\(log_{12}21=\dfrac{log_321}{log_312}=\dfrac{log_3\left(7\cdot3\right)}{log_3\left(2^2\cdot3\right)}=\dfrac{log_37+log_33}{log_34+log_33}\)

\(=\dfrac{log_37+1}{log_32^2+1}=\dfrac{log_37+1}{2\cdot log_32+1}=\dfrac{b+1}{2a+1}\)

Do 2 > 1 ⇒ hàm số y = log₂x đồng biến trên D = \(\left(0;+\infty\right)\)

\(log_2x>1\\ \Rightarrow x>2\)

\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)

\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)

c, ĐK: \(x>-7\)

\(log_3\left(x+7\right)\ge-1\\ \Leftrightarrow x+7\ge\dfrac{1}{3}\\ \Leftrightarrow x\ge-\dfrac{20}{3}\)

Kết hợp với ĐKXĐ, ta có:\(x\ge-\dfrac{20}{3}\)

d, ĐK: \(x>\dfrac{1}{2}\)

\(log_{0,5}\left(x+7\right)\ge log_{0,5}\left(2x-1\right)\\ \Leftrightarrow x+7\le2x-1\\ \Leftrightarrow x\ge8\)

Kết hợp với ĐKXĐ, ta được: \(x\ge8\)

a) \({\log _{\frac{1}{7}}}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\)               (ĐK: \(x + 1 > 0;2 - x > 0 \Leftrightarrow  - 1 < x < 2\))

\(\begin{array}{l} \Leftrightarrow {\log _{{7^{ - 1}}}}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow  - {\log _7}\left( {x + 1} \right) > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow {\log _7}{\left( {x + 1} \right)^{ - 1}} > {\log _7}\left( {2 - x} \right)\\ \Leftrightarrow {\left( {x + 1} \right)^{ - 1}} > 2 - x\\ \Leftrightarrow \frac{1}{{x + 1}} - 2 + x > 0\\ \Leftrightarrow \frac{{1 + \left( {x - 2} \right)\left( {x + 1} \right)}}{{x + 1}} > 0\\ \Leftrightarrow \frac{{1 + {x^2} - x - 2}}{{x + 1}} > 0 \Leftrightarrow \frac{{{x^2} - x - 1}}{{x + 1}} > 0\end{array}\)

Mà – 1 < x < 2 nên x + 1 > 0

\( \Leftrightarrow {x^2} - x - 1 > 0 \Leftrightarrow \left[ \begin{array}{l}x < \frac{{1 - \sqrt 5 }}{2}\\x > \frac{{1 + \sqrt 5 }}{2}\end{array} \right.\)

KHĐK ta có \(\left[ \begin{array}{l} - 1 < x < \frac{{1 - \sqrt 5 }}{2}\\\frac{{1 + \sqrt 5 }}{2} < x < 2\end{array} \right.\)

b) \(2\log \left( {2x + 1} \right) > 3\)              (ĐK: \(2x + 1 > 0 \Leftrightarrow x > \frac{{ - 1}}{2}\))

\(\begin{array}{l} \Leftrightarrow \log \left( {2x + 1} \right) > \frac{3}{2}\\ \Leftrightarrow 2x + 1 > {10^{\frac{3}{2}}} = 10\sqrt {10} \\ \Leftrightarrow x > \frac{{10\sqrt {10}  - 1}}{2}\end{array}\)

KHĐK ta có \(x > \frac{{10\sqrt {10}  - 1}}{2}\)

\(a,D=R\\ b,2x-3>0\\ \Rightarrow x>\dfrac{3}{2}\\ \Rightarrow D=(\dfrac{3}{2};+\infty)\\ c,-x^2+4>0\\ \Rightarrow x^2< 4\\ \Leftrightarrow-2< x< 2\\ \Rightarrow D=\left(-2;2\right)\)

có ai pít làm ko ,nhanh lên dùm mk nha 

Chọn D

Đáp án C.

a) \(\log_{12}12^3=3.\log_{12}12=3.1=3\)

b) \(\log_{0,5}0,25=\log_{2^{-1}}2^{-2}=\dfrac{-2}{-1}\log_22=2.1=2\)

c) \(\log_aa^{-3}=-3.\log_aa=-3.1=-3\)

a: \(log_{12}12^3=3\)

b: \(=log_{0.5}0.5^2=2\)

c: \(log_aa^{-3}=-3\)

\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)

d, Điều kiện: x > 1

\(log\left(x-1\right)< 0\\ \Leftrightarrow x-1< 1\\ \Leftrightarrow1< x< 2\)

e, Điều kiện: \(x>\dfrac{1}{2}\)

\(log_{\dfrac{1}{5}}\left(2x-1\right)\ge log_{\dfrac{1}{5}}\left(x+3\right)\\ \Leftrightarrow2x-1\ge x+3\\ \Leftrightarrow x\ge4\)

f, Điều kiện: x > 4

\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)