a: \(-6\cdot\left(-\dfrac{2}{3}\right)\cdot0.25=6\cdot\dfrac{2}{3}\cdot\dfrac{1}{4}=4\cdot\dfrac{1}{4}=1\)

b: \(\dfrac{-15}{4}\cdot\dfrac{-7}{15}\cdot\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}\cdot\dfrac{12}{5}\)

\(=\dfrac{84}{20}=\dfrac{21}{5}\)

c: \(\left(-2\dfrac{1}{5}\right)\cdot\left(-\dfrac{9}{11}\right)\cdot\left(-\dfrac{1}{14}\right)\cdot\dfrac{2}{5}\)

\(=-\dfrac{11}{5}\cdot\dfrac{2}{5}\cdot\dfrac{9}{11}\cdot\dfrac{1}{14}\)

\(=-\dfrac{11}{11}\cdot\dfrac{2}{14}\cdot\dfrac{9}{25}\)

\(=-\dfrac{9}{175}\)

\(a,=4\cdot0,25=1\\ b,=\dfrac{7}{4}\cdot\left(-\dfrac{12}{5}\right)=-\dfrac{21}{5}\\ c,=\left(-\dfrac{11}{5}\right)\left(-\dfrac{9}{11}\right)\left(-\dfrac{15}{14}\right)\cdot\dfrac{2}{5}\\ =\dfrac{9}{5}\cdot\left(-\dfrac{15}{14}\right)\cdot\dfrac{2}{5}=-\dfrac{27}{14}\cdot\dfrac{2}{5}=-\dfrac{27}{35}\\ d,=\left(-\dfrac{11}{2}\right)\left(-\dfrac{1}{2}\right)+\dfrac{4}{9}=\dfrac{11}{4}+\dfrac{4}{9}=\dfrac{115}{36}\\ e,=\dfrac{5}{4}\cdot\left(-\dfrac{8}{15}\right)-\dfrac{3}{5}-\dfrac{3}{10}=-\dfrac{2}{3}-\dfrac{3}{5}-\dfrac{3}{10}=-\dfrac{47}{30}\)

\(f,B=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(6-1\right)}=\dfrac{2\cdot6}{5\cdot3}=\dfrac{4}{5}\\ g,=\dfrac{5}{8}+\dfrac{9}{4}\cdot\dfrac{5}{3}-\dfrac{5}{24}=\dfrac{5}{8}+\dfrac{15}{4}-\dfrac{5}{24}=\dfrac{25}{6}\\ h,=\dfrac{49}{38}\cdot\left(\dfrac{152}{11}-\dfrac{57}{11}\right):\dfrac{245}{418}=\dfrac{49}{38}\cdot\dfrac{418}{245}\cdot\dfrac{95}{11}=\dfrac{95\cdot11}{5\cdot11}=19\\ k,=\dfrac{11}{30}+\dfrac{18}{35}\cdot\dfrac{35}{54}-\dfrac{18}{35}\cdot\dfrac{49}{18}-\dfrac{18}{35}\cdot\dfrac{28}{48}\\ =\dfrac{11}{30}+\dfrac{1}{3}-\dfrac{7}{5}-\dfrac{3}{10}=-1\)

https://hoc24.vn/cau-hoi/giai-het-giup-mk-vs-a-huhu.2030470567354

làm hết r mà :vv

a: \(-6\cdot\left(-\dfrac{2}{3}\right)\cdot0.25=6\cdot\dfrac{2}{3}\cdot\dfrac{1}{4}=4\cdot\dfrac{1}{4}=1\)

f: \(B=\dfrac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot5}=\dfrac{4}{5}\)

Câu 1:

\(\left(4x+3\right)\left(3x^2+x-2\right)\left(2x^2-3x-5\right)=0\\ \Leftrightarrow\left(4x+3\right)\left(3x-2\right)\left(x+1\right)\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-1\\x=\dfrac{2}{3}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow A=\left\{-1;-\dfrac{3}{4};\dfrac{2}{3};\dfrac{5}{2}\right\}\)

Câu 2:

\(\left(x^2-4\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\Leftrightarrow A=\left\{-2;2;3\right\}\\ \left|5x\right|-11\le0\Leftrightarrow\left|5x\right|\le11\Leftrightarrow-11\le5x\le11\\ \Leftrightarrow-\dfrac{11}{5}\le x\le\dfrac{11}{5}\\ \Leftrightarrow B=\left[-\dfrac{11}{5};\dfrac{11}{5}\right]\)

\(\Leftrightarrow A\cap B=\left\{-2;2\right\}\\ A\cup B=\left[-\dfrac{11}{5};3\right]\\ A\B=\left\{3\right\}\)