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\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)
\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)
\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)
\(=\sqrt{\frac{13}{16}}:5\)
\(=\frac{\sqrt{13}}{4}:5\)
\(=\frac{\sqrt{13}}{20}\)
$\textbf{a)}$
$A=\dfrac23\cdot\sqrt{81}-\left(-\dfrac34\right)\cdot\sqrt{\dfrac9{64}}+\left(\dfrac{\sqrt2}{3}\right)^2$
$=\dfrac23\cdot9+\dfrac34\cdot\dfrac38+\dfrac29$
$=6+\dfrac9{32}+\dfrac29$
$=\dfrac{1873}{288}.$
$\textbf{b)}$
$B=\left(-\sqrt{\dfrac54}\right)^2-\sqrt{\dfrac94}:(-4,5)-\sqrt{\dfrac{25}{16}}\cdot\sqrt{\dfrac{64}{9}}$
$=\dfrac54-\dfrac32:\left(-\dfrac92\right)-\dfrac54\cdot\dfrac83$
$=\dfrac54+\dfrac13-\dfrac{10}{3}$
$=\dfrac{15+4-40}{12}$
$=-\dfrac74.$
$\textbf{a)}$
$A=\dfrac{\left(\dfrac49\right)^2\cdot\left(-\dfrac9{16}\right)\cdot(-1)^{19}}{\left(\dfrac4{25}\right)^2\cdot\left(-\dfrac{25}{144}\right)^2\cdot\left(-\dfrac{49}{144}\right)^2}$
$=\dfrac{\dfrac{16}{81}\cdot\left(-\dfrac9{16}\right)\cdot(-1)}{\dfrac{16}{625}\cdot\dfrac{625}{144^2}\cdot\dfrac{49^2}{144^2}}$
$=\dfrac{\dfrac19}{\dfrac{49^2}{144^4}}$
$=\dfrac19\cdot\dfrac{144^4}{49^2}$
$=\dfrac19\cdot\dfrac{(2^4\cdot3^2)^4}{7^4}$
$=\dfrac{2^{16}\cdot3^6}{7^4}$
$=\dfrac{47775744}{2401}.$
$\textbf{b)}$
$B=\dfrac{\left(\dfrac23\right)^3\cdot\left(-\dfrac34\right)^2\cdot(-1)^{2003}}{\left(\dfrac25\right)^2\cdot\left(-\dfrac5{12}\right)^3}$
$=\dfrac{\dfrac8{27}\cdot\dfrac9{16}\cdot(-1)}{\dfrac4{25}\cdot\left(-\dfrac{125}{1728}\right)}$
$=\dfrac{-\dfrac16}{-\dfrac5{432}}$
$=\dfrac16\cdot\dfrac{432}{5}$
$=\dfrac{72}{5}.$
Ta có : \(\frac{\left(\frac{2}{5}\right)^9.10^9-\left(\frac{-9}{4}\right)^5:\left(\frac{-3}{16}\right)^{10}}{4^{12}+16^9}\)
=\(\frac{\left(\frac{2}{5}.10\right)^9-\left[\left(\frac{-3}{2}\right)^2\right]^5:\left(\frac{-3}{16}\right)^{10}}{\left(2^2\right)^{12}+\left(2^4\right)^9}\)=\(\frac{4^9-\left(\frac{-3}{2}\right)^{10}:\left(\frac{-3}{16}\right)^{10}}{2^{24}+2^{36}}\)=\(\frac{4^9-\left[\left(\frac{-3}{2}\right):\left(\frac{-3}{16}\right)\right]^{10}}{2^{24}\left(1+2^{12}\right)}\)
=\(\frac{\left(2^2\right)^9-8^{10}}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{\left(2^2\right)^9-\left(2^3\right)^{10}}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{2^{18}-2^{30}}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{2^{18}\left(1-2^{12}\right)}{2^{24}\left(1+2^{12}\right)}\)=\(\frac{1-2^{12}}{2^6\left(1+2^{12}\right)}\)
Không biết đúng hong nha!