Answer:
22. 165
23. 729
24. 720
25. 81
Solution:
22.
Adding the three equations yields 2(ab+bc+ca) = 206. Hence ab + bc + ca = 103
Subtracting each of the three equations from this, we have ab = 15, bc = 55 and ca = 33. It follows that the square of abc is the product of these three numbers. Since each of a, b and c is positive, abc = 165.
23.
Since [tex]b-9=\sqrt{a^2-6a+b} + |b-9|\geq |b-9|[/tex], we have [tex]b\geq 9[/tex].
Hence [tex]\sqrt{a^2-6a+b} =0[/tex].
If [tex]b>9[/tex], we have [tex]a^2 - 6a + b = (a-3)^2+(b-9)>0[/tex], which is a contradiction.
Hence [tex]b = 9[/tex]. From [tex]a^2 - 6a + 9 = 0[/tex], we have [tex]a = 3[/tex] so that [tex]b^a = 9^3 = 729[/tex]
24.
Since [tex]10[/tex]×[tex]11[/tex]×[tex]12[/tex]×[tex]...[/tex]×[tex]19 = 2^8[/tex]×[tex]3^4[/tex]×[tex]5^2[/tex]×11×13×17×19
Thus the maximum value of a is [tex]2^4[/tex] × [tex]3^2[/tex] × [tex]5 = 720[/tex]
25.
Since only additions and multiplications are involved, the value of the expression grows progressively. Clearly, the three multipliers, namely, the third, the fifth and the seventh numbers, should be 1, 2 and 3 but in reverse order, while the other five numbers should be in increasing order. Hence the minimum value is (((4 + 5) × 3 + 6) × 2 + 7) × 1 + 8 = 81
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