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=(-3,15).\(\left[\left(-7,2\right)+12,4-4,8\right]\)
=(-3,15).0,4
=-1,26
\(a.\) \(\frac{x}{3.15}\) \(=\) \(\frac{0.15}{7.2}\)
<=> \(x=\) \(\frac{0.15\cdot3.15}{7.2}\)
<=>\(x=0.065625\)
\(b.\) \(\frac{-2.6}{x-11}=\frac{-12}{42}\)
<=>\(x-11=\frac{-2.6\cdot-12}{42}\)
<=>\(x-11=0.742857142\)
<=>\(x=11.74285714\)
\(3\frac{4}{5}:2x=0,25:2\frac{2}{3}\)
\(\Leftrightarrow\frac{19}{5}:2x=0,25:\frac{8}{3}\)
\(\Leftrightarrow\frac{19}{5}:2x=0,25\cdot\frac{3}{8}\)
\(\Leftrightarrow\frac{19}{5}:2x=\frac{3}{32}\)
\(\Leftrightarrow2x=\frac{19}{5}:\frac{3}{32}=\frac{19}{5}\cdot\frac{32}{3}=\frac{608}{15}\)
\(\Leftrightarrow x=\frac{608}{15}:2=\frac{608}{30}=\frac{304}{15}\)
\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
a) => x . 7,2 = 0,15 . 3,15
=> x . 7,2 =0,4725
=> x = 0,4725 : 7,2 =21/320
b) => x . 25x = 6.24
=> 25 . x2 =144
=> x2 = 144 : 25 =144/25
=> x= 12/5 hoặc -12/5 hoặc 12/-5
c) => 1/5:x =8/3
=> x = 1/5 :8/3 =3/40
d) => x-1 = 6,7 . ( x+5)
=> x-1 = 6,7x+33,5
=> x-6,7x = 33,5+1 =34,5
=> x ( 1-6,7) = 34,5
=> -5,7 x = 34,5
=> x = 34,5 : (-5,7) =-115/19
\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)
\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)
\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)
\(=\frac{3^{61}}{3^{60}}\)
\(=3\)
\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)
\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)
\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)
\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)
\(=\frac{3^4}{3^3}\)
\(=3\)
Ta có :
\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
Vậy \(P=\frac{3}{11}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)
đề bài của bn sai nên mk sửa luôn nha
\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)
\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)
\(=\frac{3}{5}+\frac{3}{5}\)
\(=\frac{6}{5}\)
\(\frac{2^{10}.3^{11}+2^{10}.3^6}{2^{11}.3^{11}+2^{11}.3^6}=\frac{2^{10}\left(3^{11}+3^6\right)}{2^{11}\left(3^{11}+3^6\right)}=\frac{2^{10}}{2^{11}}=\frac{1}{2}\)