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a) \({3^{{x^2} - 4x + 5}} = 9 \Leftrightarrow {x^2} - 4x + 5 = 2 \Leftrightarrow {x^2} - 4x + 3 = 0 \Leftrightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0\)
\( \Leftrightarrow \left[ \begin{array}{l}x = 3\\x = 1\end{array} \right.\)
Vậy phương trình có nghiệm là \(x \in \left\{ {1;3} \right\}\)
b) \(0,{5^{2x - 4}} = 4 \Leftrightarrow 2x - 4 = {\log _{0,5}}4 \Leftrightarrow 2x = 2 \Leftrightarrow x = 1\)
Vậy phương trình có nghiệm là x = 1
c) \({\log _3}(2x - 1) = 3\) ĐK: \(2x - 1 > 0 \Leftrightarrow x > \frac{1}{2}\)
\( \Leftrightarrow 2x - 1 = 27 \Leftrightarrow x = 14\) (TMĐK)
Vậy phương trình có nghiệm là x = 14
d) \(\log x + \log (x - 3) = 1\) ĐK: \(x - 3 > 0 \Leftrightarrow x > 3\)
\(\begin{array}{l} \Leftrightarrow \log \left( {x.\left( {x - 3} \right)} \right) = 1\\ \Leftrightarrow {x^2} - 3x = 10\\ \Leftrightarrow {x^2} - 3x - 10 = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x - 5} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = - 2 (loại) \,\,\,\\x = 5 (TMĐK) \,\,\,\,\,\,\,\end{array} \right.\end{array}\)
Vậy phương trình có nghiệm x = 5
\(a,3^x>\dfrac{1}{243}\\ \Leftrightarrow3^x>3^{-5}\\ \Leftrightarrow x>-5\\ b,\left(\dfrac{2}{3}\right)^{3x-7}\le\dfrac{3}{2}\\ \Leftrightarrow3x-7\le1\\ \Leftrightarrow3x\le8\\ \Leftrightarrow x\le\dfrac{8}{3}\\ c,4^{x+3}\ge32^x\\ \Leftrightarrow2^{2x+6}\ge2^{5x}\\ \Leftrightarrow2x+6\ge5x\\ \Leftrightarrow3x\le6\\ \Leftrightarrow x\le2\)
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
\(ln\left(x+3\right)\ge ln\left(2x-8\right)\\ \Leftrightarrow x+3\ge2x-8\\\Leftrightarrow4< x\le11\)
\(a,\left(0,3\right)^{x-3}=1\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\\ b,5^{3x-2}=25\\ \Leftrightarrow3x-2=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\\ c,9^{x-2}=243^{x+1}\\ \Leftrightarrow3^{2x-4}=3^{5x+5}\\ \Leftrightarrow2x-4=5x+5\\ \Leftrightarrow3x=-9\\ \Leftrightarrow x=-3\)
d, Điều kiện: \(x>-1;x\ne0\)
\(log_{\dfrac{1}{x}}\left(x+1\right)=-3\\ \Leftrightarrow x+1=x^3\\ x\simeq1,325\left(tm\right)\)
e, Điều kiện: \(x>\dfrac{5}{3}\)
\(log_5\left(3x-5\right)=log_5\left(2x+1\right)\\ \Leftrightarrow3x-5=2x+1\\ \Leftrightarrow x=6\left(tm\right)\)
f, Điều kiện: \(x>\dfrac{1}{2}\)
\(log_{\dfrac{1}{7}}\left(x+9\right)=log_{\dfrac{1}{7}}\left(2x-1\right)\\ \Leftrightarrow x+9=2x-1\\ \Leftrightarrow x=10\left(tm\right)\)
\(a,0,1^{2-x}>0,1^{4+2x}\\ \Leftrightarrow2-x>2x+4\\ \Leftrightarrow3x< -2\\ \Leftrightarrow x< -\dfrac{2}{3}\)
\(b,2\cdot5^{2x+1}\le3\\ \Leftrightarrow5^{2x+1}\le\dfrac{3}{2}\\ \Leftrightarrow2x+1\le log_5\left(\dfrac{3}{2}\right)\\ \Leftrightarrow2x\le log_5\left(\dfrac{3}{2}\right)-1\\ \Leftrightarrow x\le\dfrac{1}{2}log_5\left(\dfrac{3}{2}\right)-\dfrac{1}{2}\\ \Leftrightarrow x\le log_5\left(\dfrac{\sqrt{30}}{10}\right)\)
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: \(x\ge8\)
\(y=\dfrac{1}{3}\left(m-1\right)x^3-\left(m-1\right)x^2+\left(m+3\right)x-2\)
\(y'=\)\(x^2\left(m-1\right)-2x\left(m-1\right)+m+3\)
a)\(y'=0\)\(\Leftrightarrow x^2\left(m-1\right)-2x\left(m-1\right)+m+3=0\)
Xét m=1 => pt tt: 3=0 (vô lí)
=> \(m\ne1\)
Để y'=0 có hai nghiệm pb cùng dấu
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta>0\\x_1x_2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-16m+16>0\\\dfrac{m+3}{m-1}>0\end{matrix}\right.\)\(\Rightarrow m< -3\)
b)y'=0 có hai nghiệm \(\Leftrightarrow\Delta\ge0\) \(\Leftrightarrow m\le-3\)
Theo viet có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{m-1}=2\\x_1x_2=\dfrac{m+3}{m-1}\end{matrix}\right.\)
Có x12+x22=4
\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=4\)
\(\Leftrightarrow\)\(4-\dfrac{2\left(m+3\right)}{m-1}=4\)
\(\Leftrightarrow m=-3\) (tm)
Vậy m=-3
(đúng không ạ?)
\(a,\left(\dfrac{1}{4}\right)^{x-2}=\sqrt{8}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^{2x-4}=\left(\dfrac{1}{2}\right)^{-\dfrac{3}{2}}\\ \Leftrightarrow2x-4=-\dfrac{3}{2}\\ \Leftrightarrow2x=\dfrac{5}{2}\\ \Leftrightarrow x=\dfrac{5}{4}\)
\(b,9^{2x-1}=81\cdot27^x\\ \Leftrightarrow3^{4x-2}=3^{4+3x}\\ \Leftrightarrow4x-2=4+3x\\ \Leftrightarrow x=6\)
c, ĐK: \(x-2>0\Rightarrow x>2\)
\(2log_5\left(x-2\right)=log_59\\
\Leftrightarrow log_5\left(x-2\right)^2=log_59\\
\Leftrightarrow\left(x-2\right)^2=3^2\\
\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\\
\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = 5.
d, ĐK: \(x-1>0\Leftrightarrow x>1\)
\(log_2\left(3x+1\right)=2-log_2\left(x-1\right)\\ \Leftrightarrow log_2\left(3x+1\right)\left(x-1\right)=2\\ \Leftrightarrow3x^2-2x-1=4\\ \Leftrightarrow3x^2-2x-5=0\\ \Leftrightarrow\left(3x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Vậy phương trình có nghiệm \(x=\dfrac{5}{3}\)
Bài 2:
a: \(log^2_{\dfrac{1}{3}}x-5\cdot log_3x+4=0\)
=>\(log_{\dfrac{1}{3}}^2x+5\cdot log_{\dfrac{1}{3}}x+4=0\)
=>\(\left(log_{\dfrac{1}{3}}x+1\right)\left(log_{\dfrac{1}{3}}x+4\right)=0\)
=>\(\left[{}\begin{matrix}log_{\dfrac{1}{3}}x=-1\\log_{\dfrac{1}{3}}x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=81\end{matrix}\right.\)
b: \(log_2^24x-log_{\sqrt{2}}2x=5\)
=>\(log_2^24x-2\cdot log_22x=5\)
=>\(\left(log_24x\right)^2-2\cdot log_22x=5\)
=>\(\left(1+log_22x\right)^2-2\cdot log_22x=5\)
=>\(\left(log_22x\right)^2+1=5\)
=>\(\left(log_22x\right)^2=4\)
=>\(\left[{}\begin{matrix}log_22x=2\\log_22x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=\dfrac{1}{4}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=\dfrac{1}{8}\end{matrix}\right.\)
Bài 1:
a:
ĐKXĐ: x>1
\(log_3\left(2x+1\right)-log_3\left(x-1\right)=1\)
=>\(log_3\left(\dfrac{2x+1}{x-1}\right)=1\)
=>\(\dfrac{2x+1}{x-1}=3\)
=>3(x-1)=2x+1
=>3x-3=2x+1
=>x=4(nhận)
b:
ĐKXĐ: x>2
\(log_2\left(x-1\right)+log_2\left(x-2\right)=log_5\left(125\right)\)
=>\(log_2\left[\left(x-1\right)\left(x-2\right)\right]=3\)
=>\(\left(x-1\right)\left(x-2\right)=2^3=8\)
=>\(x^2-3x-6=0\)
=>\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{2}\left(nhận\right)\\x=\dfrac{3-\sqrt{33}}{2}\left(loại\right)\end{matrix}\right.\)
c: \(log_2\left(sinx\right)+log_2\left(cosx\right)=-2\)
=>\(log_2\left(sinx\cdot cosx\right)=-2\)
=>\(log_2\left(\dfrac{1}{2}\cdot sin2x\right)=-2\)
=>\(\dfrac{1}{2}\cdot sin2x=\dfrac{1}{4}\)
=>\(sin2x=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x=\dfrac{\Omega}{6}+k2\Omega\\2x=\dfrac{5}{6}\Omega+k2\Omega\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Omega}{12}+k\Omega\\x=\dfrac{5}{12}\Omega+k\Omega\end{matrix}\right.\)
\(x\in\left(0;2\Omega\right)\)
=>\(\left[{}\begin{matrix}\dfrac{\Omega}{12}+k\Omega\in\left(0;2\Omega\right)\\\dfrac{5}{12}\Omega+k\Omega\in\left(0;2\Omega\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k+\dfrac{1}{12}\in\left(0;2\right)\\k+\dfrac{5}{12}\in\left(0;2\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}k\in\left(-\dfrac{1}{12};\dfrac{23}{12}\right)\\k\in\left(-\dfrac{5}{12};\dfrac{19}{12}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}k\in\left(0;1\right)\\k\in\left(0;1\right)\end{matrix}\right.\)
=>\(x\in\left\{\dfrac{\Omega}{12};\dfrac{13}{12}\Omega;\dfrac{5}{12}\Omega;\dfrac{17}{12}\Omega\right\}\)