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a)\(=\frac{1}{7}\)
b)\(=\frac{1}{3}\)
quy đồng \(\frac{1}{7}=\frac{3.1}{3.7}=\frac{3}{21}\)
\(\frac{1}{3}=\frac{7.1}{7.3}=\frac{7}{21}\)
\(\frac{3}{21}+\frac{7}{21}=\frac{9}{21}=\frac{3}{7}\)
C = \(\frac{1010}{\left(1010-2\right).8-994}\)
= \(\frac{1010}{1010.8-2.8-994}\)
=\(\frac{1010}{1010.\left(8-1\right)}\)
=\(\frac{1}{7}\)
D = \(\frac{1.2.3.\left(1+2^3+3^3+5^3\right)}{1.3.6.\left(1+2^3+3^3+5^3\right)}\)
= \(\frac{1}{3}\)
Quy đồng ta được 2 phân số là: \(\frac{3}{21};\frac{7}{21}\)
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Rightarrow\left(2x-9\right).80=240.39\)
\(160x-720=9360\)
\(160x=9360+720\)
\(160x=10080\)
\(x=10080:160\)
\(x=63\)
Vậy x=63
1. Ta có : \(C=\frac{1010}{1008\cdot8-994}=\frac{1010}{\left(1010-2\right)\cdot8-994}=\frac{1010}{1010\cdot8-16-994}\)
\(=\frac{1010}{1010\cdot8-1010}=\frac{1010}{1010\left(8-1\right)}=\frac{1}{7}\)
\(D=\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot3\cdot6+2\cdot6\cdot12+3\cdot9\cdot18+5\cdot15\cdot30}\)
\(D=\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{1\cdot2\cdot3\cdot3+2\cdot4\cdot6\cdot3+3\cdot6\cdot9\cdot3+5\cdot10\cdot15\cdot3}\)
\(D=\frac{1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15}{3\left(1\cdot2\cdot3+2\cdot4\cdot6+3\cdot6\cdot9+5\cdot10\cdot15\right)}=\frac{1}{3}\)
2. \(\frac{2x-9}{240}=\frac{39}{80}\)
=> \(\frac{2x-9}{240}=\frac{39\cdot3}{80\cdot3}=\frac{117}{240}\)
=> 2x - 9 = 117
=> 2x = 117 + 9 = 126
=> x = 126 : 2 = 63
C = \(\frac{1010}{1008.8-994}\)
= \(\frac{1010}{1008.7+1008-994}\)
= \(\frac{1010}{1008.7+14}\)
= \(\frac{1010}{1008.7+2.7}\)
= \(\frac{1010}{7\left(1008+2\right)}=\frac{1010}{7.1010}=\frac{1}{7}\)
D = \(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.6+2.6.12+3.9.18+5.15.30}\)
= \(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{1.3.2.3+2.6.4.3+3.9.6.3+5.15.10.3}\)
= \(\frac{1.2.3+2.4.6+3.6.9+5.10.15}{3\left(1.2.3+2.4.6+3.6.9+5.10.15\right)}=\frac{1}{3}\)
BCNN(7,3) = 21
\(\frac{1}{7}=\frac{1.3}{7.3}=\frac{3}{21}\)
\(\frac{1}{3}=\frac{1.7}{3.7}=\frac{7}{21}\)
