a) 2/3=36/54  => x=36

b)15/6=10/3  =>x=3

\(a,\frac{2}{3}=\frac{x}{54}\Rightarrow x=\frac{2\times54}{3}\Rightarrow x=36\)

\(b,\frac{10}{x}=\frac{15}{6}\Rightarrow x=\frac{10\times6}{15}\Rightarrow x=4\)

\(c,\frac{2}{3}< \frac{x}{6}< 1\Rightarrow\frac{2}{3}\times6< x< 1\times6\Rightarrow4< x< 6\)

a, 10 + 2x = 45 ÷ 43

10 + 2x = 45/43

2x = 45/43 - 10

2x = -385/43

x = -385/43 ÷ 2

x = -385/86

b, 2x - 138 = 23.32

2x - 138 = 736

2x = 736 + 138

2x = 874

x = 874 ÷ 2

x = 437

c, 231 - ( x - 6 ) = 1339 ÷ 13

231 - ( x - 6 ) = 103

x - 6 = 231 - 103

x - 6 = 128

x = 128 + 6

x = 134

d, x + 5.2 - ( 32 + 16.3 ÷ 6 - 15 ) = 0

x + 10 - 25 = 0

x + 10 = 0 + 25

x + 10 = 25

x = 25 - 10

x = 15

a) 10 + 2 . x = 45 : 43

            2 . x = 45/43 -10

            2 . x = -385/43

                 x = -385/43 : 2

                 x = -385/86

b) 2 . x - 138 = 23 . 32

    2 . x - 138 = 736

             2 . x = 736+138

             2 . x = 874

                  x = 874 : 2

                  x = 437

c) 231 - (x -6) = 1339 : 13

    231 - (x -6) = 103

               x - 6 = 231 -103

               x - 6 = 98

                    x = 98 +6

                    x = 104

d) x + 5 . 2 - (32 +16 . 3 : 6 - 15) = 0

        x + 5 . 2 - (32 + 48 : 6 - 15) = 0

               x + 5 . 2 - (32 + 8 - 15) = 0

                               x + 5 . 2 - 25 = 0

                               x + 10 - 25 = 0

                                      x + 10 = 0 + 25

                                      x + 10 = 25

                                              x = 25 - 10

                                              x = 15

\(3x\left(2x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=-1\\x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)

\(\frac{\frac{6}{5}+\frac{6}{35}-\frac{6}{125}-\frac{6}{2009}-\frac{6}{2011}}{\frac{7}{5}+\frac{7}{35}-\frac{7}{125}-\frac{7}{2009}-\frac{7}{2011}}\)

\(=\frac{6.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}{7.(\frac{1}{5}+\frac{1}{35}-\frac{1}{125}-\frac{1}{2009}-\frac{1}{2011})}\)

\(=\frac{6}{7}\)

Tìm x

\(a,3x(2x+1)=0\)

\(\Rightarrow\hept{\begin{cases}3x=0\\2x+1=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-1}{2}\end{cases}}\)

Vậy \(x=0\)hoặc \(x=\frac{-1}{2}\)

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

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

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

\(\frac{4}{3}x=\frac{2}{3}-5\)

\(\frac{4}{3}x=\frac{-13}{3}\)

\(x=\frac{-13}{3}\div\frac{4}{3}\)

\(x=\frac{-13}{4}\)

Chúc ban học tốt

a) 2x-138=2³.3²

=>2x-138=8.9

=>2x-138=72

=>2x=72+138

=>2x=210

=>x=210:2

=>x=105

b)231-(x-6)=1339:13

=>231-(x-6)=103

=>x-6=231-103

=>x-6=128

=>x=128+6

=>x=134

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{2001}{2003}\)

\(\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}\right)=\frac{1}{2}\cdot\frac{2001}{2003}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{2001}{4006}\)

\(\frac{1}{x+1}=\frac{1}{2003}\)

\(\Rightarrow x+1=2003\)

\(x=2002\)

Vậy x = 2002

Bài này lớp 6 thật à bạn. 

đặt A=(x+1)+(X+2)+(x+3)+....+(x+99)

=> A= x+1+x+2+x+3+....+x+100

=x+x+x+x+...+x+(1+2+3+4+..+99)( có 99x)

=> 99x+4950=0

=> 99x=-4950

=> x=-50