19/10/2023
Làm sao để có câu trả lời hay nhất?
19/10/2023
19/10/2023
Bài 1
a)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
-12\ : \left(\frac{3}{4} -\frac{5}{6}\right) =-12\ : \left(\frac{9}{12} -\frac{10}{12}\right)\\
=-12\ : \left( -\frac{1}{12}\right) =-12\times ( -12) =144
\end{array}$
b)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\left(\frac{1}{3} -2\frac{1}{6}\right) \ : \left( -3\frac{13}{36}\right)\\
=\left(\frac{1}{3} -\frac{13}{6}\right) \ : \left( -\frac{121}{36}\right)\\
=-\frac{11}{6} \ :\left( -\frac{121}{36}\right)\\
=\frac{6}{11}
\end{array}$
c)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
2^{3} +3.\left(\frac{1}{9}\right)^{0} -\frac{1}{2^{2}} .4+\left[( -2)^{2} \ : \frac{1}{2}\right] .8\\
=8+3-1+8.8=74
\end{array}$
d)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\mid \frac{-1}{2} \mid +\left( -\frac{1}{3}\right)^{2} \ : \mid -2\mid -\left(\frac{-2}{3}\right)^{0}\\
=\frac{1}{2} +\frac{1}{9} \ : 2-1=\frac{-4}{9}
\end{array}$
e)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
-\sqrt{64} +2\sqrt{( -5)^{2}} -7\sqrt{0,01} +8.\sqrt{\frac{25}{16}}\\
=-8+2.5-7.\frac{1}{10} +8.\frac{5}{4}\\
=-8+10-\frac{7}{10} +10=\frac{113}{10}\\
\end{array}$
f)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\left[ 3\sqrt{225} -5\sqrt{( -12)^{2}} -12.\mid \frac{-7}{6} \mid \right] .\sqrt{2\frac{1}{4}}\\
=\left[ 3.15-5.12-12.\frac{7}{6}\right] .\frac{3}{2} =-29.\frac{3}{2} =-\frac{87}{2}
\end{array}$
g)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\frac{2^{7} .9^{3}}{6^{3} -8^{2}} =\frac{2^{7} .3^{6}}{2^{3} .3^{3} -2^{6}} =\frac{2^{4} .3^{6}}{3^{3} -2^{3}}\\
\end{array}$
h)
$\displaystyle \frac{15^{8} .27^{2} .2^{24}}{6^{14} .10^{9}} =\frac{3^{8} .5^{8} .3^{6} .2^{24}}{2^{14} .3^{14} .2^{9} .5^{9}} =\frac{1}{10}$
i)
$\displaystyle \mid \frac{-1}{5} \mid -\frac{( -2)^{2}}{\mid -5\mid } -\frac{\mid -2\mid }{5} =\frac{1}{5} -\frac{4}{5} -\frac{2}{5} =-1$
k)
$\displaystyle \left( -\frac{2}{3}\right)^{3} \ : \left( -\frac{2}{3}\right)^{2} +\frac{2^{40} .3^{29}}{8^{13} .9^{15}} =\frac{-2}{3} +\frac{2^{40} .3^{29}}{2^{39} .3^{30}} =-\frac{2}{3} +\frac{2}{3} =0$
l)
$\displaystyle \begin{array}{{>{\displaystyle}l}}
\left[\left(\frac{1}{3}\right)^{2} .\left(\frac{27}{7}\right) +\sqrt{\frac{16}{49}} -3\right] \ : \frac{4}{7}\\
=\left[\frac{1}{9} .\frac{27}{7} +\frac{4}{7} -3\right] \ : \frac{4}{7}\\
=-2\ : \frac{4}{7} =-\frac{7}{2}
\end{array}$
m)
$\displaystyle \sqrt{16} .\sqrt{4} -\sqrt{25} +2.\sqrt{49} =4.2-5+2.7=17$
19/10/2023
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