answer:Let's assume the common difference of the arithmetic sequence {a_n} is d. Since a_5=11 and a_8=5, we have d= dfrac{a_8-a_5}{8-5}= dfrac{5-11}{3}=-2, thus a_1=a_5-4d=11-4times(-2)=19, therefore, the general formula for a_n is a_n=a_1+(n-1)d=19-2(n-1)=-2n+21, and the sum of the first 10 terms is S_{10}= dfrac{10(19+21-2times10)}{2}=100. Hence, the general formula for a_n is boxed{-2n+21} and the sum of the first 10 terms S_{10} is boxed{100}.

answer:11. Answer: 8000 . tan alpha=4 tan beta=frac{4}{tan alpha} Rightarrow tan alpha=2. The two legs are frac{200}{sqrt{5}} and frac{400}{sqrt{5}} respectivelytext { Area }=frac{1}{2} times frac{200}{sqrt{5}} times frac{400}{sqrt{5}}=8000 .

answer:Two, 7.2 leqslant x<8.Notice that, the required inequality is a quadratic inequality in terms of [x].begin{array}{l}text { By } 4[x]^{2}-36[x]+45 leqslant 0 Rightarrow frac{3}{2} leqslant[x] leqslant frac{15}{2} Rightarrow 2 leqslant[x] leqslant 7 Rightarrow 2 leqslant x<8 .end{array}

answer:6

answer:【Solution】Suppose f satisfies the conditions in the problem, then we havefleft(x^{u} y^{v}right) leqslant f(x)^{frac{1}{4 u}} f(y)^{frac{1}{4 v}} .Taking u=frac{1}{2}, we getfleft(x^{frac{1}{2}} y^{v}right) leqslant f(x)^{frac{1}{2}} f(y)^{frac{1}{4 v}},Taking v such that quad y^{v}=x^{frac{1}{2}} quad(x>1, y>1),i.e.,v=frac{ln x}{2 ln y} text {. }Thus,begin{array}{l}f(x) leqslant f(x)^{frac{1}{2}} cdot f(y)^{frac{ln y}{2 ln x}}, f(x)^{frac{1}{2}} leqslant f(y)^{frac{ln y}{2 ln x}}, f(x)^{ln x} leqslant f(y)^{ln y} .end{array}By symmetry, we get f(y)^{ln y} leqslant f(x)^{ln x}.From (3) and (4), we have f(x)^{ln x}=f(y)^{ln y}.This implies that f(x)^{ln x} is a constant C(C>1), i.e.,begin{array}{l}f(x)^{ln x}=C text {, } f(x)=C^{frac{1}{ln x}} quad(C>1) text {. } end{array}Conversely, if f(x)=C^{frac{1}{ln x}}(C>1), then for x, y>1, u, v>0, we havebegin{aligned}fleft(x^{u} y^{v}right) & =C^{frac{1}{ln left(x^{prime prime} y^{prime prime}right)}} & =C^{frac{1}{u ln x+v ln y}} .end{aligned}Since (u ln x+v ln y)^{2} geqslant 4 u v ln x ln y,it follows that frac{1}{u ln x+v ln y} leqslant frac{u ln x+v ln y}{4 u v ln x ln y}=frac{1}{4 v ln y}+frac{1}{4 u ln x} text {. }Thus,begin{aligned}& fleft(x^{u} y^{v}right) = & C^{frac{1}{4 ln x+v ln y}} leqslant & C^{frac{1}{4 u ln x}+frac{1}{4 v ln y}} = & C^{frac{1}{4 u ln x}} cdot C^{frac{1}{4 v ln y}} = & f(x)^{frac{1}{4 u}} cdot f(y)^{frac{1}{42}} .end{aligned}Therefore, the function that satisfies the conditions in the problem isf(x)=C^{frac{1}{ln x}} quad(C>0) .

answer:To solve, the coefficient of the x^2 term in the expansion of (x^2 - 3x + 2)^4 is calculated as C_{4}^{1} times 2^{3} + C_{4}^{2} times (-3)^{2} times 2^{2} = 248. Therefore, the answer is boxed{248}. Analyzing the origins of the x^2 term, there are two possible scenarios to calculate its coefficient. This question examines the method of finding the coefficient of a characteristic term in the expansion of a polynomial, with the key being to clarify the possible origins of the x^2 term.