(x-6)10=(x-6)8
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6 x 5 = 30
6 x 10 = 60
6 x 2 = 12
6 x 7 = 42
6 x 8 = 48
6 x 3 = 18
6 x 9 = 54
6 x 6 = 36
6 x 4 = 24
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
6 x 1 = 6 6 x 9 = 54
6 x 2 = 12 6 x 8 = 48
6 x 3 = 18 6 x 7 = 42
6 x 4 = 24 0 x 6 = 0
6 x 5 = 30 6 x 0 = 0
6 x 6 = 36 6 x 10 = 60
6 x 4 =24
6 x 1 =6
6 x 9 =54
6 x 10 =60
6 x 6 =36
6 x 3 =18
6 x 2 =12
0 x 6 =0
6 x 8 =48
6 x 5 =30
6 x 7 =42
6 x 0 =0
(X -10/1994 -1) + (X-8/1996 - 1) + (X-6/1998 - 1)+ (X-4/2000 - 1) + (X-2/2002 - 1) = (X-2002/2 - 1) + (X-2000/4 - 1) + (X-1998/6 - 1) + (X-1996/8 - 1) + (X-1994/10 - 1)
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 = x-2004/2 + x-2004/4 + x-2004/6 + x-2004/8 + x-2004/1994
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 - x-2004/2 - x-2004/4 - x-2004/6 - x-2004/8 - x-2004/1994 = 0
=> (x - 2004)(1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) = 0
Mà (1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) \(\ne\)0
=> x - 2004 = 0
=> x = 2004
Vậy x = 2004
`(3xx4xx7)/(5xx3xx4)=7/5`
`(2xx5xx6xx8)/(6xx2xx8xx9)=5/9`
`(4xx5xx6)/(3xx10xx8)= (4xx5xx6)/(3xx5xx2xx4xx2)= 6/(3xx2xx2)= 6/(6xx2)=1/2`
\(a,\dfrac{3\times4\times7}{5\times3\times4}=\dfrac{7}{5}\)
\(b,\dfrac{2\times5\times6\times8}{6\times2\times8\times9}=\dfrac{5}{9}\)
\(c,\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{2}{4}=\dfrac{1}{2}\)
b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow x+3=11\)
hay x=8
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)
\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)
Suy ra: x=5
d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)
\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)
\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)
\(\Leftrightarrow8^x=8^{20}\)
Suy ra: x=20