1. 当 $x$ 为何值时, $x+5$, $x + 2$, $x - 1$, $x - 4$ 的积不大于 $-80$?
解答: $$(x + 5)(x + 2)(x - 1)(x - 4) le-80 Rightarrow (x^2 + x - 20)(x^2 + x - 2) + 80 le 0$$ $$Rightarrow (x^2 + x)^2 - 22(x^2 + x) + 120 le 0 Rightarrow (x^2 + x - 10)(x^2 + x - 12) le 0$$ $$Rightarrow left(x - {-1 + sqrt{41} over 2} ight)left(x - {-1 - sqrt{41} over 2} ight)(x + 4)(x - 3) le 0$$ $$Rightarrow x in left[-4, {-1 - sqrt{41} over 2} ight] cup left[{-1 + sqrt{41} over 2}, 3 ight].$$
2. 求 $k$ 的取值范围, 使得下列不等式恒成立 $${2x^2 + 2kx + k over 4x^2 + 6x + 3} < 1.$$ 解答: $${2x^2 + 2kx + k over 4x^2 + 6x + 3} < 1 Rightarrow 2x^2 + 2kx + k < 4x^2 + 6x + 3$$ $$2x^2 + (6 - 2k)x + (3 - k) > 0 Rightarrow Delta = (6 - 2k)^2 - 8(3 - k) < 0$$ $$Rightarrow 4k^2 - 16k + 12 < 0 Rightarrow k^2 - 4k + 3 < 0$$ $$ Rightarrow kin (1, 3).$$
3. 解关于 $x$ 的不等式: $$x^{log_ax} > {x^{9over2} over a^2}$$ 其中 $a > 0$, $a e 1$.
解答:
令 $log_ax = t$, 则 $$x^t > {x^{9over2} over a^2} Rightarrow x^{t - {9over2}} > a^{-2}$$ $a > 1$ 时, $$log_ax^{t - {9over2}} > -2 Rightarrow tleft(t - {9over2} ight) > -2$$ $$Rightarrow 2t^2 - 9t + 4 > 0 Rightarrow t < {1over2}, t > 4$$ $$Rightarrow log_ax < {1over2}, log_ax > 4$$ $$Rightarrow x inleft(0, a^{1over2} ight)cupleft(a^4, +infty ight).$$ $0 < a < 1$ 时, $$log_ax^{t - {9over2}} < -2 Rightarrow tleft(t - {9over2} ight) < -2$$ $$Rightarrow 2t^2 - 9t + 4 < 0 Rightarrow {1over2} < t < 4$$ $$Rightarrow {1over2} < log_ax < 4$$ $$Rightarrow x inleft(a^4, a^{1over2} ight).$$ 综上, 不等式的解集为 $$egin{cases} x inleft(0, a^{1over2} ight)cupleft(a^4, +infty ight), (a > 1)\ x inleft(a^4, a^{1over2} ight), (0 < a < 1)end{cases}$$
4. 解不等式: $${1over 2^x - 1} > {1 over 1 - 2^{x - 1}}.$$ 解答:
令 $2^x = t$, 则 $${1over t-1} > {1over 1 - {1over2}t} Rightarrow {1 over t - 1} > {2 over 2 - t} Rightarrow {1over t - 1} + {2 over t - 2} > 0$$ $$Rightarrow {3t-4 over (t - 1)(t - 2)} > 0 Rightarrow (3t - 4)(t - 1)(t - 2) > 0 Rightarrow 1 < t < {4over3}, t > 2$$ $$Rightarrow 1 < 2^x < {4over3}, 2^x > 2 Rightarrow xinleft(0, 2 - log_23 ight) cup (1, +infty).$$
5. 设 $a$ 是正常数, 解不等式: $$2^{3x} - 2^x < a(2^x - 2^{-x}).$$ 解答:
令 $2^x = t$, 则 $$t^3 - t < aleft(t - {1over t} ight) Rightarrow t^4 - t^2 < a(t^2 - 1) Rightarrow (t^2 - a)(t - 1)(t + 1) < 0$$ $a = 1$ 时, $$(t+1)^2(t-1)^2 < 0 Rightarrow tinphi.$$ $a > 1$ 时, $$(t + sqrt{a})(t - sqrt{a})(t + 1)(t - 1) < 0 Rightarrow -sqrt{a} < t < -1, 1 < t < sqrt{a}$$ $$1 < 2^x < sqrt{a} Rightarrow 0 < x < {1over2}log_2a.$$ $0 < a < 1$ 时, $$(t + sqrt{a})(t - sqrt{a})(t + 1)(t - 1) < 0 Rightarrow -1 < t < -sqrt{a}, sqrt{a} < t < 1$$ $$sqrt{a} < 2^x < 1 Rightarrow {1over2}log_2a < x < 0.$$ 综上, $$egin{cases}xinphi, (a = 1)\ 0 < x < {1over2}log_2a, (a > 1)\ {1over2}log_2a < x < 0, (0 < a < 1)end{cases}$$
6. 在直角坐标平面上求满足不等式 $$2 - x^2 - y^2 - sqrt{(1 - x^2)^2 + (1 + y^2)^2} > 0$$ 的点集 $(x, y)$.
解答: $$left(1 - x^2 ight) + left(1 - y^2 ight) > sqrt{(1 - x^2)^2 + (1 + y^2)^2} ge 0$$ $$Rightarrow left[left(1 - x^2 ight) + left(1 - y^2 ight) ight]^2 > (1 - x^2)^2 + (1 + y^2)^2$$ $$Rightarrow egin{cases}left(1 - x^2 ight) + left(1 - y^2 ight) > 0\ left(1 - x^2 ight)left(1 - y^2 ight) > 0 end{cases}$$ $$Rightarrow egin{cases}1 - x^2 > 0\ 1 - y^2 > 0 end{cases} Rightarrow egin{cases}|x| < 1\ |y| < 1 end{cases}$$ 即满足题意的点集 $(x, y)$ 是位于正方形 $-1 < x < 1$, $-1 < y < 1$ 之内点.
7. 解不等式: $$10^{lg^2x} + x^{lg{1over x}} + {1over sqrt{y - y^2}} le 2sqrt2(sinpi z + cospi z).$$ 解答:
与例题类似处理方法. $$egin{cases}10^{lg^2x} + x^{lg{1over x}} = x^{lg x} + x^{-lg x} ge 2\ y-y^2 = y(1 - y) le left({y + 1-y over 2} ight)^2 = {1over 4}Rightarrow {1oversqrt{y - y^2}} ge 2\ 2sqrt2(sinpi z + cospi z) le 2sqrt2 cdot sqrt2 = 4end{cases}$$ 因此上述不等式均取等号. $$egin{cases}x = 1\ y = {1over2}\ z = 2k + {1over4}, (kinmathbf{Z})end{cases}$$
8. 解关于 $x$ 的不等式: $$sqrt{a^2 - 2x^2} > x + a.$$ 解答:
分类讨论.
$x + a < 0$ 时, $$egin{cases}x + a <0\ a^2 - 2x^2 ge 0Rightarrow x^2 le {1over2}a^2 end{cases}$$ 若 $a = 0$, 则无解.
若 $a > 0$, 则 $$egin{cases}x < -a\ -{sqrt2 over 2}a le x le {sqrt2over2}aend{cases} Rightarrow xinphi$$ 若 $a < 0$, 则 $$egin{cases}x < -a\ {sqrt2 over 2}a le x le -{sqrt2over2}aend{cases} Rightarrow xinleft[{sqrt2over2}a, -{sqrt2over2}a ight]$$ $x + a ge 0$ 时, $$egin{cases}x + a ge 0\ a^2 - 2x^2 ge 0\ a^2 - 2x^2 > (x + a)^2 end{cases} Rightarrow egin{cases}x ge -a\ x^2 le {1over2}a^2\ 3x^2+2ax < 0 Rightarrow x(3x + 2a) < 0end{cases}$$ 若 $a = 0$, 则无解.
若 $a > 0$, 则 $$egin{cases}x ge -a\ -{sqrt2 over 2}a le x le {sqrt2over2}a\ -{2over3}a < x < 0end{cases} Rightarrow x in left(-{2over3}a, 0 ight)$$ 若 $a < 0$, 则 $$egin{cases}x ge -a\ {sqrt2 over 2}a le x le -{sqrt2over2}a\ 0 < x < -{2over3}aend{cases} Rightarrow x in phi$$ 综上, $$egin{cases} x in left(-{2over3}a, 0 ight), (a > 0)\ xinleft[{sqrt2over2}a, -{sqrt2over2}a ight], (a < 0)\ xinphi, (a = 0)end{cases}$$