\(\left(-0.2\right)^4.\left(-0.2\right)^5\)
H24
Những câu hỏi liên quan
\(\left\{\left[\left(6.2:0.31-\frac{5}{6}\cdot0.9\right)\cdot0.2+0.15\right]:0.2\right\}:\left[\left(2+1\frac{4}{11}\cdot0.22:0.1\right)\cdot\frac{1}{33}\right]\)
B = \(\dfrac{120-\left(-0.5\right).\left(-40\right).\left(-5\right).\left(-0.2\right).20.0.25}{5+10+15+...+2015}\)
\(B=\dfrac{120-\left(-0,5\right)\cdot\left(-40\right)\cdot\left(-5\right)\cdot\left(-0,2\right)\cdot20\cdot0\cdot25}{5+10+15+...+2015}\\ =\dfrac{120-0}{5+10+15+...+2015}\\ =\dfrac{120}{5+10+15+...+2015}\\ ĐặtA=5+10+15+2015\\ A=\left(2015+5\right)\cdot\left[\left(2015-5\right):5+1\right]:2\\ A=407030\\ VậyB=\dfrac{120}{407030}=\dfrac{6}{203515}\)
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Thực hiện phép tính: \(\frac{\left(\frac{3}{4}\right)^{12}.\left(-2\right)^0.2^{12}}{\left(-2\right)^3.4^5.\left(-0,75\right)^{12}}=?\)
\(\left(\sqrt{8}-3\sqrt{2}+2\sqrt{5}\right)\left(\sqrt{2}+10\sqrt{0.2}\right)\)
=(2căn 5-căn 2)(2căn 5+căn 2)
=20-2=18
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\(\orbr{-20.83\times0.2}+\left(-9.17\right).0.2]\div[2.47\times0.5\times\left(-3.53\right)\times0.5]\)
Tính
a)\(\frac{\left(0,6\right)^5}{\left(0.2\right)^6}\)
b)\(\frac{2^7.9^3}{6^5.8^2}\)
c)\(\left(\frac{3}{7}+\frac{1}{2}\right)^2\)
d)\(\left(\frac{3}{4}-\frac{5}{6}\right)^2\)
e)\(\left(\frac{-10}{3}\right)^5.\left(\frac{-6}{5}\right)^4\)
Help me!
\(\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0.2\right)\) \(\frac{5}{18}-1.456:\frac{7}{15}+4.5\cdot\frac{4}{5}\)
\(\left(\frac{9}{25}-2,18\right):\left(3\frac{4}{5}+0,2\right)\)
\(=\left(\frac{9}{25}-\frac{109}{50}\right):\left(\frac{19}{5}+\frac{1}{5}\right)\)
\(=\frac{-91}{50}:4=\frac{-91}{200}\)
\(\frac{5}{18}-1,456:\frac{7}{15}+4,5.\frac{4}{5}\)
\(=\frac{5}{18}-\frac{182}{125}.\frac{15}{7}+\frac{18}{5}\)
\(=\frac{5}{18}-\frac{78}{25}+\frac{18}{5}\)
\(=\frac{341}{450}\)
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a)\(\dfrac{4^2.4^3}{2^{10}}\)
b)\(\dfrac{\left(0,6\right)^5}{\left(0.2\right)^6}\)
c)\(\dfrac{2^7.9^3}{6^5.8^2}\)
d)\(\dfrac{6^3+3.6^2+3^3}{-13}\)
a,
\(\dfrac{4^2\cdot4^3}{2^{10}}=\dfrac{4^5}{2^{10}}=\dfrac{\left(2^2\right)^5}{2^{10}}=\dfrac{2^{10}}{2^{10}}=1\)
b,
\(\dfrac{\left(0,6\right)^5}{\left(0,2\right)^6}=\dfrac{\left(0,2\cdot3\right)^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{\left(0,2\right)^5\cdot3^5}{\left(0,2\right)^5\cdot0,2}=\dfrac{243}{0,2}=\dfrac{243}{\dfrac{1}{5}}=243\cdot5=1215\)
c,
\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^6\cdot2\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{3}{2^4}=\dfrac{3}{16}\)
d,
\(\dfrac{6^3+3\cdot6^2+3^3}{-13}=\dfrac{\left(2\cdot3\right)^3+3\cdot\left(2\cdot3\right)^2+3^3}{-13}=\dfrac{2^3\cdot3^3+3\cdot2^2\cdot3^2+3^3}{-13}=\dfrac{2^3\cdot3^3+2^2\cdot3^3+3^3}{-13}\dfrac{3^3\left(2^3+2^2+1\right)}{-13}=\dfrac{3^3\cdot13}{-13}=-3^3=-27\)
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a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
c) \(\dfrac{3}{4}-\left|x+0.5\right|=\dfrac{1}{5}\)
d) \(\left(x+0.2\right)^2+0,75=1\)
a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
\(\Leftrightarrow3-2x=-1\)
\(\Leftrightarrow-2x=-1-3\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)
\(\Leftrightarrow200x+24=120x-105\)
\(\Leftrightarrow80x=-129\)
\(\Leftrightarrow x=-\dfrac{129}{80}\)
Vậy \(x=-\dfrac{129}{80}\)
c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)
\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)
d) \(\left(x+0,2\right)^2+0,75=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2+\dfrac{3}{4}=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=1-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)
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a, \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
=>\(-\dfrac{1}{2}x=-\dfrac{1}{4}-\dfrac{3}{4}\)
=>\(-\dfrac{1}{2}x=-1\)
=>\(x=-1:(-\dfrac{1}{2})\)
=>\(x=2\)
vậy ...........
b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)
=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)
=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)
=>\(x=-\dfrac{192}{80}\)
vậy...................
c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)
=>\(\left|x+0,5\right|=\dfrac{11}{20}\)
=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
vậy ....... hoặc.....
d,\((x+0,2)^2+0,75=1\)
=>\(\left(x+0,2\right)^2=1-0,75\)
=>\(\left(x+0,2\right)^2=0,25\)
=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)
vậy..........................
HỌC TỐT NHA !!!!!
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