Đáp án:

$41)\quad D.\, S = \dfrac{11}{5}$

$42) \quad B.\, 169$

$43) \quad B.\,S = (3;+\infty)\cup (-2;-1)$

Giải thích các bước giải:

$\begin{array}{l}41)\quad \log_2(3x -2) > \log_2(6-5x)\\ \Leftrightarrow \begin{cases}3x - 2 >0\\6 - 5x >0\\3x - 2 > 6 - 5x\end{cases}\\ \Leftrightarrow \begin{cases}\dfrac23 < x < \dfrac65\\x > 1\end{cases}\\ \Leftrightarrow 1 < x < \dfrac65\\ \Leftrightarrow x \in \left(1;\dfrac65\right)\\ \Rightarrow \begin{cases}a = 1\\b = \dfrac65\end{cases}\\ \Rightarrow S = a + b = 1 + \dfrac65 = \dfrac{11}{5} \\ 42)\quad \ln(2x + 3) \geq\ln(2017 - 4x)\\ \Leftrightarrow \begin{cases}2x + 3 >0\\2017 - 4x >0\\2x + 3 \geq 2017 - 4x\end{cases}\\ \leftrightarrow \begin{cases}-\dfrac32 < x < \dfrac{2017}{4}\\x \geq \dfrac{1007}{3}\end{cases}\\ \Leftrightarrow 335,6 \approx \dfrac{1007}{3} \leq x \leq \dfrac{2017}{4} \approx 504,5\\ \text{Ta có:}\\ \quad x \in \Bbb Z^+ \Rightarrow x = \underbrace{\{336;337;\dots;503;504\}}_{\text{169 giá trị x}}\\ 43)\quad \log_{\tfrac{\pi}{4}}(x^2+1) < \log_{\tfrac{\pi}{4}}(2x + 4)\\ \Leftrightarrow \begin{cases}x^2 + 1 >0\\2x + 4 >0\\x^2 + 1 > 2x + 4\end{cases}\\ \Leftrightarrow \begin{cases}x > -2\\\left[\begin{array}{l}x >3\\x <-1\end{array}\right.\end{cases}\\ \Leftrightarrow \left[\begin{array}{l}x >3\\-2 ​< x <-1\end{array}\right.\\\Leftrightarrow S = (3;+\infty)\cup (-2;-1)\end{array}$

Câu 41: 

`D = (2/3; 6/5)`

`log_{2} (3x - 2) > log_{2} (6 - 5x)`

`-> 3x - 2 > 6 - 5x`

`-> 8x > 8`

`-> x > 1`

`-> x in (1; 6/5)`

`-> S = 1 + 6/5 = 11/5`

`->` Chọn `D`

Câu 42:

`D = (-3/2; 2017/4)`

`ln (2x + 3) >= ln (2017 - 4x)`

`-> 2x + 3 >= 2017 - 4x`

`-> 6x >= 2014`

`-> x >= 1007/3`

`-> x in (335,7; 504, 25)`

Mà: `x in ZZ`

`-> x = {336, 337, ...., 504)`

`->` số số hạng là: `504 - 336 + 1 = 169` số 

`->` Chọn `B`

Bài 43:

`D = (-2; +infty)`

$log_{\dfrac{\pi}{4}} (x^2 + 1) < log_{\dfrac{\pi}{4}} (2x + 4)$

`-> x^2 + 1 > 2x + 4`

`-> x^2 - 2x - 3 > 0`

`-> x in (-infty; -1) cup (3; +infty)`

`-> x in (-2; -1) cup (3; +infty)`