giải các phương trình sau
a) \(\log_3\left(2x-5\right)=3\)
b) \(\log_4x^2=2\)
c) \(\log_7\left(3x-1\right)=\log_7\left(2x+5\right)\)
d) \(\ln\left(4x^2+2x-3\right)=\ln\left(3x^2-3\right)\)
e) \(\log\left(2x+3\right)=log\left(1-3x\right)\)
giải các bất phương trình sau
a) \(log\left(x-5\right)< 2\)
b) \(log_2\left(2x-3\right)>4\)
c) \(log_3\left(2x+5\right)\le3\)
d) \(log_4\left(4x-5\right)\ge2\)
e) \(log_3\left(1-3x\right)>3\)
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
=>\(-3x>26\)
=>\(x< -\dfrac{26}{3}\)
giải các phương trình sau
a) \(\log_5\left(4x-3\right)=2\)
b) \(\log_2x^2=2\)
c) \(\log_5\left(2x+1\right)=\log_5\left(-2x+3\right)\)
d) \(\ln\left(x^2-6x+7\right)=\ln\left(x-3\right)\)
e) \(\log\left(5x-1\right)=log\left(4-2x\right)\)
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
=>\(x=\dfrac{5}{7}\left(nhận\right)\)
giải các bất phương trình sau
a) \(log\left(x-2\right)< 3\)
b) \(log_2\left(2x-1\right)>3\)
c) \(log_3\left(-x-1\right)\le2\)
d) \(log_2\left(2x-3\right)\ge2\)
e) \(log_3\left(2x-7\right)>2\)
a: \(log\left(x-2\right)< 3\)
=>\(\left\{{}\begin{matrix}x-2>0\\log\left(x-2\right)< log9\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x-2>0\\x-2< 9\end{matrix}\right.\Leftrightarrow2< x< 11\)
b: \(log_2\left(2x-1\right)>3\)
=>\(\left\{{}\begin{matrix}2x-1>0\\log_2\left(2x-1\right)>log_29\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-1>0\\2x-1>9\end{matrix}\right.\Leftrightarrow2x-1>9\)
=>2x>10
=>x>5
c: \(log_3\left(-x-1\right)< =2\)
=>\(\left\{{}\begin{matrix}-x-1>0\\log_3\left(-x-1\right)< =log_39\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-x-1>0\\-x-1< =9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x>1\\-x< =10\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -1\\x>=-10\end{matrix}\right.\Leftrightarrow-10< =x< -1\)
d: \(log_2\left(2x-3\right)>=2\)
=>\(\left\{{}\begin{matrix}2x-3>0\\log_2\left(2x-3\right)>=log_24\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-3>0\\2x-3>=4\end{matrix}\right.\)
=>2x-3>=4
=>2x>=7
=>\(x>=\dfrac{7}{2}\)
e: \(log_3\left(2x-7\right)>2\)
=>\(\left\{{}\begin{matrix}2x-7>0\\log_3\left(2x-7\right)>log_39\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>\dfrac{7}{2}\\2x-7>9\end{matrix}\right.\)
=>2x-7>9
=>2x>16
=>x>8
a.
\(log\left(x-2\right)< 3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\x-2< 10^3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>2\\x< 1002\end{matrix}\right.\) \(\Rightarrow2< x< 1002\)
b.
\(log_2\left(2x-1\right)>3\Leftrightarrow\left\{{}\begin{matrix}2x-1>0\\2x-1>2^3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{9}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{9}{2}\)
c.
\(log_3\left(-x-1\right)\le2\Rightarrow\left\{{}\begin{matrix}-x-1>0\\-x-1\le3^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x< -1\\x\ge-10\end{matrix}\right.\) \(\Rightarrow-10\le x< -1\)
d.
\(log_2\left(2x-3\right)\ge2\Leftrightarrow\left\{{}\begin{matrix}2x-3>0\\2x-3\ge2^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x>\dfrac{7}{2}\end{matrix}\right.\) \(\Rightarrow x>\dfrac{7}{2}\)
e,
\(log_3\left(2x-7\right)>2\Leftrightarrow\left\{{}\begin{matrix}2x-7>0\\2x-7>3^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{7}{2}\\x>8\end{matrix}\right.\) \(\Rightarrow x>8\)
Lời giải:
a. ĐK: $x>2$
$\log(x-2)<3$
$\Leftrightarrow x-2< 10^3$
$\Leftrightarrow x< 1002$
Vậy $2< x< 1002$
b. ĐK: $x> \frac{1}{2}$
$\log_2(2x-1)>3$
$\Leftrightarrow 2x-1> 2^3$
$\Leftrightarrow 2x> 9$
$\Leftrightarrow x> \frac{9}{2}$
Vậy $x> \frac{9}{2}$
c. ĐK: $x< -1$
$\log_3(-x-1)\leq 2$
$\Leftrightarrow -x-1\leq 3^2=9$
$\Leftrightarrow x+1\geq -9$
$\Leftrightarrow x\geq -10$
Vậy $-10\leq x< -1$
d. ĐK: $x> \frac{3}{2}$
$\log_2(2x-3)\geq 2$
$\Leftrightarrow 2x-3\geq 2^2=4$
$\Leftrightarrow x\geq \frac{7}{2}$
Vậy $x\geq \frac{7}{2}$
e. ĐK: $x> \frac{7}{2}$
$\log_3(2x-7)>2$
$\Leftrightarrow 2x-7> 3^2=9$
$\Leftrightarrow x> 8$
Vậy $x>8$
Giải các bất phương trình sau
a/ (x+1).(x-1).(3x-6)>0
b/ \(\dfrac{x+3}{x-2}\le0\)
c/ \(\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0\)
d/ \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
e/ \(\dfrac{2x^2+x}{1-2x}\ge1-x\)
f/ \(\dfrac{\left(2+x\right)^5.\left(x+1\right).\left(3-x\right)^{11}}{\left(2-x\right).\left(1-x\right)^{20}}\le0\)
a) \(\left(x+1\right)\left(x-1\right)\left(3x-6\right)>0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x>2\end{matrix}\right.\)
b) \(\dfrac{x+3}{x-2}\le0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow-3\le x< 2\)
d) \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
\(\Leftrightarrow\dfrac{2x-5}{3x+2}-\dfrac{3x+2}{2x-5}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5\right)^2-\left(3x+2\right)^2}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{\left(2x-5+3x+2\right)\left(2x-5-3x-2\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
\(\Leftrightarrow\dfrac{-\left(5x-3\right)\left(x+7\right)}{\left(3x+2\right)\left(2x-5\right)}< 0\)
Lập bảng xét dấu ta được kết quả :
\(Bpt\Leftrightarrow\left[{}\begin{matrix}-7< x< -\dfrac{2}{3}\\\dfrac{5}{3}< x< \dfrac{5}{2}\end{matrix}\right.\)
Giải các bất phương trình sau
a/ (x+1).(x-1).(3x-6)>0
b/ \(\dfrac{x+3}{x-2}\le0\)
c/ \(\dfrac{\left(2x-5\right).\left(x+2\right)}{-4x+3}\ge0\)
d/ \(\dfrac{2x-5}{3x+2}< \dfrac{3x+2}{2x-5}\)
e/ \(\dfrac{2x^2+x}{1-2x}\ge1-x\)
f/ \(\dfrac{\left(2+x\right)^5.\left(x+1\right).\left(3-x\right)^{11}}{\left(2-x\right).\left(1-x\right)^{20}}\le0\)
Giải các phương trình sau :
a) \(\ln\left(4x+2\right)-\ln\left(x-1\right)=\ln x\)
b) \(\log_2\left(3x+1\right)\log_3x=2\log_2\left(3x+1\right)\)
c) \(2^{\log_3x^2}.5^{\log_3x}=400\)
d) \(\ln^3x-3\ln^2x-4\ln x+12=0\)
a) Điều kiện: \(\left\{{}\begin{matrix}4x+2>0\\x-1>0\\x>0\end{matrix}\right.\)
Hay là: \(x>1\)
Khi đó biến đổi pương trình như sau:
\(\ln\dfrac{4x+2}{x-1}=\ln x\)
\(\Leftrightarrow\dfrac{4x+2}{x-1}=x\)
\(\Leftrightarrow4x+2=x\left(x-1\right)\)
\(\Leftrightarrow x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{5+\sqrt{33}}{2}\\x_2=\dfrac{5-\sqrt{33}}{2}\left(loại\right)\end{matrix}\right.\)
Vậy nghiệm của phương trình là: \(x=\dfrac{5+\sqrt{33}}{2}\)
b) Điều kiện: \(\left\{{}\begin{matrix}3x+1>0\\x>0\end{matrix}\right.\)
Hay là: \(x>0\)
Biến đổi phương trình như sau:
\(\log_2\left(3x+1\right)\log_3x-2\log_2\left(3x+1\right)=0\)
\(\Leftrightarrow\log_2\left(3x+1\right)\left(\log_3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\log_2\left(3x+1\right)=0\\\log_3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=2^0\\x=3^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=9\end{matrix}\right.\)
Vậy nghiệm là x = 9.
c) Điều kiện: x > 0.
Khi đó biến đổi phương trình như sau:
\(2^{\log_3x^2}.5^{\log_3x}=400\)
\(\Leftrightarrow2^{2\log_3x}.5^{\log_3x}=400\)
\(\Leftrightarrow\left(2^2.5\right)^{\log_3x}=400\)
\(\Leftrightarrow20^{\log_3x}=20^2\)
\(\Leftrightarrow\log_3x=2\)
\(\Leftrightarrow x=3^2=9\) (thỏa mãn)
Giải các phương trình sau:
a) \(\log \left( {x + 1} \right) = 2;\)
b) \(2{\log _4}x + {\log _2}\left( {x - 3} \right) = 2;\)
c) \(\ln x + \ln \left( {x - 1} \right) = \ln 4x;\)
d) \({\log _3}\left( {{x^2} - 3x + 2} \right) = {\log _3}\left( {2x - 4} \right).\)
a, ĐK: \(x+1>0\Leftrightarrow x>-1\)
\(log\left(x+1\right)=2\\ \Leftrightarrow x+1=10^2\\ \Leftrightarrow x+1=100\\ \Leftrightarrow x=99\left(tm\right)\)
b, ĐK: \(\left\{{}\begin{matrix}x-3>0\\x>0\end{matrix}\right.\Rightarrow x>3\)
\(2log_4x+log_2\left(x-3\right)=2\\ \Leftrightarrow log_2x+log_2\left(x-3\right)=2\\ \Leftrightarrow log_2\left(x^2-3x\right)=2\\ \Leftrightarrow x^2-3x=4\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
c, ĐK: \(x>1\)
\(lnx+ln\left(x-1\right)=ln4x\\ \Leftrightarrow ln\left[x\left(x-1\right)\right]-ln4x=0\\ \Leftrightarrow ln\left(\dfrac{x-1}{4}\right)=0\\ \Leftrightarrow\dfrac{x-1}{4}=1\\ \Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\)
d, ĐK: \(\left\{{}\begin{matrix}x^2-3x+2>0\\2x-4>0\end{matrix}\right.\Rightarrow x>2\)
\(log_3\left(x^2-3x+2\right)=log_3\left(2x-4\right)\\ \Leftrightarrow x^2-3x+2=2x-4\\ \Leftrightarrow x^2-5x+6=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=3\left(tm\right)\end{matrix}\right.\)
Giải các bất phương trình, hệ phương trình
a) \(\dfrac{x^2-4x+3}{2x-3}\ge x-1\)
b) \(3x^2-\left|4x^2+x-5\right|>3\)
c)\(4x-\left|2x^2-8x-15\right|\le-1\)
d)\(x+3-\sqrt{21-4x-x^2}\ge0\)
e)\(\left\{{}\begin{matrix}x\left(x+5\right)< 4x+2\\\left(2x-1\right)\left(x+3\right)\ge4x\end{matrix}\right.\)
f)\(\dfrac{1}{x^2-5x+4}\le\dfrac{1}{x^2-7x+10}\)
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2