= \(\frac{17}{8}:\frac{25}{14}-\left(15-\frac{40}{3}\right):\frac{25}{6}\)

= \(\frac{17}{8}.\frac{14}{25}-\left(\frac{45}{3}-\frac{40}{3}\right).\frac{6}{25}\)

= \(\frac{119}{100}-\frac{5}{3}.\frac{6}{25}\)  =  \(\frac{119}{100}-\frac{2}{5}\)

=  \(\frac{119}{100}-\frac{40}{100}=\frac{79}{100}\)

Chúc bạn Hk tốt!!!!!

=\(\frac{79}{100}\)

Nhưng mk cần các cách làm ra bạn ơi

Đề bài là:Tính các giá trị biểu thức sau ạ

a: \(=\left(9-\dfrac{13}{18}\right):\dfrac{325}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)

\(=\dfrac{149}{18}\cdot\dfrac{27}{325}-\dfrac{17}{8}\cdot\dfrac{40}{51}\)

\(=\dfrac{447}{650}-\dfrac{5}{3}=-\dfrac{1909}{1950}\)

b: \(=\dfrac{48}{64}+\left(\dfrac{4}{5}-2-\dfrac{4}{15}\right):\dfrac{11}{3}\)

\(=\dfrac{3}{4}+\dfrac{-22}{15}\cdot\dfrac{3}{11}=\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{15-8}{20}=\dfrac{7}{20}\)

\(=\dfrac{17}{8}:\dfrac{51}{40}-\left(15-13-\dfrac{1}{3}\right):\dfrac{25}{6}\)

\(=\dfrac{5}{3}-\dfrac{5}{3}\cdot\dfrac{6}{25}=\dfrac{5}{3}\cdot\dfrac{19}{25}=\dfrac{19}{15}\)

\(2\dfrac{1}{8}:1\dfrac{11}{40}-\left(15-13\dfrac{1}{3}\right):4\dfrac{1}{6}\)

\(=\dfrac{17}{8}.\dfrac{40}{51}-\left(15-\dfrac{40}{3}\right).\dfrac{6}{25}\)

\(=\dfrac{17}{8}.\dfrac{40}{51}-\dfrac{5}{3}.\dfrac{6}{25}\)

\(=\dfrac{5}{3}-\dfrac{2}{5}\)

\(=\dfrac{19}{15}\)

\(E=\dfrac{\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{2002}-1\right)\left(\dfrac{1}{2003}-1\right)}{\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot...\cdot\dfrac{9999}{10000}}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{100^2}\right)}\)

\(=\dfrac{\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{2002}\right)\left(1-\dfrac{1}{2003}\right)}{\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{100}\right)\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{100}{101}\cdot\dfrac{101}{102}\cdot...\cdot\dfrac{2002}{2003}}{\left(1+\dfrac{1}{2}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)}\)

\(=\dfrac{100}{2003}:\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\right)\)

\(=\dfrac{100}{2003}:\left(\dfrac{101}{2}\right)=\dfrac{100}{2003}\cdot\dfrac{2}{101}=\dfrac{200}{202303}\)

đấy nhá

b) Ta có: \(\left|x\right|-\dfrac{3}{4}=\dfrac{5}{3}\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{3}+\dfrac{3}{4}=\dfrac{20}{12}+\dfrac{9}{12}=\dfrac{29}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{12}\\x=-\dfrac{29}{12}\end{matrix}\right.\)

c) Ta có: \(\left|2x-\dfrac{1}{3}\right|+\dfrac{5}{6}=1\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{1}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{6}\\2x-\dfrac{1}{3}=\dfrac{-1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{1}{6}+\dfrac{1}{3}=\dfrac{1}{2}\\2x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{1}{12}\end{matrix}\right.\)

\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow\)x-2-5(x+1)=15

\(\Leftrightarrow\) x-2-5x-5=15

\(\Leftrightarrow\)x-5x=15+2+5

\(\Leftrightarrow\)-4x=22

\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)

vậy

nhớ like nha

`c)(15/(sqrt6+1)+4/(sqrt6-2)-12/(3-sqrt6))*(sqrt6+11)`

`=((15(sqrt6-1))/(6-1)+(4(sqrt6+2))/(6-4)-(12(3+sqrt6))/(9-6))*(sqrt6+11)`

`=(3(sqrt6-1)+2(sqrt6+2)-4(3+sqrt6))*(sqrt6+11)`

`=(3sqrt6-3+2sqrt6+4-12-4sqrt6)*(sqrt6+11)`

`=(sqrt6-11)(sqrt6+11)`

`=6-121=-115`

c) Ta có: \(\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+11\right)\)

\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

=6-121=-115

\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)