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binh rồi căn thì cứ chuyển bỏ dấu âm đi nó tương tự dấu giá trị tuyệt đối thôi
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a) \(\frac{-6}{21}.\frac{3}{2}=-\frac{3}{7}\) b) \(\left(-3\right).\left(\frac{-7}{12}\right)=\frac{21}{12}=\frac{7}{4}\)
c) \(\left(\frac{11}{12}:\frac{33}{16}\right).\frac{3}{5}=\frac{11}{12}.\frac{16}{33}.\frac{3}{5}=\frac{4}{15}\)
d) \(\sqrt{\left(-7\right)^2}+\sqrt{\frac{2}{16}}=7+\sqrt{\frac{1}{8}}\)
c) \(\frac{1}{2}.\sqrt{100}-\sqrt{\frac{1}{16}}+\left(\frac{1}{3}\right)^0=\frac{1}{2}.10-\frac{1}{4}+1=5\frac{3}{4}\)
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a. \(\left(2-\frac{3}{4}\right)^2:\frac{11}{16}=\frac{5}{4}^2.\frac{11}{16}=\frac{25}{16}.\frac{16}{11}=\frac{25}{11}\)
b. \(2^3.\frac{7}{20}+\frac{7}{10}=8.\frac{7}{20}+\frac{7}{10}=\frac{14}{5}+\frac{7}{10}=\frac{7}{2}\)
c. \(\sqrt{3^2+4^2}-\sqrt{1^3+2^3+3^3}=\sqrt{9+16}-\sqrt{1+8+27}\)
\(=\sqrt{25}-\sqrt{36}=5-6=-1\)
d. \(21^3:\left(-7\right)^3=\left(21:\left(-7\right)\right)^3=-3^3=-27\)
a) \(\left(2-\frac{3}{4}\right)^2\div\frac{11}{16}=\left(\frac{5}{4}\right)^2.\frac{16}{11}=\frac{25}{16}.\frac{16}{11}=\frac{25}{11}\)
b) \(2^3.\frac{7}{20}+\frac{7}{10}=8.\frac{7}{20}+\frac{7}{10}=\frac{14}{5}+\frac{7}{10}=\frac{7}{2}\)
c) \(\sqrt{3^2+4^2}-\sqrt{1^3+2^3+3^3}=\sqrt{9+16}-\sqrt{1+8+27}\)
\(=\sqrt{25}-\sqrt{36}=5-6=-1\)
d) \(\frac{21^3}{\left(-7\right)^3}=\frac{9261}{-343}=-27\)