Giải các phương trình sau:
a) \(\left(\dfrac{1}{9}\right)^{x+1}>\dfrac{1}{81}\)
b) \(\left(\sqrt[4]{3}\right)^x\le27.3^x\)
c) \(log_2\left(x+1\right)\le log_2\left(2-4x\right)\)
a) \(\left(\dfrac{1}{9}\right)^{x+1}>\dfrac{1}{81}\);
b) \(\left(\sqrt[4]{3}\right)^x\le27.3^x\);
c) \(log_2\left(x+1\right)\le log_2\left(2-4x\right)\).
\(a,\left(\dfrac{1}{9}\right)^{x+1}>\dfrac{1}{81}\\ \Leftrightarrow\left(\dfrac{1}{9}\right)^{x+1}>\left(\dfrac{1}{9}\right)^2\\ \Leftrightarrow x+1< 2\\ \Leftrightarrow x< 1\)
\(b,\left(\sqrt[4]{3}\right)^x\le27\cdot3^x\\ \Leftrightarrow3^{\dfrac{x}{4}}\le3^{x+3}\\ \Leftrightarrow\dfrac{x}{4}\le3=x\\ \Leftrightarrow-\dfrac{3}{4}x\le3\\ \Leftrightarrow x\ge-4\)
c, ĐK: \(\left\{{}\begin{matrix}x+1>0\\2-4x>0\end{matrix}\right.\Leftrightarrow-1< x< \dfrac{1}{2}\)
\(log_2\left(x+1\right)\le log_2\left(2-4x\right)\\ \Leftrightarrow x+1\le2-4x\\ \Leftrightarrow5x\le1\\ \Leftrightarrow x\le\dfrac{1}{5}\)
Kết hợp với ĐKXĐ, ta được: \(-1< x\le\dfrac{1}{5}\)
tổng tất cả các nghiệm pt:
a, \(log_2\left(x+1\right)+log_2x=1\)
b, \(log_{\dfrac{1}{3}}^2\left(4x\right)-5log_3\left(2x\right)=5\)
c, \(log_2\left(x-1\right)+log_2\left(x-2\right)=log_5125\)
a:
ĐKXĐ: x+1>0 và x>0
=>x>0
=>\(log_2\left(x^2+x\right)=1\)
=>x^2+x=2
=>x^2+x-2=0
=>(x+2)(x-1)=0
=>x=1(nhận) hoặc x=-2(loại)
c: ĐKXĐ: x-1>0 và x-2>0
=>x>2
\(PT\Leftrightarrow log_2\left(x^2-3x+2\right)=3\)
=>\(\Leftrightarrow x^2-3x+2=8\)
=>x^2-3x-6=0
=>\(\left[{}\begin{matrix}x=\dfrac{3+\sqrt{33}}{2}\left(nhận\right)\\x=\dfrac{3-\sqrt{33}}{2}\left(loại\right)\end{matrix}\right.\)
tìm tập xác định của hàm số sau
a) \(y=log_2\left(2x-4\right)\)
b) \(y=log_2\left(2x+8\right)\)
c) \(y=log_3\left(4-x\right)\)
d) \(y=log_2\dfrac{1}{x+4}\)
d) \(y=log_3\left(x-3\right)\left(x+9\right)\)
ĐKXĐ:
a.
\(2x-4>0\Rightarrow x>2\Rightarrow D=\left(2;+\infty\right)\)
b.
\(2x+8>0\Rightarrow x>-4\Rightarrow D=\left(-4;+\infty\right)\)
c.
\(4-x>0\Rightarrow x< 4\Rightarrow D=\left(-\infty;4\right)\)
d.
\(\dfrac{1}{x+4}>0\Rightarrow x>-4\Rightarrow D=\left(-4;+\infty\right)\)
e.
\(\left(x-3\right)\left(x+9\right)>0\Rightarrow\left[{}\begin{matrix}x>3\\x< -9\end{matrix}\right.\) \(\Rightarrow D=\left(-\infty;-9\right)\cup\left(3;+\infty\right)\)
a: ĐKXĐ: 2x-4>0
=>2x>4
=>x>2
b: ĐKXĐ: 2x+8>0
=>2x>-8
=>x>-4
c: ĐKXĐ: 4-x>0
=>-x>-4
=>x<4
d: ĐKXĐ: \(\dfrac{1}{x+4}>0\)
=>x+4>0
=>x>-4
e: ĐKXĐ: \(\left(x-3\right)\left(x+9\right)>0\)
=>\(\left[{}\begin{matrix}x-3>0\\x+9< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< -9\end{matrix}\right.\)
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 phương trình sau :
a) \(\left(\dfrac{1}{2}\right)^{\log_{\dfrac{1}{3}}\left(x^2-3x+1\right)}\)
b) \(4x^2+3.3^{\sqrt{x}}+x.3^{\sqrt{x}}< 2x^2.3^{\sqrt{x}}+2x+6\)
c) \(\log_x4.\log_2\dfrac{5-12x}{12x-8}\ge2\)
1) Giải phương trình:
\(4\log_2^2x+x\log_2\left(x+2\right)=2\log_2x\left[x+\log_2\left(x+2\right)\right]\)
2) Tìm tất cả bộ hai số thực \(\left(x;y\right)\) thỏa mãn đẳng thức:
\(x^{\log_2x}+4^y+\left(x-5\right)2^{y+1}+57=18x\)
Giải các bất phương trình sau:
a) \(\left(x^2+3x-4\right)\left(3-2x\right)< 0\)
\(\dfrac{x^2+3x+4}{x^2-2}\ge0\)
\(\dfrac{x\left(x^2+4x+4\right)}{x^2-1}\ge0\)
b) \(\dfrac{3x-2}{2-x}\le-x\)
c) \(\dfrac{x-3}{x+1}>\dfrac{x+4}{x+2}\)
d) \(\dfrac{x+2}{x-2}-\dfrac{x+3}{x-2}>1\)
e) \(|2x-3|>x+1\)
f) \(|2x-5|\le x+1\)
g) \(x-4-|x^2+3x-4|>0\)
h) \(\left|x^2+4x+3\right|>\left|x^2-4x-5\right|\)
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
=>\(-3x>26\)
=>\(x< -\dfrac{26}{3}\)
Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
Bài 2: Giải các phương phương trình sau:
a) \(\dfrac{3\left(x-3\right)}{4}\)+\(\dfrac{4x-10,5}{4}\)=\(\dfrac{3\left(x+1\right)}{5}\)+6
b) \(\dfrac{2\left(3x+1\right)+1}{4}\)-5=\(\dfrac{2\left(3x-1\right)}{5}\)-\(\dfrac{3x+2}{10}\)
Mik đang cần gấp nha!!❤
Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
<=> 6.6 - 0.9x = 2,6 + 0,1x - 4
<=> - 0.9x - 0,1x = -6.6 -1,4
<=> -x = -8
<=> x = 8
Vậy x = 8
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
<=> 3,6 - x - 0,5 = x - 5,5 + x
<=> - x - 3,1 = -5,5
<=> - x = -2.4
<=> x = 2.4
Vậy x = 2.4