{{x + 12}}{2} - \dfrac{{5\left( {x - 2} \right)}}{6} = \dfrac{x}{3} + 11\\ \Leftrightarrow 4\left( {x + 5} \right) + 3\left( {x + 12} \right) - \left[ {5\left( {x - 2} \right)} \right] = 2x + 66\\ \Leftrightarrow 4x + 20 + 3x + 36 - 5x + 10 = 2x + 66\\ \Leftrightarrow 2x + 66 = 2x + 66\\ \Leftrightarrow 0x = 0\left( {VSN} \right)\\ \Leftrightarrow x = 0 \)

\(d)\dfrac{x-10}{1994}+\dfrac{x-8}{1996}+\dfrac{x-6}{1998}+\dfrac{x-4}{2000}+\dfrac{x-2}{2002}=\dfrac{x-2002}{2}+\dfrac{x-2000}{4}+\dfrac{x-1998}{6}+\dfrac{x-1996}{8}+\dfrac{x-1994}{10}\\ \Leftrightarrow \dfrac{x-10}{1994}-1+\dfrac{x-8}{1996}-1+\dfrac{x-6}{1998}-1+\dfrac{x-4}{2000}-1+\dfrac{x-2}{2002}-1=\dfrac{x-2002}{2}-1+\dfrac{x-2000}{4}-1+\dfrac{x-1998}{6}-1+\dfrac{x-1996}{8}-1+\dfrac{x-1994}{10}-1\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}=\dfrac{x-2004}{2}+\dfrac{x-2004}{4}+\dfrac{x-2004}{6}+\dfrac{x-2004}{8}+\dfrac{x-2004}{10}\\ \Leftrightarrow \dfrac{x-2004}{1994}+\dfrac{x-2004}{1996}+\dfrac{x-2004}{1998}+\dfrac{x-2004}{2000}\dfrac{x-2004}{2002}-\dfrac{x-2004}{2}-\dfrac{x-2004}{4}-\dfrac{x-2004}{6}-\dfrac{x-2004}{8}-\dfrac{x-2004}{10}=0\\ \Leftrightarrow \left(x-2004\right)\left(\dfrac{1}{1994}+\dfrac{1}{1996}+\dfrac{1}{1998}+\dfrac{1}{2000}+\dfrac{1}{2002}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}=0\right)\\ \Leftrightarrow x-2004=0\\ \Leftrightarrow x=2004\)

ve 2 bang 0 

con ve 1 o can tinh du co lon 

vi 0 nhan bao nhieu cung bang o

\(=\)\(\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right)\times\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)

\(=\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right)\times0\)

\(=0\)

\(\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)

=\(\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right).\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)

=\(\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right).\left(\frac{1}{12}-\frac{1}{12}\right)\)

=\(\left(\frac{67}{111}+\frac{2}{33}-\frac{15}{117}\right).0\)

=0

Bài 1 :

    \(a+b=3.\left(a-b\right)=\)\(2\frac{a}{b}\)

\(\Rightarrow a+b=3.\left(a-b\right)\)

\(\Rightarrow a+b=3a-3b\)

\(\Rightarrow3a-3b-a-b=0\)

\(\Rightarrow2a-4b=0\)

\(\Rightarrow2.\left(a-2b\right)=0\)

\(\Rightarrow\hept{\begin{cases}a-2b=0\\a=2b\end{cases}}\)

Ta có : \(a+b=\frac{2a}{b}\)

Thay \(a=2b\) vào 

\(\Rightarrow2b+b=\frac{2.23}{b}\)

\(\Rightarrow3b=\frac{4b}{b}\Rightarrow3b=4\)

\(\Rightarrow b=\frac{4}{3}\Rightarrow a=2.\frac{4}{3}=\frac{8}{3}\)

Vậy \(a=\frac{8}{3}\) và \(b=\frac{4}{3}\)

Chúc bạn học tốt ( -_- )

Bài 2 :

\(B=50+\frac{50}{3}+\frac{25}{3}+\frac{20}{4}+\frac{10}{5}+\frac{100}{6.7}+...+\)\(\frac{100}{98.99}+\frac{1}{99}\)

\(B=\frac{100}{2}+\frac{100}{6}+\frac{100}{12}+\frac{100}{20}+\frac{100}{30}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{9900}\)

\(B=\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+\frac{100}{4.5}+\frac{100}{5.6}+\frac{100}{6.7}+...+\frac{100}{98.99}+\frac{100}{99.100}\)

\(B=100.\frac{100}{2}+\frac{100}{2}-\frac{1}{3}+\frac{100}{3}-\frac{100}{4}+\frac{100}{4}-\frac{100}{5}+\frac{100}{5}-\frac{100}{6}+\frac{100}{6}\)\(-\frac{100}{7}+...+\frac{100}{98}+\frac{100}{99}+\frac{100}{99}-1\)

\(B=100-1\)

\(B=99\)

Chúc bạn học tốt ( -_- )