(3/20 + 1/2 - 1/5 ) X 12/49 = 1/7

3.1/3 + 2/9 = 11/9

\(\left(\frac{3}{20}+\frac{1}{2}-\frac{1}{15}\right).\frac{12}{49}\)

\(=\left(\frac{9}{60}+\frac{30}{60}-\frac{4}{60}\right).\frac{12}{49}\)

\(=\frac{35}{60}.\frac{12}{49}\)

\(=\frac{1}{7}\)

1/10

1/10

1/10

a) \(\frac{3}{2}.\frac{7}{2}+\frac{5}{3}.\frac{1}{2}=\frac{21}{2}+\frac{5}{6}=\frac{63}{6}+\frac{5}{6}=\frac{68}{6}=\frac{34}{3}\)

b) \(\frac{1}{9}.\frac{2}{145}-\frac{13}{3}.\frac{2}{145}+\frac{2}{145}=\frac{2}{145}\left(\frac{1}{9}-\frac{39}{9}+\frac{9}{9}\right)=\frac{2}{145}\left(-29\right)=-0.4\)

c) \(\left(-\frac{91}{12}\right):\frac{15}{7}-\frac{1}{18}:\frac{15}{7}+\frac{20}{9}:\frac{15}{7}=\frac{7}{15}\left(-\frac{91}{12}-\frac{1}{18}+\frac{20}{9}\right)=\frac{7}{15}\left(-\frac{65}{12}\right)=-\frac{91}{36}\)

A=\(2^2-9^3+4^{-2}.16-2.5^2\)
\(=4-729+1-50=-774\)
B=\(\left(2^3.2\right).\dfrac{1}{2}+3^{-2}.3^2-7.1+5\)
\(B=2^4.\dfrac{1}{2}+1-7+5=8+1-7+5=7\)
 

 C = 2^-3 + (5²)³.5^-3 + 4^-3.16 - 2.3² - 105.(\(\dfrac{24}{51}\))⁰

C =  \(\dfrac{1}{8}\) + 5⁶.5^-3 + 4^-3.4² - 2.9 - 105.1

C =  \(\dfrac{1}{8}\) + 5³ + \(\dfrac{1}{4}\) - 18 - 105

C =  (\(\dfrac{1}{8}\) + \(\dfrac{1}{4}\))  - (105 - 125 + 18)

C = \(\dfrac{3}{8}\) - (-20 + 18)

C = \(\dfrac{3}{8}\)  + 2

C = \(\dfrac{19}{8}\)

D = 2^-3 . \(\dfrac{1}{2^{-2}}\).\(\dfrac{2}{3}\) + 4^-2.8 - 7. (\(\dfrac{17}{23}\))⁰ + 19

D = 2^-1.\(\dfrac{2}{3}\) + \(\dfrac{1}{16}\).8 - 7.1 + 19

D = \(\dfrac{1}{3}\) + \(\dfrac{1}{2}\) - 7 + 19

D = \(\dfrac{5}{6}\)  + (19 - 7)

D = \(\dfrac{5}{6}\) + 12

D = \(\dfrac{77}{6}\)

giải giúp ii

\(1,\left(3x+2\right)\left(5-x^2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

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

3/2

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