Bài 1:

Ta có:

\(\left(\dfrac{ab}{2}-\dfrac{6ab}{7}\right):\dfrac{5b^2}{14}=\left(\dfrac{7ab}{14}-\dfrac{12ab}{14}\right).\dfrac{14}{5b^2}\)

\(=\dfrac{-5ab}{14}.\dfrac{14}{5b^2}=\dfrac{-a}{b}\)(1)

Thay \(a=\dfrac{2007}{2010};b=\dfrac{2011}{2010}\) vào (1) ta được:

\(\dfrac{-\dfrac{2007}{2010}}{\dfrac{2011}{2010}}=-\dfrac{2007}{2011}\)

Vậy......................

Chúc bạn học tốt!!!

Bài 2:

\(\left(-1\dfrac{1}{2}:\dfrac{3}{-4}\right).\left(-4\dfrac{1}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -\dfrac{1}{2}.\dfrac{3}{4}:\dfrac{1}{8}+1\)

\(\Rightarrow2.\left(-\dfrac{9}{2}\right)-\dfrac{1}{4}< \dfrac{x}{8}< -3+1\)

\(\Rightarrow-\dfrac{37}{4}< \dfrac{x}{8}< -2\)

\(\Rightarrow\dfrac{-74}{8}< \dfrac{x}{8}< -\dfrac{16}{8}\)

\(\Rightarrow-74< x< -16\)

\(\Rightarrow x\in\left\{-73;-72;-71;....;-18;-17\right\}\)

Vậy..............................

Chúc bạn học tốt!!!

Bài 3:

\(\dfrac{x+3}{2x-3}< 0\)

Vì \(x+3>2x-3\) nên để \(\dfrac{x+3}{2x-3}< 0\) thì:

\(x+3< 0\Rightarrow x< -3\)

Vậy.................

Chúc bạn học tốt!!!

\(\dfrac{5}{8}:\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{5}{8}\times\dfrac{4}{3}-\dfrac{1}{6}=\dfrac{5}{6}-\dfrac{1}{6}=\dfrac{4}{6}=\dfrac{2}{3}\)

\(\left(\dfrac{3}{14}+\dfrac{1}{2}\right)\times\dfrac{7}{5}=\left(\dfrac{3}{14}+\dfrac{7}{14}\right)\times\dfrac{7}{15}=\dfrac{10}{14}\times\dfrac{7}{15}=\dfrac{5}{7}\times\dfrac{7}{15}=\dfrac{5}{12}=\dfrac{1}{3}\)

a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

\(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{6}=\dfrac{-6}{6}\cdot\dfrac{1}{x-2}=\dfrac{-1}{x-2}\)

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

A= 4/7.

Biết có cái

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

Lời giải:

Cách 1: Tính giá trị từng biểu thức trong ngoặc

A=

Cách 2: Bỏ dấu ngoặc rồi nhóm các số hạng thích hợp

A =

= (6-5-3) -

= -2 -0 - = - (2 + ) = -2

cách 1:

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{6}{1}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{5}{1}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{3}{1}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{18}{3}-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(\dfrac{15}{3}+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(\dfrac{9}{3}-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{16}{3}+\dfrac{1}{2}\right)-\left(\dfrac{20}{3}-\dfrac{3}{2}\right)-\left(\dfrac{2}{3}+\dfrac{5}{2}\right)\)

= \(\left(\dfrac{32}{6}+\dfrac{3}{6}\right)-\left(\dfrac{40}{6}-\dfrac{9}{6}\right)-\left(\dfrac{4}{6}+\dfrac{15}{6}\right)\)

= \(\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)

= \(-\dfrac{15}{6}=-\dfrac{5}{2}\)

cách 2:

A = \(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)

= \(\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)-\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)

= \(\left(-2\right)-0+\dfrac{1}{2}\)

= \(-\dfrac{5}{2}\)

a) đặt mẫu chứng là x-2

a) = (\(-\dfrac{141}{20}\)- \(\dfrac{1}{4}\)) : (-5) + \(\dfrac{1}{15}\) - \(\dfrac{1}{15}\)

    = \(-\dfrac{73}{10}\) : - 5

    = \(\dfrac{73}{50}\)

b) = \(\left(\dfrac{3}{25}-\dfrac{28}{25}\right)\). \(\dfrac{7}{3}\) : \(\left(\dfrac{7}{2}-\dfrac{11}{3}.14\right)\)

    = \(-\dfrac{7}{3}\) . \(-\dfrac{6}{287}\)

    = \(\dfrac{2}{41}\)

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)