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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
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1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)
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a) \(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Rightarrow\left(x-3\right).7=\left(x+5\right).5\)
\(\Rightarrow7x-21=5x+25\)
\(\Rightarrow7x-5x=21+25\)
\(\Rightarrow2x=46\)
\(\Rightarrow x=23\)
Vậy \(x=23\)
b) \(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Rightarrow\left(x-1\right).\left(x+1\right)=7.9\)
\(\Rightarrow\left(x-1\right)x-\left(x+1\right)=7.9\)
\(\Rightarrow x^2-x-x-1=63\)
\(\Rightarrow x^2-1=63\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=8\) hoặc \(x=-8\)
Vậy \(x=8\) hoặc \(x=-8\)
c) \(\frac{x+4}{20}=\frac{5}{x+4}\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow x+4=\pm10\)
+) \(x+4=10\Rightarrow x=6\)
+) \(x+4=-10\Rightarrow x=-16\)
Vậy \(x\in\left\{6;-16\right\}\)
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Ta có : \(\frac{\left(4^x\right)^2}{2^x}=8\)
\(\Rightarrow4^{2x}=8.2^x\)
\(\Rightarrow4^{2x}=2^3.2^x\)
\(\Rightarrow\left(2^2\right)^{2x}=2^{x+3}\)
\(\Rightarrow2^{4x}=2^{x+3}\)
=> 4x = x + 3
=> 3x = 3
=> x = 1
Vậy x = 1.
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\(\frac{2}{3}x+\frac{5}{7}=\frac{3}{10}\)
\(\Rightarrow\frac{2}{3}x=\frac{3}{10}-\frac{5}{7}\)
\(\Rightarrow\frac{2}{3}x=-\frac{29}{70}\)
\(\Rightarrow x=-\frac{29}{70}:\frac{2}{3}\)
\(\Rightarrow x=-\frac{87}{140}\)
tíc mình nha
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1/ 2x = 45.46
=> 2x = 45 + 6
=> 2x = 411
=> 2x = (22)11
=> 2x = 222
=> x = 22
vậy_
2/ 2x = 46.163
=> 2x = (22)6.(24)3
=> 2x = 212.212
=> 2x = 212 + 12
=> 2x = 224
=> x = 24
3/ 2x = 45.162
=> 2x = (22)5.(24)2
=> 2x = 210.28
=> 2x = 210 + 8
=> 2x = 218
=> x = 18
vậy_
\(\frac{1}{2^x}=4^5.4^3=4^{5+3}=4^8\)
\(\Rightarrow1=4^8.2^x=2^{2.8+x}=2^{16+x}\)
ta có 1 < 21 => 216+x < 21
=> 216+x = 20
=> 16+x=0
=> x= -16
| x2 +|x-1| |=x2 +5
=> x2 + |x-1| = x2 + 5
=> |x - 1| = 5
=> x - 1 = 5 hoặc x - 1 = -5
=> x = 6 hoặc x = -4
Vậy S = { -4; 6 }
#Châu's ngốc
Bài giải
\(\left|x^2+ | x-1\text{ }|\right|=x^2+5\)
Mà \(x^2+5\ge5\)nên :
\(x^2+\left|x-1\right|=x^2+5\)
\(\left|x-1\right|=x^2+5-x^2\)
\(\left|x-1\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=6\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{-4\text{ ; }6\right\}\)