a) `A= 81. [(12 - 12/7 - 12/289 -12/85)/(4 - 4/7 - 4/289 -4/85) : (5+ 5/13 + 5/169 + 5/91)/(6+ 6/13 + 6/169 + 6/91)] : 158158158/711711711`
`A= 81 [ (12(1- 1/7 - 1/289 - 1/85))/(4(1 - 1/7 - 1/289 - 1/85)) : (5( 1 + 1/13 + 1/169 + 1/91))/(6(1 + 1/13 + 1/169 + 1/91))] : (158.1001001)/(711. 1001001)`
`A= 81 ( 12/4 : 5/6) : 158/711`
`A = 81 ( 3. 6/5) : 2/9`
`A= 81 . 18/5 : 2/9`
`A= 1458/5 : 2/9`
`A= 1458/5 . 9/2`
`A= 324/5`
Vậy `A= 324/5`
b)
1) `-2/3( x- 1/4) =1/3 ( 2x-1)`
`=> -2/3 x + 1/6 = 1/3 . 2x - 1/3`
`=> -2/3 x + 1/6= 2/3 x - 1/3`
`=> -2/3 x - 2/3 x= -1/3- 1/6`
`=> -4/3 x = -1/2`
`=> x= -1/2 : (-4/3)`
`=> x = 1/2 . 3/4`
`=> x=3/8`
Vậy `x= 3/8`
2) `1/5 . 2^x + 1/3 . 2^(x+1) = 1/5. 2^7 + 1/3 . 2^8`
`=> 1/5 . 2^x + 1/3 . 2^x . 2 = 1/5 . 2^7 + 1/3 . 2^7 .2`
`=> 2^x( 1/5 + 2/3) = 2^7( 1/5 + 2/3)`
`=> 2^x = 2^7`
`=>x=7`
Vậy `x =7`
c) Gọi `ƯCLN(a;b) = d` thì `a= dm; b= dn` `(m;n = 1)`
`BCNNN(a;b) = ab : d = dm . dn : d = dmn`
Ta có: `dmn + d =15`
`=> d(mn +1) =15`
Vì `mn ge 0 => mn + 1 ge 1`
`=> d= 1; mn +1 = 15`
`=> mn = 15 -1 =14`
Từ `mn = 14`
`=> m= 7 ; n =2`
`m = 1; n= 14`
`m= 14; n= 1`
`m= 2; n=7`
Vậy các cặp số `(m:n)` là `( 7;2) ; ( 2;7) ; ( 1;14) ; (14;1)`
d) Với mọi `x` ta luôn có: `|x +1 | ge 0 ; |x-2| ge0; |x+7| ge 0`
`=> |x+1| + |x-2| + |x+7| ge 0`
`=> 5x-10 ge 0`
`=> 5x ge 10`
`=> x ge 2`
Ta có:`|x+1| + |x-2| + |x+7| = 5x -10`
`=> x+1 + x -2 + x+ 7= 5x-10`
`=> 3x + 6 = 5x -10`
`=> 3x - 5x = -10 -6`
`=> -2x = -16`
`=> x= 8` (thỏa mãn)
Vậy `x=8`
