\(=\sqrt{5}-1-\sqrt{5}-1-3\sqrt{5}=-3\sqrt{5}-2\)

77/45

7/6

29/10

\(\dfrac{3}{5}+\dfrac{4}{3}\times\dfrac{5}{6}=\dfrac{3}{5}+\dfrac{10}{9}=\dfrac{27}{45}+\dfrac{50}{45}=\dfrac{77}{45}\)

\(2-\dfrac{7}{6}+\dfrac{1}{3}=\dfrac{12}{6}-\dfrac{7}{6}+\dfrac{2}{6}=\dfrac{7}{6}\)

\(\dfrac{7}{6}\times3-\dfrac{3}{5}=\dfrac{7}{2}-\dfrac{3}{5}=\dfrac{35}{10}-\dfrac{6}{10}=\dfrac{29}{10}\)

giải chi tiết đi

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

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

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

Đề bài yêu cầu gì?

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3>=-5\\2x-3< =5\end{matrix}\right.\Leftrightarrow-1< =x< =4\)

Giải:

B=1,4 .15/49 - (4/5+2/3) :2 1/5

B=7/5.15/49 - 22/15 : 11/5

B=3/7 - 2/3

B=-5/21

Chúc bạn học tốt!

Bạn nên đánh lại đề bài nhé.

a) \(\dfrac{3}{5}\times x=\dfrac{4}{7}\)

            \(x=\dfrac{4}{7}\div\dfrac{3}{5}\)

            \(x=\dfrac{20}{21}\)

b) \(\dfrac{1}{8}\div x=\dfrac{1}{5}\)

           \(x=\dfrac{1}{8}\div\dfrac{1}{5}\)

          \(x=\dfrac{5}{8}\)

20/21

5/8

20/21

5/8

\(=\left(\dfrac{3+8}{12}\right)\cdot\dfrac{4}{5}=\dfrac{11}{12}\cdot\dfrac{4}{5}=\dfrac{44}{60}=\dfrac{11}{15}\)