(-2/3x-3/5)(3/-2-10/3)=2/5
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\(\left(\frac{-2}{3x}-\frac{3}{5}\right)\left(\frac{3}{-2}-\frac{10}{3}\right)\)
\(=\left[-\left(\frac{2}{3x}+\frac{3}{5}\right)\right]\left[-\left(\frac{3}{2}+\frac{10}{3}\right)\right]\)
\(=\left(\frac{2}{3x}+\frac{3}{5}\right)\left(\frac{3}{2}+\frac{10}{3}\right)\)
\(=\left(\frac{10}{15x}+\frac{9x}{15x}\right)\left(\frac{9}{6}+\frac{20}{6}\right)\)
\(=\frac{10+9x}{15x}.\frac{9+20}{6}\)
\(=\frac{29.\left(10+9x\right)}{90}\)
a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)
(\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)
- \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)
\(x\) = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))
\(x=\) - \(\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)
\(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)
3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)
3\(x\) - 3,7 = - \(\dfrac{19}{2}\)
3\(x\) = - \(\dfrac{19}{2}\) + 3,7
3\(x\) = - \(\dfrac{29}{5}\)
\(x\) = - \(\dfrac{29}{5}\) : 3
\(x\) = - \(\dfrac{29}{15}\)
Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\)
\(10^3.100^2.1000^5\)
=\(10^3.10^5.10^{15}\)
=\(10^{23}\)
b) \(16.64.8^2:\left(4^3.2^5.16\right)\)
=\(2^4.2^6.2^6:\left(2^6.2^5.2^4\right)\)
=\(2^{10}.2^6:\left(2^{11}.2^4\right)\)
=\(2^{16}:2^{15}\)
=2
c) \(\left(20.2^4+12.2^4-48.2^2\right):8^2\)
= \(\left[2^4.\left(20+12\right)-48.2^2\right]:8^2\)
= \(\left[16.32-48.4\right]:64\)
= \(\left[512-192\right]:64\)
= \(320:64\)
= \(5\)
Câu d thì mình chưa hiểu đề bài thì bạn viết lại hộ mình để mình giải cho
=6/15x(-9/6-20/6)
=6/15x[-(9/6+20/6)]
=6/15x(-29/6)
=-174/90
=-29/15