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a: \(\dfrac{1.12}{2.8}=\dfrac{-4}{-10};\dfrac{1.12}{-4}=\dfrac{2.8}{-10};\dfrac{2.8}{1.12}=\dfrac{-10}{-4};\dfrac{-4}{1.12}=\dfrac{-10}{2.8}\)
b: \(\dfrac{0.16}{0.4}=\dfrac{0.32}{0.8};\dfrac{0.16}{0.32}=\dfrac{0.4}{0.8};\dfrac{0.4}{0.16}=\dfrac{0.8}{0.32};\dfrac{0.32}{0.16}=\dfrac{0.8}{0.4}\)
1/
a, \(A=\dfrac{2}{3}+\dfrac{3}{4}.\left(-\dfrac{4}{9}\right)=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
b, \(B=2\dfrac{3}{11}.\dfrac{11}{12}.\left(-2,2\right)=\dfrac{25}{11}.\dfrac{11}{12}.\left(-\dfrac{11}{5}\right)=-\dfrac{55}{12}\)
c, \(C=\left(\dfrac{3}{4}-0,2\right):\left(0,4-\dfrac{4}{5}\right)=\left(\dfrac{3}{4}-\dfrac{1}{5}\right):\left(\dfrac{2}{5}-\dfrac{4}{5}\right)=\dfrac{11}{20}:\left(-\dfrac{2}{5}\right)=-\dfrac{11}{8}\)
2/
a, \(\dfrac{11}{12}-x=\dfrac{2}{3}+\dfrac{1}{4}\\ \Rightarrow\dfrac{11}{12}-x=\dfrac{11}{12}\\ \Rightarrow x=0\)
b, \(2x\left(x-\dfrac{1}{7}\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-\dfrac{1}{7}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{1}{7}\end{matrix}\right.\)
c, \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\\ \Rightarrow\dfrac{1}{4}:x=-\dfrac{7}{20}\\ \Rightarrow x=-\dfrac{5}{7}\)
vì IaI=\(\frac{1}{2}\Rightarrow a=\frac{1}{2};a=\frac{-1}{2}\)nên ta có :
với a=\(\frac{1}{2}\)thì A= 2.\(\frac{1}{2}-3.\left(-2\right)=1+6=7\)
với a=\(\frac{-1}{2}\)thì A=2\(\frac{-1}{2}-3.\left(-2\right)=-1+6=5\)
Vậy A= 7 hoặc A=5
Ta có:
\(\left(\frac{3}{15}-\frac{12}{16}-\frac{7}{14}\right)\cdot\left(-1\frac{5}{14}\right)\)
\(=\left(\frac{1}{5}-\frac{3}{4}-\frac{1}{2}\right)\cdot\left(-\frac{19}{14}\right)\)
\(=\left(\frac{4}{20}-\frac{15}{20}-\frac{10}{20}\right)\cdot\left(-\frac{19}{14}\right)\)
\(=\left(-\frac{21}{20}\right)\cdot\left(-\frac{19}{14}\right)\)
\(=\frac{57}{40}\)
a) Ta có:
\(3^3+3\cdot\left(-\frac{1}{2}\right)^3-0,75\)
\(=27+3\cdot\left(-\frac{1}{8}\right)-\frac{3}{4}\)
\(=27+\left(-\frac{3}{8}\right)-\frac{6}{8}\)
\(=27+\left[\left(-\frac{3}{8}\right)+\left(-\frac{6}{8}\right)\right]\)
\(=27+\left(-\frac{9}{8}\right)\)
\(=\frac{207}{8}\)
B
D