a: \(=\dfrac{5}{14}\left(\dfrac{2}{9}+\dfrac{7}{9}+1\right)=\dfrac{5}{14}\cdot2=\dfrac{5}{7}\)

b: \(=\dfrac{6}{8}\left(\dfrac{6}{7}-6\right)=\dfrac{6}{8}\cdot\dfrac{-36}{7}=\dfrac{-216}{56}=\dfrac{-27}{7}\)

Học sinh nhẩm và ghi kết quả như sau:

7 x 5 = 35

7 x 6 = 42

7 x 2 = 14

7 x 4 = 28

35 : 7 =5

42 : 7 = 6

14 : 7 = 2

28: 7 =4

35 : 5 = 7

42 : 6 = 7

14 : 2 = 7

28 : 4 = 7

|7 - \(\dfrac{3}{4}\)\(x\)| - \(\dfrac{3}{2}\) = \(\dfrac{1}{\dfrac{1}{2}}\)

|7 - \(\dfrac{3}{4}x\)|  - \(\dfrac{3}{2}\) = 2

|7 - \(\dfrac{3}{4}\)\(x\)| = 2 + \(\dfrac{3}{2}\)

|7 - \(\dfrac{3}{4}x\)| = \(\dfrac{7}{2}\)

\(\left[{}\begin{matrix}7-\dfrac{3}{4}x=\dfrac{7}{2}\\7-\dfrac{3}{4}x=-\dfrac{7}{2}\end{matrix}\right.\) 

\(\left[{}\begin{matrix}\dfrac{3}{4}x=7-\dfrac{7}{2}\\\dfrac{3}{4}=7+\dfrac{7}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{7}{2}\\\dfrac{3}{4}x=\dfrac{21}{2}\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=14\end{matrix}\right.\)

 5  - |\(x-3\)| = 5

       |\(x-3\)| = 5 - 5

      |\(x-3\)| = 0

      \(x-3\) = 0 

      \(x\) = 3

|\(\dfrac{7}{2}\) - 42| + 14 = 14 ( vô lý xem lại đề bài nhé em)

\(\dfrac{2}{3}\) - |\(x+2\)| = 2

     |\(x+2\)| = \(\dfrac{2}{3}\) - 2

    |\(x+2\)| = - \(\dfrac{4}{3}\)  

   |\(x+2\)| ≥ 0 ∀ \(x\)

Vậy \(x\in\) \(\varnothing\)

có đúng ko

\(\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{11}\cdot\dfrac{1}{7}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{7}\cdot\dfrac{1}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)

\(=\dfrac{5}{7}\cdot\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\)

\(=\dfrac{5}{7}\cdot\dfrac{-7}{11}=\dfrac{-5}{11}\)

=4/5(3/7+6/7-2/7)=4/5

4/5.(3/7+6/7-4/14)

=4/5.1

=4/5

4/5

a: Đ

b: S

c: S

d: 88/66=44/33=4/3

a, = \(\frac{11}{21}\)x \(\left(\frac{7}{4}+\frac{5}{4}\right)\)

   = \(\frac{11}{21}\)x 3

  = \(\frac{33}{21}\)

b) = \(\left(\frac{23}{14}-\frac{9}{14}\right)\)x \(\frac{6}{13}\)

    = 1 x \(\frac{6}{13}\)

    = \(\frac{6}{13}\)

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