\(=\dfrac{-11}{7}+\dfrac{1}{16}+\dfrac{6}{7}=\dfrac{-5}{7}+\dfrac{1}{16}=\dfrac{-80+7}{112}=\dfrac{-73}{112}\)

ko chép đề

\(=\frac{18.17.5}{5.21.17}=\frac{18}{21}=\frac{6}{7}\)

\(\frac{3.7.13.2}{13.7.7}=\frac{6}{7}\)

sorry bạn mik làm tắt

6/7 và 6/7 bằng nhau nha

\(\frac{36.85.20}{25.84.34}=\frac{2^2.3^2.5.17.2^2.5}{5^2.2^2.3.7.2.17}=\frac{2^4.3^2.5^2.17}{2^3.3.5^2.7.17}=\frac{6}{7}\)                           1

\(\frac{30.63.65.8}{117.200.49}=\frac{2.3.5.3^2.7.5.13.2^3}{3^2.13.2^3.5^2.7^2}=\frac{2^4.3^3.5^2.7.13}{2^3.3^2.5^2.7^2.13}=\frac{6}{7}\)                2

Từ 1 và hai suy ra \(\frac{6}{7}=\frac{6}{7}\)

Nên hai phép tính bằng nhau

\(B=\frac{36.85.20}{25.84.34}=\)\(\frac{6.6.5.17.4.5}{5.5.14.6.17.2}\)\(=\frac{6}{7}\)

\(C=\frac{30.63.65.8}{117.200.49}=\frac{6.5.7.9.5.13.23}{13.9.5.5.8.7.7}=\frac{6}{7}\)

\(\Rightarrow\)2 biểu thức B = C

thks n`, các bn trả lời đi, mk sẽ like vẻn vẹn 3 cái nữa thui đó!

vì cả B và C đều bằng 6/7

nên B = C

(\(\frac{121}{122}\). \(\frac{123}{125}\)) .(\(\frac{1995}{1996}\).\(\frac{17}{16}\)- \(\frac{21}{25}\):\(\frac{16}{17}\)).(\(\frac{42}{30}\).\(\frac{75}{23}\)-\(\frac{19}{23}\).\(\frac{210}{38}\))

= (121/122 . 123/125) . ( 1995/1996 . 17/16 - 21/25 : 16/ 17) . 0

= 0

\(\begin{array}{l}0,49 = {\left( {0,7} \right)^2};\\\,\frac{1}{{32}} =\frac{1^5}{2^5}={\left( {\frac{1}{2}} \right)^5};\\\,\frac{{ - 8}}{{125}} =\frac{(-2)^3}{5^3}= {\left( {\frac{{ - 2}}{5}} \right)^3};\end{array}\)

\(\frac{{16}}{{81}} =\frac{4^2}{9^2}= {\left( {\frac{4}{9}} \right)^2} (hoặc \,\frac{{16}}{{81}} =\frac{2^4}{3^4}= {\left( {\frac{2}{3}} \right)^4});\\\,\frac{{121}}{{169}} =\frac{11^2}{13^2}= {\left( {\frac{{11}}{{13}}} \right)^2}\)

\(=2\frac{4}{\sqrt{5}}-5\frac{3}{5\sqrt[]{5}}-6\frac{11}{3\sqrt{5}}\)

\(=\frac{2.4.15-5.3.3-6.11.5}{15\sqrt{5}}\)

\(=\frac{-255}{15\sqrt{5}}=\frac{-17\sqrt{5}}{5}\)