d) 7/18 . x - 2/3 = 5/18

    7/18 . x         = 5/18+2/3

                         = 17/18 : 7/18

                         = 17/7

e) 4/9 - 7/8 . x = -2/3

            7/8 . x = 4/9 - -2/3

                       = 10/9 : 7/8

                       = 80/63

f) 1/6 + -5/7 : x = -7/18

            -5/7 : x = -7/18 - 1/6

            -5/7 : x = -5/9

                        = -5/7 : -5/9

                        = 9/7

Sau đó bạn thử lại kết quả nha!

a: \(B=\left(-\dfrac{1}{5}-\dfrac{5}{7}+\dfrac{-3}{35}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{2}\right)+\dfrac{1}{41}\)

\(=\dfrac{-7-25-3}{35}+\dfrac{3+2+1}{6}+\dfrac{1}{41}=\dfrac{42}{41}-1=\dfrac{1}{41}\)

Đề không sai đâu !!

Bài a làm gì có z

Thay a,b,c lần lượt vào biểu thức...

Tính được kết quả:

a) A= \(-\frac{7}{10}\)

b) B= \(-\frac{2}{7}\)

c) C= 0

a) Thay a= \(-\frac{6}{5}\)vào BT A ta có:

\(\left(-\frac{6}{5}\right).\frac{1}{2}-\left(-\frac{6}{5}\right).\frac{2}{3}+\left(-\frac{6}{5}\right).\frac{3}{4}\)= \(-\frac{7}{10}\)

Các bài dưới lần lượt thế thôi bạn

Ta làm như sau:

   \(\frac{6}{18}\)+\(\frac{6}{54}\)+\(\frac{6}{108}\)+...+\(\frac{6}{990}\)

=\(\frac{6}{3.6}\)+\(\frac{6}{6.9}\)+\(\frac{6}{9.12}\)+...\(\frac{6}{30.33}\)

=2 (\(\frac{3}{3.6}\)+\(\frac{3}{6.9}\)+\(\frac{3}{9.12}\)+...+\(\frac{3}{30.33}\)

=2 (\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))

=2 ( \(\frac{1}{3}-\frac{1}{33}\))

=2.\(\frac{10}{33}\)=\(\frac{2.10}{33}\)=\(\frac{20}{33}\)

\(\frac{6}{18}+\frac{6}{54}+\frac{6}{108}+...+\frac{6}{990}\)

=\(\frac{6}{3.6}+\frac{6}{6.9}+\frac{6}{9.12}+...+\frac{6}{30.33}\)

= 2.(\(\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\))

=2.(\(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\))

=2.[\(\frac{1}{3}+\left(\frac{-1}{6}+\frac{1}{6}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)+...+\left(\frac{-1}{30}+\frac{1}{30}\right)+\frac{-1}{33}\)]

=2.\(\left[\frac{1}{3}+\frac{-1}{33}\right]\)

=2.\(\left[\frac{11}{33}+\frac{-1}{33}\right]\)

=2.\(\frac{10}{33}\)

=\(\frac{20}{33}\)

\(\frac{5.\left(2^2.3^2\right)^9\left(2^2\right)^6-2.\left(2^2.3\right)14.3^{ }^4}{22}\)

khó quá                

HREYHRFGT

daica nghiêm túc đi

a) \(\frac{1}{n}\) - \(\frac{1}{n+1}\) = \(\frac{n+1}{n\left(n+1\right)}\) - \(\frac{n}{n\left(n+1\right)}\) = \(\frac{1}{n\left(n+1\right)}\) = \(\frac{1}{n}\) . \(\frac{1}{n+1}\) =>đpcm

b) A= \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\)+...+\(\frac{1}{8}\) - \(\frac{1}{9}\) +\(\frac{1}{9}\)

= \(\frac{1}{2}\) + \(\frac{1}{9}\)= \(\frac{11}{18}\)