ĐGTD ĐH Bách khoa - Tư duy Toán học - Bất phương trình có mũ

$$\begin{gathered} e^x+e^{-x}<\frac{5}{2}\iff e^{2x}+1<\frac{5}{2}{e}^x\iff 2e^{2x}-5e^x+2<0 \\ \iff \left(e^x-2\right)\left(2e^x-1\right)<0\iff \frac{1}{2}<e^x<2\iff -\ln 2<x<\ln 2 \end{gathered}$$

Đáp án cần chọn là: B

Vì $0<\frac{1}{3}<1$ nên ta có

$$\begin{gathered} {\left(\frac{1}{3}\right)}^{\sqrt{x^2-3x-10}}>{\left(\frac{1}{3}\right)}^{x-2}\iff \sqrt{x^2-3x-10}<x-2 \\ \iff \left\{\begin{array}{c}{x}^2-3x-10<{(x-2)}^2\\ x^2-3x-10\ge 0\\ x-2>0\end{array}\right.\iff 5\le x<14 \\ \Rightarrow x=\left\{\mathrm{5,6,7,8,9,10,11,12,13}\right\} \end{gathered}$$

Đáp án cần chọn là: C

Ta có

$\begin{array}{l}{\left(\frac{1}{5}\right)}^{x^2-2\text{x}}\ge \frac{1}{125}\iff {\left(\frac{1}{5}\right)}^{x^2-2\text{x}}\ge {\left(\frac{1}{5}\right)}^3\\ \iff x^2-2\text{x}\le 3\iff x^2-2\text{x}-3\le 0\iff -1\le \text{x}\le 3\end{array}$

Số nghiệm nguyên là 5.

Đáp án cần chọn là: D

${\left(x^2+x+1\right)}^x<1$

Lấy loganepe hai vế ta có $\ln {\left(x^2+x+1\right)}^x<\ln 1\left(\ast \right)$

Vì

$$\begin{gathered} x^2+x+1={(x+\frac{1}{2})}^2+34>0\Rightarrow (\ast )\iff xln \\ (x^{\mathit{2}}+x+1)<0\iff [\begin{array}{c}\left\{\begin{array}{c}x\mathrm{<}\mathrm{0}\\ ln\mathrm{(}{x}^{\mathrm{2}}\mathrm{+}x\mathrm{+}\mathrm{1}\mathrm{)}\mathrm{>}\mathrm{0}\end{array}\right.\\ \left\{\begin{array}{c}x\mathrm{>}\mathrm{0}\\ ln\mathrm{(}{x}^{\mathrm{2}}\mathrm{+}x\mathrm{+}\mathrm{1}\mathrm{)}\mathrm{<}\mathrm{0}\end{array}\right.\end{array} \\ \iff \left[\begin{array}{c}\left\{\begin{array}{c}x<0\\ x^2+x+1>1\end{array}\right.\\ \left\{\begin{array}{c}x>0\\ x^2+x+1<1\end{array}\right.\end{array}\right.\iff \left[\begin{array}{c}\left\{\begin{array}{c}x<0\\ x^2+x>0\end{array}\right.\\ \left\{\begin{array}{c}x>0\\ x^2+x<0\end{array}\right.\end{array}\right.\iff \left[\begin{array}{c}\left\{\begin{array}{c}x<0\\ \left[\begin{array}{c}x>0\\ x<-1\end{array}\right.\end{array}\right.\\ \left\{\begin{array}{c}x>0\\ -1<x<0\end{array}\right.\end{array}\right.\iff x<-1 \end{gathered}$$

Vậy tập nghiệm của bất phương trình là $\left(-\infty ;-1\right)$

Đáp án cần chọn là: C

$\begin{array}{l}{\left(2^n+3^n\right)}^{2020}<{\left(2^{2020}+3^{2020}\right)}^n\\ \iff \ln {\left(2^n+3^n\right)}^{2020}<\ln {\left(2^{2020}+3^{2020}\right)}^n\\ \iff 2020\ln \left(2^n+3^n\right)<n\ln \left(2^{2020}+3^{2020}\right)\\ \iff \frac{\ln \left(2^n+3^n\right)}{n}<\frac{\ln \left(2^{2020}+3^{2020}\right)}{2020}\end{array}$

Xét hàm đặc trưng $f\left(x\right)=\frac{\ln \left(2^x+3^x\right)}{x}\left(x\in {\mathbb{N}}^{\ast }\right)$ ta có:

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\frac{(2^x+3^x)\text{'}}{2^x+3^x}.x-ln(2^x+3^x)}{x^2}\forall x\in {\mathbb{N}}^{\ast } \\ f\text{'}\left(x\right)=\frac{(2^{x}ln2+3^{x}ln3)x-(2^x+3^x).ln(2^x+3^x)}{x^2(2^x+3^x)}\forall x\in {\mathbb{N}}^{\ast } \\ =\frac{2^{x}ln2.x-2^{x}ln(2^x+3^x)+3^{x}ln3.x-3^{x}ln(2^x+3^x)}{x^2(2^x+3^x)}\forall x\in {\mathbb{N}}^{\ast } \\ f\text{'}\left(x\right)=\frac{2^x(xln2-ln(2^x+3^x\left)\right)+3^x(xln3-ln(2^x+3^x\left)\right)}{x^2(2^x+3^x)}\forall x\in {\mathbb{N}}^{\ast } \\ f\text{'}\left(x\right)=\frac{2^x[ln{2}^x-ln(2^x+3^x\left)\right]+3^x[ln{3}^x-ln(2^x+3^x\left)\right]}{x^2(2^x+3^x)}\forall x\in {\mathbb{N}}^{\ast } \end{gathered}$$

Vì $\left\{\begin{array}{c}{2}^x<2^x+3^x\Rightarrow ln{2}^x<ln(2^x+3^x)\\ 3^x<2^x+3^x\Rightarrow ln{3}^x<ln(2^x+3^x)\end{array}\right.\Rightarrow f\text{'}\left(x\right)<0\forall x\in \mathbb{N}\ast$

⇒ Hàm số y = f(x) nghịch biến trên N*

Lại có: f(n)<f(2020) => n>2020

Kết hợp điều kiện đề bài ta có $2020<n\le \mathrm{9999,}n\in {\mathbb{N}}^{\ast }$

Vậy có $\frac{9999-2021}{1}+1=7979$ giá trị của n thỏa mãn yêu cầu bài toán.

Đáp án cần chọn là: D