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Ta có 1008. 8 – 994 = 1008. 7 + 1008 – 994
= 1008. 7 + 14
= 7. (1008 + 2) = 7. 1010
Vậy C = 1010/(7.1010) = 1/7
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
\(\frac{1}{7}=\frac{1.3}{7.3}=\frac{3}{21}\)
\(\frac{1}{3}=\frac{1.7}{3.7}=\frac{7}{21}\)
a, - 151515 / 232323 = -15/23
b, 1.2.3 + 2.4.6 + 4.8.12 / 1.3.5 + 2.4.6 + 4.12.20
= 146/341
Ta co:
\(\frac{5+a}{8a}\)= \(\frac{-1}{12}\)
<=> 12(5+a)=-8a
<=>60+12a=-8a
<=>-20a=60
<=>a=-3
nhớ mik nha:)
theo bài ra ta có \(\frac{a+5}{8a}\)=\(\frac{-1}{12}\)suy ra (5+a).12=-8a suy ra 60+12a+8a=0 suy ra 20a=-60 suy ra a=-3
vậy..............
a)\(=\frac{1}{7}\)
b)\(=\frac{1}{3}\)
quy đồng \(\frac{1}{7}=\frac{3.1}{3.7}=\frac{3}{21}\)
\(\frac{1}{3}=\frac{7.1}{7.3}=\frac{7}{21}\)
\(\frac{3}{21}+\frac{7}{21}=\frac{9}{21}=\frac{3}{7}\)
\(\frac{1.2.3+2.4,6+4.8.12+7.14.21}{1.3.5+2.6.10+4.12.20+7.21.35}\)
\(=\frac{1\left(1.2.3\right)+2\left(1.2.3\right)+4\left(1.2.3\right)+7\left(1.2.3\right)}{1\left(1.3.5\right)+2\left(1.3.5\right)+4\left(1.2.3\right)+7\left(1.2.3\right)}\)
\(=\frac{6\left(1+2+4+7\right)}{15\left(1+2+4+7\right)}=\frac{6}{15}=\frac{3}{5}\)
\(A=\frac{2.6.10+4.12.20+6.18.30+...+20.60.100}{1.2.3+2.4.6+3.6.9+...+10.20.30}\)
=> \(A=\frac{2^3.1.3.5+4^3.1.3.5+6^3.1.3.5+...+20^3.1.3.5}{1.2.3+2^3.1.2.3+3^3.1.2.3+...+10^3.1.2.3}\)
=> \(A=\frac{1.3.5\left(2^3+4^3+6^3+...+20^3\right)}{1.2.3\left(1+2^3+3^3+...+10^3\right)}=\frac{1.5.2^3.\left(1+2^3+3^3+...+10^3\right)}{1.2.\left(1+2^3+3^3+...+10^3\right)}\)
=> \(A=5.2^2=20\)
Đáp số: A=20
\(\frac{2\cdot6\cdot10+4\cdot12\cdot20+...+20\cdot60\cdot100}{1\cdot2\cdot3+2\cdot4\cdot6+...+10\cdot20\cdot30}=\frac{10\cdot2\left(1\cdot2\cdot3+2\cdot4\cdot6+...+10\cdot20\cdot30\right)}{1\cdot2\cdot3+2\cdot4\cdot6+...+10\cdot20\cdot30}\)
\(=20\)