a) \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)

b) \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=-\frac{2560}{3}\)
e) \(\left(1+\frac{2}{3}-\frac{1}{4}\right).\left(\frac{4}{5}-\frac{3}{4}\right)^2=\frac{17}{12}.\left(\frac{1}{20}\right)^2=\frac{17}{12}.\frac{1}{400}=\frac{17}{4800}\)

f) \(2:\left(\frac{1}{2}-\frac{2}{3}\right)^3=2:\left(-\frac{1}{6}\right)^3=2:-\frac{1}{216}=-432\)

a, \(\frac{169}{196}\)

b, \(\frac{1}{144}\)

c, \(\frac{1}{100}\)

d, \(\frac{-2560}{3}\)

\(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)

\(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(-\frac{1}{12}\right)^2=\frac{1}{144}\)

\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^8.2^8}{5^{10}.42^{10}}=\frac{1}{100}\)

\(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\left[\left(-\frac{10}{3}\right).\left(-\frac{6}{5}\right)\right]^4.\left(-\frac{10}{3}\right)=4^4.\left(-\frac{10}{3}\right)=-\frac{2560}{3}\)

a. \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6+7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{13^2}{14^2}=\frac{169}{196}\)

b. \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9-10}{12}\right)^2=\left(-\frac{1}{12}\right)^2=\frac{\left(-1\right)^2}{12^2}=\frac{1}{144}\)

c. \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{1}{100}\)

d.\(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\frac{\left(-10\right)^5}{3^5}.\frac{\left(-6\right)^4}{5^4}=\frac{\left(-2.5\right)^5.\left(-2.3\right)^4}{3^5.5^4}=\frac{\left(-2\right)^5.5^5.\left(-2\right)^4.3^4}{3^5.5^4}=\frac{\left(-2\right)^9.5}{3}=\frac{-512.5}{3}=-\frac{2560}{3}\)

a, \(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{6}{14}+\frac{7}{14}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)

b, \(\left(\frac{3}{4}-\frac{5}{6}\right)^2=\left(\frac{9}{12}-\frac{10}{12}\right)^2=\left(-\frac{1}{12}\right)^2=\frac{1}{144}\)

c, \(\frac{5^4.20^4}{25^5.4^5}=\frac{\left(5.20\right)^4}{\left(25.4\right)^5}=\frac{100^4}{100^5}=\frac{100^4}{100^4.100}=\frac{1}{100}\)

d, \(\left(\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\left(\frac{10}{3}\right)^4.\frac{10}{3}.\left(-\frac{6}{5}\right)^4=\left(\frac{10}{3}.-\frac{6}{5}\right)^4.\frac{10}{3}=\left(-4\right)^4.\frac{10}{3}=256.\frac{10}{3}=853\frac{1}{3}\)

Cũng dễ mà, làm bài này nhìu rùi.

Bài làm

\(a,\left(\frac{3}{7}+\frac{1}{2}\right)^2\)

\(=\left(\frac{3}{7}\right)^2+\left(\frac{1}{2}\right)^2\)

\(=\frac{9}{49}+\frac{1}{4}\)

\(=\frac{36}{196}+\frac{49}{196}\)

\(=\frac{85}{196}\)

\(b,\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

\(=\left(-\frac{1}{12}\right)^2\)

\(=\frac{1}{144}\)

\(c,\frac{5^4.20^4}{25^5.4^5}\)

\(=\frac{5^4.\left(5.4\right)^4}{\left(5.5\right)^5.4^5}\)

\(=\frac{5^4.5^4.4^4}{5^5.5^5.4^5}\)

\(=\frac{1}{5.5.4}\)

\(=\frac{1}{100}\)

~ Check đúng cho minh nha. ~

# Học tốt #

\(a,\left(\frac{3}{7}+\frac{1}{2}\right)^2\)

\(< =>\left(\frac{6}{14}+\frac{7}{14}\right)^2\)

\(< =>\left(\frac{13}{14}\right)^2\)

\(< =>\frac{169}{196}\)

\(b,\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

\(< =>\left(\frac{9}{12}-\frac{10}{12}\right)^2\)

\(< =>\left(\frac{-1}{12}\right)^2\)

\(< =>\frac{-1}{144}\)

\(c,\frac{5^4\cdot20^4}{25^5\cdot4^5}\)

\(< =>\frac{25^2\cdot\left(4\right)^4\cdot\left(5\right)^4}{25^5\cdot4^5}\)

\(< =>\frac{1\cdot1\cdot\left(5\right)^4}{25^3\cdot4}\)

\(< =>\frac{1\cdot25^2}{25^3\cdot4}\)

\(< =>\frac{1}{25\cdot4}\)

\(< =>\frac{1}{100}\)

a,(3/7)^2+2.3/7.1/2+(1/2)^2

=9/49+3/7+1/4

=169/196

b,(3/4)^2-2.3/4.5/6+(5/6)^2

=9/16-5/4+25/36

=1/144

a)

\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)                         

b)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ =  - 2 - \frac{5}{7}.\frac{7}{{15}}\\ =  - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)

c)

\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)             

d)

\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)

a)

\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\=  - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)                          

b)

\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)                  

d)

\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ =  - \frac{3}{4}\end{array}\).