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\(\frac{27.18+27.103-27.102}{15.33+33.12}=\frac{27\left(18+103-102\right)}{33\left(15+12\right)}\)
\(=\frac{27.19}{33.27}=\frac{19}{33}\)
k nha
\(\frac{27.18+27.103-27.102}{15.33+33.12}=\frac{27.\left(18+103-102\right)}{33.\left(15+12\right)}=\frac{27.19}{33.27}=\frac{19}{33}\)
\(\frac{27\cdot18+27\cdot103-120\cdot27}{15\cdot33+33\cdot12}\)
\(=\frac{27\left(18+103-120\right)}{33\cdot\left(15+12\right)}\)
\(=\frac{27}{33\cdot3}=\frac{27}{99}\)
\(=\frac{3}{11}\)
Đề bài : Tính
\(\frac{27.18+27.103-120.27}{15.33+33.12}\)
\(=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)
\(=\frac{27}{33.27}\)
\(=\frac{1}{33}.\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\frac{1}{10.11}+7.\frac{1}{11.12}+7.\frac{1}{12.13}+...+7.\frac{1}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=\frac{6}{70}\)
\(=\frac{3}{35}\)
\(M=\frac{32}{323}\) \(N=\frac{86}{589}\) \(\frac{M}{N}=\frac{496}{731}\)
\(A=\frac{4}{23.27}+\frac{5}{27.32}+\frac{6}{32.38}+\frac{8}{38.46}\)
\(A=\frac{1}{23}-\frac{1}{27}+\frac{1}{27}-\frac{1}{32}+\frac{1}{32}-\frac{1}{38}+\frac{1}{38}-\frac{1}{46}\)
\(A=\frac{1}{23}-\frac{1}{46}\)
\(A=\frac{1}{46}\)
=1/23-1/27+1/27-1/32+1/32-1/38+1/38-1/46
=1/32-1/46
xiin lỗi mk ko tính ra kết quả được,mk có chút việc bận
tk mk nha ,thanks nha
\(2\cdot31\cdot12+4\cdot6\cdot42+8\cdot27\cdot3\)
\(=\left(2\cdot12\right)\cdot31+\left(4\cdot6\right)\cdot42+\left(8\cdot3\right)\cdot27\)
\(=24\cdot31+24\cdot42+24\cdot27\)
\(=24\cdot\left(31+42+27\right)\)
\(=24\cdot100=2400\)
\(C=\frac{1.5.3.2+2.10.6.2+4.20.12.2+...+9.45.27.2}{1.3.5+2.6.10+4.12.20+...+9.27.45}\)
\(C=\frac{2\left(1.5.3+2.10.6+4.20.12+...+9.45.27\right)}{1.3.5+2.6.10+4.12.20+...+9.27.45}\) = 2
H=\(\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}=\frac{1.2.3+2^3.\left(1.2.3\right)+3^3.\left(1.2.3\right)+5^3.\left(1.2.3\right)}{1.3.6+2^3.\left(1.3.6\right)+3^3.\left(1.3.6\right)+5^3.\left(1.3.6\right)}=\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}=\frac{2}{6}=\frac{1}{3}\)
=27.(18+103-120)/33.(15+12)
= 27.1/33.27
=1/33
\(=\frac{27\left(18+103-120\right)}{33\left(15+12\right)}\)
\(=\frac{27\times1}{33\times27}\)
\(=\frac{1}{33}\)