câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)

\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)

\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)

\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)

\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)

\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)

\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)

\(B=1^{2010}-1^{2009}=1-1=0\)

câu 2

a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)

\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)

\(\Leftrightarrow2x=\dfrac{31}{12}\)

\(\Leftrightarrow x=\dfrac{31}{24}\)

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

\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)

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

\(\Leftrightarrow x=-\dfrac{5}{6}\)

f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)

a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)

b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)

c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)

\(A=\dfrac{5}{9}.\dfrac{22}{-3}+\dfrac{5}{9}.\dfrac{5}{-3}=\dfrac{5}{9}\left(\dfrac{22}{-3}+\dfrac{5}{-3}\right)=\dfrac{5}{9}.\dfrac{22+5}{-3}=\dfrac{5}{9}.\dfrac{27}{-3}\)

\(A=\dfrac{5.27}{9.-3}=\dfrac{5.27}{-27}=-5\)

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

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

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

\(B=-2+\dfrac{0}{2}+\left(\dfrac{-1}{2}\right)=-2+0-\dfrac{1}{2}=\dfrac{-5}{2}\)

5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)

=\(4+6-3+5\)

=\(12\)

2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)

=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)

=\(\dfrac{11}{25}.\left(-100\right)\)

=\(-44\)

Lời giải:

a)

\(\frac{\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}}{\frac{8}{3}-\frac{8}{5}+\frac{8}{7}-\frac{8}{9}+\frac{8}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}{8\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\frac{1}{11}\right)}\) \(=\frac{2}{8}=\frac{1}{4}\)

b)

\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{50}-1\right)\left(\frac{1}{51}-1\right)\)

\(=\frac{1-2}{2}.\frac{1-3}{3}.\frac{1-4}{4}....\frac{1-50}{50}.\frac{1-51}{2}=\frac{(-1)(-2)(-3)...(-49)(-50)}{2.3.4....50.51}\)

\(=\frac{(-1)^{50}.1.2.3....49.50}{2.3.4...50.51}=\frac{1}{51}\)

a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)

= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)

= \(\dfrac{-1}{24}-\dfrac{5}{8}\)

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

b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)

= \(8\dfrac{5}{8}\)

c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)

= \(-49\)

d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)

= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)

= 0

e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)

=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)

=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)

=\(-\dfrac{516}{7}\)

a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)

=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)

=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)

=\(\dfrac{119}{240}\)

a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)

\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)

\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)

\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)

a: \(=\left|\dfrac{3}{2}-\dfrac{7}{3}\right|^2+\dfrac{1}{4}=\dfrac{17}{18}\)

b: \(=\left|1-2-\dfrac{1}{3}\right|+\dfrac{5}{6}=1+\dfrac{1}{3}+\dfrac{5}{6}=\dfrac{13}{6}\)

c: \(=\left|\dfrac{3}{2}-\dfrac{7}{4}\right|-\dfrac{7}{4}=-\dfrac{3}{2}\)

d: =x-5+8-x=3

A=\(\dfrac{5}{4}\).(5-\(\dfrac{4}{3}\)).(\(-\dfrac{1}{11}\))

= \(\dfrac{5}{4}\).\(\dfrac{11}{3}\).(\(-\dfrac{1}{11}\))

=\(\dfrac{5}{4}\).[\(\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\text{]}\)

=\(\dfrac{5}{4}.\dfrac{1}{3}\)

=\(\dfrac{5}{12}\) (1)

B=\(\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\) =\(\dfrac{3}{4}:\text{[}\left(-12\right).\left(-\dfrac{2}{3}\right)\text{]}\)

=\(\dfrac{3}{4}:8\) =\(\dfrac{3}{4}.\dfrac{1}{8}\)=\(\dfrac{3}{32}\)(2)

C=\(\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\) =\(\dfrac{5}{4}:\text{[}\left(-15\right).\left(-\dfrac{2}{5}\right)\text{]}\)

=\(\dfrac{5}{4}:6=\dfrac{5}{4}.\dfrac{1}{6}=\dfrac{5}{24}\left(3\right)\)

D=(-3).\(\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\) =(-3).\(\left(-\dfrac{7}{12}\right)\):(-7)=\(\dfrac{7}{4}:\left(-7\right)\)=\(\dfrac{7}{4}\).\(\left(\dfrac{-1}{7}\right)\)=\(\dfrac{-1}{4}\) (4)

Từ (1),(2),(3)và(4)=>Ta có thể sắp xếp các kết quả trên theo thứ tự tăng dần là:

(Bạn tự làm nhé! mình bận đi học rồi)