\(a,x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x-\frac{61}{8}=\frac{5}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{10}{8}+\frac{61}{8}=\frac{71}{8}=8\frac{7}{8}\)

\(b,x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x+\frac{43}{5}=\frac{37}{4}\)

=> \(x=\frac{37}{4}-\frac{43}{5}=\frac{13}{20}\)

\(c,\left[x-7\frac{5}{8}\right]:\frac{1}{2}=3\)

=> \(\left[x-\frac{61}{8}\right]=3\cdot\frac{1}{2}\)

=> \(\left[x-\frac{61}{8}\right]=\frac{3}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}=\frac{12}{8}+\frac{61}{8}=\frac{73}{8}=9\frac{1}{8}\)

d, \(\frac{x}{1\cdot3}+\frac{x}{3\cdot5}+\frac{x}{5\cdot7}+...+\frac{x}{97\cdot99}=99\)

=> \(\frac{x}{2}\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{97\cdot99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\left[1-\frac{1}{99}\right]=99\)

=> \(\frac{x}{2}\cdot\frac{98}{99}=99\)

=> \(\frac{98x}{198}=99\)

=>  98x = 99 . 198

=> 98x = 19602

=> x = 19602 : 98 = 9801/49

a) \(x-7\frac{5}{8}=1\frac{1}{4}\)

=> \(x=\frac{5}{4}+\frac{61}{8}\)

=> \(x=\frac{71}{8}\)

b) \(x+7\frac{5}{8}=9\frac{1}{4}\)

=> \(x=\frac{37}{4}-\frac{61}{8}\)

=> \(x=\frac{13}{8}\)

c) \(\left(x-7\frac{5}{8}\right):\frac{1}{2}=3\)

=> \(x-\frac{61}{8}=3.\frac{1}{2}\)

=> \(x-\frac{61}{8}=\frac{3}{2}\)

=> \(x=\frac{3}{2}+\frac{61}{8}\)

=> \(x=\frac{73}{8}\)

d) \(\frac{x}{1.3}+\frac{x}{3.5}+...+\frac{x}{97.99}=99\)

=> \(x.\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\right)=99\)

=> \(\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\right)=99\)

=> \(x\left(1-\frac{1}{99}\right)=99:\frac{1}{2}\)

=> \(x.\frac{98}{99}=198\)

=> \(x=198:\frac{98}{99}=\frac{9801}{49}\)

\(b,\text{ }x+7\frac{5}{8}=9\frac{1}{4}\)

\(x=9\frac{1}{4}-\left(7+\frac{5}{8}\right)\)

\(x=9+\frac{1}{4}-7-\frac{5}{8}\)

\(x=\left(9-7\right)+\left(\frac{1}{4}-\frac{5}{8}\right)\)

\(x=2+-\frac{3}{8}\)

\(x=2\frac{-3}{8}\)

a)15/8+3/4-5/12

=45+18-10/24

=53/24

b)11/24.12/33+5/6

=11.12/12.2.11.3+5/6

=1/6+5/6

=6/6=1

c)15/8+7/24:5/8

=15/8+7/24.8/5

=15/8+7.8/3.8.5

=15/8+7/15

=đề sai, nếu đúng thì như này

=8/15+7/15

=15/15=1

hbhvgvvggvgbhhgvbnj

mà đây là môn toán mà

55

\(6\frac{2}{7}+7\frac{3}{5}+8\frac{6}{9}+9\frac{1}{4}+\frac{3}{5}+\frac{6}{7}=\frac{1}{3}+\frac{3}{4}+1967\)

đề bài như vậy phải ko, đề còn yêu cầu j ko?

a) chứng tỏ : abcabc chia hết cho 11

Ta có 123123:11=11193

Vậy abcabc chia hết cho 11

b)\(\frac{9\cdot15\cdot21\cdot12\cdot20}{5\cdot6\cdot45\cdot18\cdot4}=\frac{9\cdot3\cdot5\cdot3\cdot7\cdot2\cdot2\cdot3\cdot2\cdot2\cdot5}{5\cdot2\cdot3\cdot5\cdot3\cdot3\cdot2\cdot3\cdot3\cdot2\cdot2}\)\(=\frac{7\cdot5}{3}=\frac{35}{3}\)

bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\) 

     \(x\times\dfrac{11}{16}=2\) 

     \(x=2:\dfrac{11}{16}\) 

    \(x=\dfrac{32}{11}\)

Bài 1 : 

 \(\dfrac{x}{16}\times\left(2017-1\right)=2\)

          \(\dfrac{x}{16}\times2016=2\)

                      \(\dfrac{x}{16}=\dfrac{2}{2016}\)

                         \(x=\dfrac{2}{2016}\times16\)

                         \(x=\dfrac{1}{63}\)

1- (5\(\dfrac{4}{9}\) +x+7\(\dfrac{7}{18}\)) : 15\(\dfrac{3}{4}\) = 0 

1- (\(\dfrac{49}{9}+x+\dfrac{133}{18}\)) : \(\dfrac{63}{4}=0\) 

 (\(\dfrac{49}{9}+\dfrac{133}{18}\)+\(x\) ) : \(\dfrac{63}{4}\) = 1 - 0 

       (\(\dfrac{77}{6}\) + \(x\) ) : \(\dfrac{63}{4}\) = 1

         \(\dfrac{77}{6}+x\)         = 1 x \(\dfrac{63}{4}\) 

          \(\dfrac{77}{6}\) + \(x\)          = \(\dfrac{63}{4}\)

                  \(x\)            = \(\dfrac{63}{4}\) - \(\dfrac{77}{6}\)

                  \(x=\) \(\dfrac{35}{12}\)