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a: \(6^x=5\)
=>\(x=log_65\)
b: \(7^{3-x}=5\)
=>\(3-x=log_75\)
=>\(x=3-log_75\)
c: \(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\)
=>\(\left(\dfrac{3}{5}\right)^{x-2}=\left(\dfrac{3}{5}\right)^3\)
=>x-2=3
=>x=5
d: \(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\)
=>\(\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)
=>x=-1
a.
\(6^x=5\Rightarrow x=log_65\)
b.
\(7^{3-x}=5\Rightarrow3-x=log_75\)
\(\Rightarrow x=3-log_75\)
c.
\(\left(\dfrac{3}{5}\right)^{x-2}=\dfrac{27}{125}\Rightarrow x-2=log_{\dfrac{3}{5}}\left(\dfrac{27}{125}\right)\)
\(\Rightarrow x-2=3\Rightarrow x=5\)
d.
\(\left(\dfrac{4}{5}\right)^x=\dfrac{5}{4}\Rightarrow\left(\dfrac{4}{5}\right)^x=\left(\dfrac{4}{5}\right)^{-1}\)
\(\Rightarrow x=-1\)
a: \(log\left(x-5\right)< 2\)
=>\(\left\{{}\begin{matrix}x-5>0\\log\left(x-5\right)< log4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-5>0\\x-5< 4\end{matrix}\right.\Leftrightarrow5< x< 9\)
b: \(log_2\left(2x-3\right)>4\)
=>\(log_2\left(2x-3\right)>log_216\)
=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>16\end{matrix}\right.\)
=>2x-3>16
=>2x>19
=>\(x>\dfrac{19}{2}\)
c: \(log_3\left(2x+5\right)< =3\)
=>\(log_3\left(2x+5\right)< =log_327\)
=>\(\left\{{}\begin{matrix}2x+5>0\\2x+5< =27\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-\dfrac{5}{2}\\x< =11\end{matrix}\right.\)
=>\(-\dfrac{5}{2}< x< =11\)
d: \(log_4\left(4x-5\right)>=2\)
=>\(log_4\left(4x-5\right)>=log_416\)
=>4x-5>=16 và 4x-5>0
=>4x>=21 và 4x>5
=>4x>=21
=>\(x>=\dfrac{21}{4}\)
e: \(log_3\left(1-3x\right)>3\)
=>\(log_3\left(1-3x\right)>log_327\)
=>\(\left\{{}\begin{matrix}1-3x>0\\1-3x>27\end{matrix}\right.\)
=>1-3x>27
=>\(-3x>26\)
=>\(x< -\dfrac{26}{3}\)
a: ĐKXĐ: \(4x-3>0\)
=>x>3/4
\(log_5\left(4x-3\right)=2\)
=>\(log_5\left(4x-3\right)=log_525\)
=>4x-3=25
=>4x=28
=>x=7(nhận)
b: ĐKXĐ: \(x\ne0\)
\(log_2x^2=2\)
=>\(log_2x^2=log_24\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-2\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{3}{2}\right\}\)
\(\log_52x+1=\log_5-2x+3\)
=>2x+1=-2x+3
=>4x=2
=>\(x=\dfrac{1}{2}\left(nhận\right)\)
d: ĐKXD: \(x\notin\left\{3\right\}\)
\(ln\left(x^2-6x+7\right)=ln\left(x-3\right)\)
=>\(x^2-6x+7=x-3\)
=>\(x^2-7x+10=0\)
=>(x-2)(x-5)=0
=>\(\left[{}\begin{matrix}x=2\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\notin\left\{\dfrac{1}{5};2\right\}\)
\(log\left(5x-1\right)=log\left(4-2x\right)\)
=>5x-1=4-2x
=>7x=5
=>\(x=\dfrac{5}{7}\left(nhận\right)\)
a.
\(2^x=2^{3x-1}\Leftrightarrow x=3x-1\)
\(\Rightarrow x=\dfrac{1}{2}\)
b.
\(7^{x-5}=49\Leftrightarrow x-5=log_749=2\)
\(\Rightarrow x=7\)
c.
\(3^{5x-3}=1\Rightarrow5x-3=log_31=0\)
\(\Rightarrow x=\dfrac{3}{5}\)
d.
\(\left(\dfrac{1}{7}\right)^{5x}=7^{x+6}\Leftrightarrow7^{-5x}=7^{x+6}\)
\(\Leftrightarrow-5x=x+6\)
\(\Rightarrow x=-1\)
a: \(27^{2-x}< =9\)
=>\(\left(3^3\right)^{2-x}< =3^2\)
=>\(3^{6-3x}< =3^2\)
=>6-3x<=2
=>-3x<=-4
=>\(x>=\dfrac{4}{3}\)
b: \(7^{3-x}< 49\)
=>\(7^{3-x}< 7^2\)
=>3-x<2
=>-x<2-3=-1
=>x>1
c: \(27^{3-x}>9\)
=>\(\left(3^3\right)^{3-x}>3^2\)
=>\(3^{9-3x}>3^2\)
=>9-3x>2
=>-3x>-7
=>\(x< \dfrac{7}{3}\)
d: \(2^{3-x}< 2^3\)
=>3-x<3
=>-x<0
=>x>0
e: \(27^{3-x^2}< 27^{x+1}\)
=>\(3-x^2< x+1\)
=>\(-x^2-x+2< 0\)
=>\(x^2+x-2>0\)
=>(x+2)(x-1)>0
=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)
a: \(5^x=4\)
=>\(x=log_54\)
b: \(5^{2-x}=8\)
=>\(2-x=log_58\)
=>\(x=2-log_58\)
c: \(\left(\dfrac{1}{3}\right)^{x+4}=243\)
=>\(3^{-x-4}=3^5\)
=>-x-4=5
=>-x=9
=>x=-9
d: \(\left(\dfrac{2}{3}\right)^x=\dfrac{3}{2}\)
=>\(\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^{-1}\)
=>x=-1
a: \(2^{x^2-2x+1}=1\)
=>\(2^{\left(x-1\right)^2}=2^0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
b: \(7^{x^2+7x}=5764801\)
=>\(7^{x^2+7x}=7^8\)
=>\(x^2+7x=8\)
=>\(x^2+7x-8=0\)
=>(x+8)(x-1)=0
=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)
c: \(6^{x^2+12x}=6^{7x}\)
=>\(x^2+12x=7x\)
=>\(x^2+5x=0\)
=>x(x+5)=0
=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)
=>\(3^{-x+1}=3^{2x-5}\)
=>-x+1=2x-5
=>-x-2x=-5-1
=>-3x=-6
=>x=2
e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)
=>\(5^{-3x-5}=5^{2x+1}\)
=>-3x-5=2x+1
=>-5x=6
=>\(x=-\dfrac{6}{5}\)
ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le3\\x\ne1\end{matrix}\right.\)
\(\dfrac{\sqrt{x+1}\left(\sqrt{x+1}+\sqrt{3-x}\right)}{2\left(x-1\right)}>x-\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x+1+\sqrt{-x^2+2x+3}}{x-1}>2x-1\)
- TH1: Với \(x>1\) BPT tương đương:
\(x+1+\sqrt{-x^2+2x+3}>\left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}>2x^2-4x\)
Đặt \(\sqrt{-x^2+2x+3}=t\ge0\Rightarrow2x^2-4x=-2t^2+6\)
BPt trở thành: \(t>-2t^2+6\Leftrightarrow2t^2+t-6>0\)
\(\Rightarrow t>\dfrac{3}{2}\Rightarrow-x^2+2x+3>\dfrac{9}{4}\Rightarrow1< x< \dfrac{2+\sqrt{7}}{2}\)
TH2: với \(x< 1\) BPT tương đương:
\(x+1+\sqrt{-x^2+2x+3}< \left(2x-1\right)\left(x-1\right)\)
\(\Leftrightarrow\sqrt{-x^2+2x+3}< 2x^2-4x\)
Tương tự như trên, đặt \(t=\sqrt{-x^2+2x+3}\ge0\) ta được \(0\le t< \dfrac{3}{2}\)
\(\Rightarrow-x^2+2x+3< \dfrac{9}{4}\) \(\Rightarrow-1\le x< \dfrac{2-\sqrt{7}}{2}\)
Vậy nghiệm của BPT là: \(\left[{}\begin{matrix}-1\le x< \dfrac{2-\sqrt{7}}{2}\\1< x< \dfrac{2+\sqrt{7}}{2}\end{matrix}\right.\)
d: cos^2x=1
=>sin^2x=0
=>sin x=0
=>x=kpi
a: =>sin 4x=cos(x+pi/6)
=>sin 4x=sin(pi/2-x-pi/6)
=>sin 4x=sin(pi/3-x)
=>4x=pi/3-x+k2pi hoặc 4x=2/3pi+x+k2pi
=>x=pi/15+k2pi/5 hoặc x=2/9pi+k2pi/3
b: =>x+pi/3=pi/6+k2pi hoặc x+pi/3=-pi/6+k2pi
=>x=-pi/2+k2pi hoặc x=-pi/6+k2pi
c: =>4x=5/12pi+k2pi hoặc 4x=-5/12pi+k2pi
=>x=5/48pi+kpi/2 hoặc x=-5/48pi+kpi/2
a.
ĐKXĐ: \(x>0\)
\(log_5x>6\Rightarrow x>6^5\Rightarrow x>7776\)
b.
ĐKXĐ: \(x>0\)
\(log_7x< 2\Rightarrow\left\{{}\begin{matrix}x>0\\x< 7^2\end{matrix}\right.\) \(\Rightarrow0< x< 49\)
c.
\(log_2x\le3\Rightarrow\left\{{}\begin{matrix}x>0\\x\le3^2\end{matrix}\right.\) \(\Rightarrow0< x\le9\)
d.
\(log_{\dfrac{1}{3}}x>27\Rightarrow\left\{{}\begin{matrix}x>0\\x< \left(\dfrac{1}{3}\right)^{27}\end{matrix}\right.\)
\(\Rightarrow0< x< \dfrac{1}{3^{27}}\)