a.\(\dfrac{1}{3}\) + x  = \(\dfrac{5}{6}\)

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

      x = \(\dfrac{1}{2}\)

b. | x-1| - \(\dfrac{2}{5}\) = \(\dfrac{11}{10}\) 

   | x-1|        = \(\dfrac{11}{10}\) + \(\dfrac{2}{5}\)

  |x-1|        = \(\dfrac{3}{2}\)

\(\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=-\dfrac{3}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{3}{2}+1\\x=-\dfrac{3}{2}+1\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c, \(\dfrac{1}{3}\) + \(\dfrac{2}{3}\) ( \(\dfrac{x}{2}\) + 3) = 1

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

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

                   \(\dfrac{x}{2}\) + 3 = 1

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

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

                     \(x\) = -4

d, \(\dfrac{x+2}{3}\) = \(\dfrac{27}{x+2}\)

(x+2)² = 27.3

(x+2) =9²

\(\left[{}\begin{matrix}x+2=9\\x+2=-9\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x=-11\end{matrix}\right.\)

1: x=3/4-1/2=3/4-2/4=1/4

2: x-1/5=2/11

=>x=2/11+1/5=21/55

3: x-5/6=16/42-8/56

=>x-5/6=8/21-4/28=5/21

=>x=5/21+5/6=15/14

4: x/5=5/6-19/30

=>x/5=25/30-19/30=6/30=1/5

=>x=1

5: =>|x|=1/3+1/4=7/12

=>x=7/12 hoặc x=-7/12

6: x=-1/2+3/4

=>x=3/4-1/2=1/4

11: x-(-6/12)=9/48

=>x+1/2=3/16

=>x=3/16-1/2=-5/16

1)x= 1/4

2)x= 2/11+ 1/5

   x= 21/55

3)x - 5/6 = 5/21

   x         = 5/21+5/6

   x         = 15/14

4)x/5 = 5/6 + -19/30

   x:5 = 1/5

   x    = 1/5.5

   x    = 1

5) |x| - 1/4 = 6/18

    |x|           = 6/18 - 1/4

    |x|            =7/12

⇒x= 7/12 hoặc -7/12

6)x = -1/2 +3/4

   x= 1/4

7) x/15 = 3/5 + -2/3

   x:15  = -1/15

  x        = -1/15. 15

  x        = -1

8)11/8 + 13/6 = 85/x  

       85/24      = 85/x

  ⇒      x           = 24

9) x - 7/8 = 13/12

   x          = 13/12 + 7/8

   x          = 47/24

10)x - -6/15 = 4/27  

     x            = 4/27 + (-6/15)

    x             = -34/135

11) -(-6/12)+x = 9/48

                    x= 9/48 - 6/12

                    x = -5/16

12) x - 4/6 = 5/25 + -7/15

      x -4/6  =  -4/15

     x           = -4/15 + 4/6

    x             = 2/5

a.-1,75-(-\(\dfrac{1}{9}\)-2\(\dfrac{1}{8}\))
-1,75-\(\dfrac{1}{9}+\dfrac{17}{8}\)
\(-\dfrac{7}{4}-\dfrac{1}{9}+\dfrac{17}{8}\)
\(\dfrac{-126}{72}-\dfrac{8}{72}+\dfrac{153}{72}\)
=\(\dfrac{19}{72}\)

b.\(\dfrac{-1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
\(\dfrac{-1}{12}-\dfrac{21}{8}+\dfrac{1}{3}\)
\(\dfrac{-2}{24}-\dfrac{63}{24}+\dfrac{64}{24}\)
=\(\dfrac{-1}{24}\)

Bài 2:

\(\dfrac{a+b}{a-b}=\dfrac{c+a}{c-a}\)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a-b}{c-a}=\dfrac{a+b+a-b}{c+a+c-a}=\dfrac{a}{c}\) (T/c dãy tỷ số = nhau)

\(\Rightarrow\dfrac{a+b}{c+a}=\dfrac{a}{c}\Rightarrow c\left(a+b\right)=a\left(c+a\right)\)

\(\Rightarrow ac+bc=ac+a^2\Rightarrow a^2=bc\)

c: Ta có: \(\dfrac{2}{5}\cdot\left[\left(\dfrac{3}{5}\right)^2:\left(-\dfrac{1}{5}\right)^2-7\right]\cdot\left(1000\right)^0\cdot\left|-\dfrac{11}{15}\right|\)

\(=\dfrac{2}{5}\cdot\left(\dfrac{9}{25}:\dfrac{1}{25}-7\right)\cdot1\cdot\dfrac{11}{15}\)

\(=\dfrac{2}{5}\cdot\dfrac{11}{15}\cdot2\)

\(=\dfrac{44}{75}\)

Cảm ơn bạn lần nữa! 

Bài 2: 

a: \(\left(3x-3\right)^2-\left(5x-3\right)^2=0\)

\(\Leftrightarrow\left(3x-3-5x+3\right)\left(3x-3+5x-3\right)=0\)

\(\Leftrightarrow-2x\left(8x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\)

CÁC BẠN CỐ GẮNG GIÚP  MÌNH VỚI NHÉ

a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)

=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)

=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)

Vậy x = 2/5

b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)

=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)

=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)

Vậy x = -6/11

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

=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)

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

=> \(\frac{2}{3}-\frac{4}{3}x=5\)

=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)

Vậy x = -13/4

d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)

=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)

=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)

=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)

=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)

=> \(3\left(229+45x\right)=4\left(61+30x\right)\)

=> \(687+135x=244+120x\)

=> \(687+135x-244-120x=0\)

=> \(\left(687-244\right)+\left(135x-120x\right)=0\)

=> \(443+15x=0\)

=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)

Vậy x = -443/15

a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)

=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)

b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)

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

=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1

c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)

=> x - 1= 2 => x = 3

d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)

=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)

2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)

Vì 3³³ > 3³² => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)

b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}

nhjeu wa bạn giải 1 mjk luôn đi