giúp mk vs mn ơi , mai cô giáo ktra mk r

a: \(=\dfrac{8}{9}\cdot\dfrac{9}{4}\cdot\dfrac{12}{19}\cdot\dfrac{19}{24}=\dfrac{1}{2}\cdot2=1\)

b: \(=\dfrac{5}{16}\cdot\dfrac{17}{15}\cdot\dfrac{8}{17}=\dfrac{5}{16}\cdot\dfrac{8}{15}=\dfrac{40}{240}=\dfrac{1}{6}\)

c: \(=\dfrac{4}{13}\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\dfrac{3}{26}=\dfrac{4}{13}-\dfrac{3}{26}=\dfrac{5}{26}\)

c: \(=\dfrac{3}{4}\left(\dfrac{6}{11}+\dfrac{5}{11}\right)-\dfrac{1}{5}=\dfrac{3}{4}-\dfrac{1}{5}=\dfrac{11}{20}\)

\(I=\dfrac{5}{4}+\dfrac{-1}{3}-\dfrac{5}{-24}=\dfrac{9}{8}\)

\(J=\dfrac{-19}{-9}+\dfrac{4}{-11}-\dfrac{-2}{3}=\dfrac{239}{99}\)

\(K=\dfrac{-5}{6}-\dfrac{7}{12}+\dfrac{-3}{4}=-\dfrac{13}{6}\)

\(L=\dfrac{-3}{20}+\dfrac{1}{5}-\dfrac{-5}{3}=\dfrac{103}{60}\)

e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)

f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)

g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)

Câu 1:

\(=\dfrac{15}{34}+\dfrac{19}{34}-1-\dfrac{15}{17}+\dfrac{1}{3}+\dfrac{3}{5}\)

\(=-\dfrac{15}{17}+\dfrac{14}{15}=\dfrac{13}{255}\)

Câu 2: 

\(=\dfrac{5^4\cdot5^4\cdot2^8}{4^4\cdot6^4\cdot3^2\cdot5}=\dfrac{5^7}{6^4\cdot3^2}\)

\(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}\)

\(A=\frac{19}{24}+\frac{-1}{2}+\frac{-1}{3}+\frac{-7}{24}\)

\(A=\left(\frac{19}{24}+\frac{-7}{24}\right)+\frac{-1}{2}+\frac{-1}{3}\)

\(A=\frac{1}{2}+\frac{-1}{2}+\frac{-1}{3}\)

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

\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{15}\)

\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}+\frac{-3}{7}+\frac{-1}{3}\)

\(B=\left(\frac{49}{168}+\frac{140}{168}+\frac{42}{168}\right)+\left(\frac{-72}{168}+\frac{-56}{168}\right)\)

\(B=\frac{231}{168}+\frac{-128}{168}=\frac{103}{168}\)

Có: \(A=\frac{-1}{3}=\frac{\left(-1\right)\cdot56}{3\cdot56}=\frac{-56}{168}\)

Mặt khác: \(-56< 103\)

\(\Rightarrow\)\(\frac{-56}{168}< \frac{103}{168}\)

\(hay\) \(A< B\)

a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)

\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)

\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)

b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)

\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)

=459/940

19) \(\sqrt{19-x}=19\)

\(\Rightarrow\sqrt{19-x}=\sqrt{19^2}\)

\(\Rightarrow19-x=19^2\)

\(\Rightarrow19-19^2=x\)

\(\Rightarrow x=19\left(1-19\right)=-19.18=-342\)

21) \(\sqrt{x-1}=\dfrac{1}{3}\)

\(\Rightarrow\sqrt{x-1}=\sqrt{\left(\dfrac{1}{3}\right)^2}\)

\(\Rightarrow x-1=\dfrac{1}{3^2}\)

\(x=\dfrac{1+9}{9}=\dfrac{10}{9}\)

24)\(\sqrt{2x+\dfrac{5}{4}}=\dfrac{3}{2}\)

\(\Rightarrow\sqrt{2x+\dfrac{5}{4}}=\sqrt{\left(\dfrac{3}{2}\right)^2}\)

\(\Rightarrow2x+\dfrac{5}{4}=\left(\dfrac{3}{2}\right)^2=\dfrac{9}{4}\)

\(\Rightarrow2x=\dfrac{9-5}{4}=1\)

\(\Rightarrow x=0,5\)

25) \(\sqrt{\dfrac{x}{3}-\dfrac{7}{6}}=\dfrac{1}{6}\)

\(\Rightarrow\sqrt{\dfrac{2x-7}{6}}=\sqrt{\left(\dfrac{1}{6}\right)^2}\)

\(\Rightarrow\dfrac{2x-7}{6}=\left(\dfrac{1}{6}\right)^2=\dfrac{1}{36}\)

\(\Rightarrow\dfrac{12x-42}{36}=\dfrac{1}{36}\)

\(\Rightarrow12x-42=1\)

\(\Rightarrow12x=43\)

\(\Rightarrow x=\dfrac{43}{12}\)

x=\(\dfrac{4}{15}\) : \(\dfrac{-2}{3}\)

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

a: Ta có: \(x\cdot\dfrac{-2}{3}=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-2}{5}\)

b: Ta có: \(x\cdot\dfrac{-7}{19}=\dfrac{-13}{24}\)

\(\Leftrightarrow x=\dfrac{13}{24}:\dfrac{7}{19}=\dfrac{247}{168}\)

\(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

Biến đổi tử số 

\(19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}\)

= 1 + \(\left(1+\dfrac{18}{2}\right)+\left(1+\dfrac{17}{3}\right)+\left(1+\dfrac{16}{4}\right)+...+\left(1+\dfrac{1}{19}\right)\)

= \(\dfrac{20}{20}+\dfrac{20}{2}+\dfrac{20}{3}+...+\dfrac{1}{19}\)

= 20 x \(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)\)

Vậy \(A=\dfrac{19+\dfrac{18}{2}+\dfrac{17}{3}+\dfrac{16}{4}+...+\dfrac{1}{19}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}\)

= \(\dfrac{20\times\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{20}}=20\)

Vậy A = 20

c.ơn nhìu a