Tính giá trị các biểu thức:
\(4\frac{7}{9}:\left(\frac{-5}{18}\right)+\frac{48}{9}.\left(-3\frac{3}{5}\right)\)
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Tính giá trị của biểu thức:
4\(\dfrac{7}{9}\):\(\left(\dfrac{-5}{18}\right)\)+\(\dfrac{48}{9}\).(-3\(\dfrac{3}{5}\))
=\(\dfrac{43}{9}\).\(\dfrac{-18}{5}\)+\(\dfrac{48}{9}\).\(\dfrac{-12}{5}\)
=\(\dfrac{43.\left(-18\right)}{9.5}\)+\(\dfrac{48.\left(-12\right)}{9.5}\)
=\(\dfrac{43.\left(-2\right)}{1.5}\)+\(\dfrac{48.\left(-4\right)}{3.5}\)
=\(\dfrac{-86}{5}\)+\(\dfrac{-192}{15}\)
=\(\dfrac{-258}{15}\)+\(\dfrac{-192}{15}\)
=-30
\(4\dfrac{7}{9}:\left(\dfrac{-5}{18}\right)+\dfrac{48}{9}.\left(-3\dfrac{3}{5}\right)\)
\(\dfrac{43}{9}:\left(\dfrac{-5}{18}\right)+\dfrac{48}{9}.\left(\dfrac{-18}{5}\right)=\dfrac{43}{9}.\left(\dfrac{-18}{5}\right)+\dfrac{48}{9}.\left(\dfrac{-18}{5}\right)=\dfrac{-18}{5}.\left(\dfrac{43}{9}+\dfrac{48}{9}\right)=\dfrac{-18}{5}.9=\dfrac{-162}{5}\)
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
3) C thiếu đề
4) \(D=\frac{1}{9}-\left|\frac{-5}{23}\right|-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{5}{23}+\frac{5}{23}-\frac{1}{9}-\frac{25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=\frac{1}{9}-\frac{1}{9}-\frac{5}{23}+\frac{5}{23}+\frac{-25}{7}+\frac{50}{4}-\frac{7}{30}\)
\(D=0+0+\frac{125}{14}-\frac{7}{30}\)
\(D=\frac{913}{105}\)
a)
\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)
b)
\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)
c)
\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)
d)
Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)
Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)
Ta có : \(A=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)
\(\Rightarrow A=\frac{58}{7}-\frac{487}{63}=\frac{5}{9}\)
P/s:Câu B tương tự nhé
Tiếp B của @Phạm Tuấn Đạt
\(B=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(\Rightarrow B=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(B=\left(\frac{92}{9}-\frac{56}{9}\right)+\frac{13}{5}\)
\(B=\frac{36}{9}+\frac{13}{5}\)
\(B=4+\frac{13}{5}\)
\(B=\frac{20}{5}+\frac{13}{5}=\frac{33}{5}\)
=-36,4 nha nếu đúng thì tích hộ mk